Sample Exercises

Hint

Diameter of an oxygen molecule,d = 3 Å
Radius, \(r=\frac{d}{2}=\frac{3}{2}=1.5 Å=1.5 \times 10^{-8} \mathrm{~cm}\)
The actual volume occupied by 1 mole of oxygen gas at \(S T P=22400 \mathrm{~cm}^3\)
Molecular volume of oxygen gas, \(V=\frac{4}{3} \pi r^3 . N\)
Where, \(N\) is Avogadro’s number \(=6.023 \times 10^{23}\) molecules \(/\) mole
\(\therefore V=\frac{4}{3} \times 3.14 \times\left(1.5 \times 10^{-8}\right)^3 \times 6.023 \times 10^{23}=8.51 \mathrm{~cm}^3\)
The ratio of the molecular volume to the actual volume of oxygen \(=\frac{8.51}{22400}\) \(=3.8 \times 10^{-4}\approx 4 \times 10^{-4}\)

Hint

The ideal gas equation relating pressure \((P)\), volume \((V)\), and absolute temperature \((T)\) is given as:
\(
P V=n R T
\)
Where,
\(R\) is the universal gas constant \(=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)
\(
\begin{aligned}
& n=\text { Number of moles }=1 \\
& T=\text { Standard temperature }=273 \mathrm{~K} \\
& P=\text { Standard pressure }=1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{Nm}^{-2} \\
& \therefore V=\frac{n R T}{T} \\
& =\frac{1 \times 8.314 \times 273}{1.013 \times 10^5} \\
& =0.0224 \mathrm{~m}^3 \\
& =22.4 \text { litres }
\end{aligned}
\)
Hence, the molar volume of a gas at STP is 22.4 litres.

Hint

The value of the ratio \(\frac{\mathrm{PV}}{\mathrm{T}}\), where the two curves meet, is \(\mu R\). This is because the ideal gas equation is given as,
\(
\begin{aligned}
& P V=\mu R T \\
& \frac{P V}{T}=\mu \mathrm{R}
\end{aligned}
\)
where,
\(P\) is the pressure,
\(T\) is the temperature,
\(V\) is the volume,
\(\mu\) is the number of moles,
\(R\) is the universal constant.
Molecular mass of oxygen \(=32.0 \mathrm{~g}\)
Mass of oxygen \(=1 \times 10^{-3} \mathrm{~kg}=1 \mathrm{~g}\)
\(
\begin{aligned}
& R=8.314 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1} \\
& \therefore\left(\frac{\mathrm{PV}}{\mathrm{T}}\right)=\left(\frac{1}{32}\right) \times 8.314 \\
& =0.26 \mathrm{~J} \mathrm{~K}^{-1}
\end{aligned}
\)
Therefore, the value of the ratio \(\left(\frac{P V}{T}\right)\), where the curves meet on the \(y\)-axis, is \(0.26 \mathrm{~J} \mathrm{~K}^{-1}\).

Hint

If we obtain similar plots for \(1.00 \times 10^{-3} \mathrm{~kg}\) of hydrogen, then we will not get the same value of \(\left(\frac{\mathrm{PV}}{\mathrm{T}}\right)\) at the point where the curves meet the \(y\)-axis.

This is because the molecular mass of hydrogen \((2.02 \mathrm{u})\) is different from that of oxygen \((32.0 \mathrm{u})\)
We have,
\(
\begin{aligned}
\left(\frac{\mathrm{PV}}{\mathrm{T}}\right) & =0.26 \mathrm{~J} \mathrm{~K}^{-1} \\
R & =8.314 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}
\end{aligned}
\)
Molecular mass \((M)\) of \(\mathrm{H}_2=2.02 \mathrm{u}\)
\(\left(\frac{\mathrm{PV}}{\mathrm{T}}\right)=\mu R\), at constant temperature
where, \(\mu=m / M\)
\(
\begin{aligned}
& \begin{aligned}
m & =\text { Mass of } \mathrm{H}_2 \\
\therefore m & =\left(\frac{\mathrm{PV}}{\mathrm{T}}\right) \times\left(\frac{M}{R}\right) \\
& =0.26 \times 2.02 / 8.31 \\
& =6.3 \times 10^{-2} \mathrm{~g} \\
& =6.3 \times 10^{-5} \mathrm{~kg}
\end{aligned} \\
& \text { Hence, } 6.3 \times 10^{-5} \mathrm{~kg} \text { of } \mathrm{H}_2 \text { will yield the same value of }\left(\frac{\mathrm{PV}}{\mathrm{T}}\right) .
\end{aligned}
\)

Hint

Initial Volume, \(V_1=30\) litre \(=30 \times 10^3 \mathrm{~cm}^3\)
\(
=30 \times 10^3 \times 10^{-6} \mathrm{~m}^3=30 \times 10^{-3} \mathrm{~m}^3
\)
Initial Pressure, \(P_1=15 \mathrm{~atm}\)
\(=15 \times 1.013 \times 10^5 \mathrm{Nm}^{-2}\)
Initial temperature, \(T_1=(27+273) K=300 K\)
The initial number of moles
\(
\mu_1=\frac{P_1 V_1}{R} T_1=\frac{15 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}{8.31 \times 300}=18.3
\)
Final Pressure, \(P_2=11\) atm
\(=11 \times 1.013 \times 10^5 \mathrm{Nm}^{-2}\)
Final Volume, \(V_2=30\) litre \(=30 \times 10^{-3} \mathrm{~cm}^3\)
Final temperature, \(T_2=17+273=290 K\)
Final number of moles, \(\mu_2=\frac{P_2 V_2}{R T_2}=\frac{11 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}{8.31 \times 290}=13.9\)
Number of moles taken out of cylinder \(=18.3-13.9=4.4\)
Mass of gas taken out of cylinder \(=4.4 \times 32 \mathrm{~g}=140.8 \mathrm{~g}=0.141 \mathrm{~kg}\)

Hint

Volume of the bubble inside, \(V_1=1.0 \mathrm{~cm}^3=1 \times 10^{-6} \mathrm{~m}^3\)
Pressure on the bubble, \(P_1=\) Pressure of water + Atmospheric pressure
\(
\begin{aligned}
& =p g h+1.01 \times 10^5=1000 \times 9.8 \times 40+1.01 \times 10^5 \\
& =3.92 \times 10^5+1.01 \times 10^5=4.93 \times 10^5 \mathrm{~Pa}
\end{aligned}
\)
Temperature, \(T_1=12^{\circ} \mathrm{C}=273+12=285 \mathrm{~K}\)
Also, pressure outside the lake, \(P_2=1.01 \times 10^5 \mathrm{~nm}^{-2}\)
Temperature, \(T_2=35^{\circ} \mathrm{C}=273+35=308 \mathrm{~K}\), Volume \(V_2=?\)
Now \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)
\(
\therefore V_2=\frac{P_1 V_1}{T_1} \cdot \frac{T_2}{P_2}=\frac{4.93 \times 10^5 \times 1 \times 10^{-6} \times 308}{285 \times 1.01 \times 10^5}=5.3 \times 10^{-6} \mathrm{~m}^{3}
\)

Hint

Here, Volume of room, \(V=25.0 \mathrm{~m}^3\), temperature, \(T=27^{\circ} \mathrm{C}=300 \mathrm{~K}\) and
Pressure, \(P=1 \mathrm{~atm}=1.01 \times 10^5 \mathrm{~Pa}\)
According to gas equation, \(P V=\mu R T=\mu N_A \cdot k_B T\)
Hence, the total number of air molecules in the volume of the given gas
\(
\begin{aligned}
& N=\mu . N_A=P \frac{V}{K_B} T \\
& \therefore N=\frac{1.01 \times 10^5 \times 25.0}{\left(1.38 \times 10^{-23}\right) \times 300}=6.1 \times 10^{26}
\end{aligned}
\)

Hint

(i) At room temperature, \(\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\)
Average thermal energy \(=(3 / 2) \mathrm{kT}\)
Where \(\mathrm{k}\) is Boltzmann constant \(=1.38 \times 10^{-23} \mathrm{~m}^2 \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}\)
\(
\begin{aligned}
& \therefore \frac{3}{2} k T=\frac{3}{2} \times 1.38 \times 10^{-38} \times 300 \\
& =6.21 \times 10^{-21} \mathrm{~J}
\end{aligned}
\)
Hence, the average thermal energy of a helium atom at room temperature \(\left(27^{\circ} \mathrm{C}\right)\) is \(6.21 \times 10^{-21} \mathrm{~J}\).
(ii) On the surface of the sun, \(T=6000 \mathrm{~K}\)
Average thermal energy \(=\frac{3}{2} k T\)
\(
\begin{aligned}
& =\frac{3}{2} \times 1.38 \times 10^{-38} \times 6000 \\
& =1.241 \times 10^{-19} \mathrm{~J}
\end{aligned}
\)
Hence, the average thermal energy of a helium atom on the surface of the sun is \(1.241 \times 10^{-19} \mathrm{~J}\).
(iii) At temperature, \(\mathrm{T}=10^7 \mathrm{~K}\)
Average thermal energy
\(
=\frac{3}{2} k T
\)
\(
\begin{aligned}
& =\frac{3}{2} \times 1.38 \times 10^{-23} \times 10^7 \\
& =2.07 \times 10^{-16} \mathrm{~J}
\end{aligned}
\)
Hence, the average thermal energy of a helium atom at the core of a star is \(2.07 \times\) \(10^{-16} \mathrm{~J}\).

Hint

All the three vessels have the same capacity, they have the same volume. Hence, each gas has the same pressure, volume, and temperature.
According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro’s number, \(\mathrm{N}=\) \(6.023 \times 10^{23}\).

The root mean square speed \(\left(\mathrm{V}_{\mathrm{rms}}\right)\) of a gas of mass \(\mathrm{m}\), and temperature \(\mathrm{T}\), is given by the relation:
\(
\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}
\)
where,
\(\mathrm{k}\) is Boltzmann constant
For the given gases, \(\mathrm{k}\) and \(\mathrm{T}\) are constants.
Hence, \(V_{\mathrm{rms}}\) depends only on the mass of the atoms, i.e., \(\mathrm{V}_{\mathrm{rms}} \propto(1 / \mathrm{m})^{1 / 2}\).
Therefore, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.

Hint

Let \(\mathrm{C}\) and \(\mathrm{C}^{\prime}\) be the rms velocity of argon and a helium gas atoms at temperature \(\mathrm{T}   \mathrm{K}\) and \(\mathrm{T}^{\prime}   \mathrm{K}\) respectively Here, \(\mathrm{M}=39.9 ; \mathrm{M}^{\prime}=4.0 ; \mathrm{T}=? ; \mathrm{T}^{\prime}=-20+273=253 \mathrm{~K}\)
Now, \(C=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 R T}{39.9}}\) and \(C ^{\prime}=\sqrt{\frac{3 R T^{\prime}}{{M}^{\prime}}}=\frac{\sqrt{3 R \times 253}}{4}\)
Since \(C=C^{\prime}\)
Therefore \(\sqrt{\frac{3 R T}{39.9}}=\frac{\sqrt{3 R \times 253}}{4}\) or \(T=\frac{39.9 \times 253}{4}=2523.7 \mathrm{~K}\)

Hint

As the gas leaks out, the volume and the temperature of the remaining gas do not change. The number of moles of the gas in the vessel is given by \(n=\frac{p V}{R T}\). The number of moles in the vessel before the leakage is
\(
n_1=\frac{p_1 V}{R T}
\)
and that after the leakage is
\(
n_2=\frac{p_2 V}{R T} \text {. }
\)
Thus, the amount leaked is
\(
\begin{aligned}
n_1-n_2 & =\frac{\left(p_1-p_2\right) V}{R T} \\
& =\frac{(200-125) \times 10^3 \mathrm{~N} \mathrm{~m}^{-2} \times 8.0 \times 10^{-3} \mathrm{~m}^3}{\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(300 \mathrm{~K})} \\
& =0.24 \mathrm{~mol}^{-1} .
\end{aligned}
\)

Hint

We have \(p=\frac{n R T}{V}\).
The pressure due to oxygen is
\(
p_1=\frac{(0.1 \mathrm{~mol})\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K})}{\left(2000 \times 10^{-6} \mathrm{~m}^{-3}\right)}=1.25 \times 10^5 \mathrm{~Pa} .
\)
Similarly, the pressure due to carbon dioxide is
\(
p_2=2.50 \times 10^5 \mathrm{~Pa} \text {. }
\)
The total pressure in the vessel is
\(
\begin{aligned}
p & =p_1+p_2 \\
& =(1.25+2.50) \times 10^5 \mathrm{~Pa}=3.75 \times 10^5 \mathrm{~Pa} .
\end{aligned}
\)

Hint

Suppose there are \(n_1\) moles of hydrogen and \(n_2\) moles of oxygen in the mixture. The pressure of the mixture will be
\(
p=\frac{n_1 R T}{V}+\frac{n_2 R T}{V}=\left(n_1+n_2\right) \frac{R T}{V}
\)
or, \(100 \times 10^3 \mathrm{~Pa}=\left(n_1+n_2\right) \frac{\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K})}{2000 \times 10^{-6} \mathrm{~m}^{-3}}\)
or, \(\quad n_1+n_2=0.08 \mathrm{~mol} \dots(i)\).

The mass of the mixture is
\(
n_1 \times 2 \mathrm{~g} \mathrm{~mol}^{-1}+n_2 \times 32 \mathrm{~g} \mathrm{~mol}^{-1}=0.76 \mathrm{~g}
\)
or, \(\quad n_1+16 n_2=0.38 \mathrm{~mol} \dots(ii)\).
from (i) and (ii),
\(
n_1=0.06 \mathrm{~mol} \text { and } n_2=0.02 \mathrm{~mol} \text {. }
\)
The mass of hydrogen \(=0.06 \times 2 \mathrm{~g}=0.12 \mathrm{~g}\) and the mass of oxygen \(=0.02 \times 32 \mathrm{~g}=0.64 \mathrm{~g}\).

Hint

Using \(p V=\) constant for the shorter arm,
\((80 \mathrm{~cm})(50 \mathrm{~cm})=p_1(50 \mathrm{~cm}-10 \mathrm{~cm})\)
or, \(\quad p_1=100 \mathrm{~cm} \dots(i)\).
Using \(p V=\) constant for the longer arm,
\(
(80 \mathrm{~cm})(100 \mathrm{~cm})=p_2\left(100-l_0\right) \mathrm{cm} \text {. } \dots(ii)
\)
From the figure,
\(
p_1=p_2+\left(l_0-10\right) \mathrm{cm}
\)
Thus by (i),
or, \(\quad p_2=110 \mathrm{~cm}-l_0 \mathrm{~cm}\).
Putting in (ii),
\(\left(110-l_0\right)\left(100-l_0\right)=8000\)
or, \(\quad l_0^2-210 l_0+3000=0\)
or, \(\quad l_0=15^{-5} 5\).
The required length is \(15-5 \mathrm{~cm}\).

Hint

The process is shown in a \(p-V\) diagram in the figure below. The process starts from \(A\) and goes through \(A B C A\).

(i) Applying \(p V=n R T\) at \(A\) and \(B\),
and \(\quad\left(2 p_0\right) V_0=n R T_B\).
Thus, \(\quad T_B=2 T_0\).
This is the temperature in the isothermal part \(B C\).
(ii) As the process \(B C\) is isothermal, \(T_C=T_B=2 T_0\).
Applying \(p V=n R T\) at \(A\) and \(C\),
and \(\quad p_0 V_C=n R\left(2 T_0\right)\)
or, \(\quad V_C=2 V_0\).

Hint

Let the initial pressure, volume and temperature in each vessel be \(p_0(=105 \mathrm{kPa}), V_0\) and \(T_0(=300 \mathrm{~K})\). Let the number of moles in each vessel be \(n\). When the first vessel is maintained at temperature \(T_0\) and the other is maintained at \(T^{\prime}=400 \mathrm{~K}\), the pressures change. Let the common pressure become \(p^{\prime}\) and the number of moles in the two vessels become \(n_1\) and \(n_2\). We have

\(
\begin{aligned}
p_0 V_0 & =n R T_0 \dots(i) \\
p^{\prime} V_0 & =n_1 R T_0 \dots(ii) \\
p^{\prime} V_0 & =n_2 R T^{\prime} \dots(iii) \\
n_1+n_2 & =2 n \dots(iv)
\end{aligned}
\)
Putting \(n, n_1\) and \(n_2\) from (i), (ii) and (iii) in (iv),
or, or,
\(
\begin{aligned}
\frac{p^{\prime} V_0}{R T_0}+\frac{p^{\prime} V_0}{R T^{\prime}} & =2 \frac{p_0 V_0}{R T_0} \\
p^{\prime}\left(\frac{T^{\prime}+T_0}{T_0 T^{\prime}}\right) & =\frac{2 p_0}{T_0} \\
p^{\prime} & =\frac{2 p_0 T^{\prime}}{T^{\prime}+T_0} \\
& =\frac{2 \times 105 \mathrm{kPa} \times 400 \mathrm{~K}}{400 \mathrm{~K}+300 \mathrm{~K}}=120 \mathrm{kPa}
\end{aligned}
\)

Hint

(i) The number of moles of hydrogen \(=14 \mathrm{~g} / 2 \mathrm{~g}=7\) and the number of moles of oxygen \(=96 \mathrm{~g} / 32 \mathrm{~g}=3\). The total number of moles in the vessel \(=7+3=10\). The pressure is 1 atm \(=1.0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}\) and the temperature \(=273 \mathrm{~K}\).
Now
\(
p V=n R T \dots(i)
\)
or,
\(
\begin{aligned}
V & =\frac{n R T}{p} \\
& =\frac{(10 \mathrm{~mol}) \times\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(273 \mathrm{~K})}{1.0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}} \\
& =0.23 \mathrm{~m}^3 .
\end{aligned}
\)
(ii) When electric spark is passed, hydrogen reacts with oxygen to form water \(\left(\mathrm{H}_2 \mathrm{O}\right)\). Each gram of hydrogen reacts with eight grams of oxygen. Thus, \(96 \mathrm{~g}\) of oxygen will be totally consumed together with \(12 \mathrm{~g}\) of hydrogen. The gas left in the vessel will be \(2 \mathrm{~g}\) of hydrogen which is \(n^{\prime}=1\) mole.
Neglecting the volume of the water formed,
\(
p^{\prime} V=n^{\prime} R T \dots(ii)
\)
From (i) and (ii),
\(
\begin{aligned}
\frac{p^{\prime}}{p} & =\frac{n^{\prime}}{n}=\frac{1}{10} \\
p^{\prime} & =p \times 0.10 \\
& =0.10 \mathrm{~atm}
\end{aligned}
\)

Hint

Let the pressure of the air in the barometer be p. We have,
\(
\begin{aligned}
p \times 50 \mathrm{~cm}^3 & =(75 \mathrm{~cm} \text { of mercury }) \times\left(2.0 \mathrm{~cm}^3\right) \\
\text { or, } \quad p & =3.0 \mathrm{~cm} \text { of mercury. }
\end{aligned}
\)
The atmospheric pressure is equal to the pressure due to the mercury column plus the pressure due to the air inside. Thus, the mercury column descends by \(3.0 \mathrm{~cm}\).

Hint

The pressure of oxygen in the space above the mercury level \(=5 \mathrm{~cm}\) of mercury
\(
\begin{aligned}
& =0.05 \mathrm{~m} \times 13600 \mathrm{~kg} \mathrm{~m}^{-3} \times 10 \mathrm{~m} \mathrm{~s}^{-2} \\
& =6800 \mathrm{~N} \mathrm{~m}^{-2}
\end{aligned}
\)
The volume of oxygen \(=\left(2 \mathrm{~cm}^2\right) \times(25 \mathrm{~cm}+5 \mathrm{~cm})\) \(=60 \mathrm{~cm}^3=6 \times 10^{-5} \mathrm{~m}^{-3}\).
The temperature \(\quad=(273+27) \mathrm{K}=300 \mathrm{~K}\).
The amount of oxygen is
\(
n=\frac{p V}{R T}
\)
\(
\begin{aligned}
& =\frac{\left(6800 \mathrm{~N} \mathrm{~m}^{-2}\right) \times 6 \times 10^{-5} \mathrm{~m}^{-3}}{\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(300 \mathrm{~K})} \\
& =16.4 \times 10^{-5} \mathrm{~mol} .
\end{aligned}
\)
The mass of oxygen is
\(
\begin{aligned}
& \left(16.4 \times 10^{-5} \mathrm{~mol}\right) \times\left(32 \mathrm{~g} \mathrm{~mol}^{-1}\right) \\
= & 5.24 \times 10^{-3} \mathrm{~g} .
\end{aligned}
\)

Hint

Let \(l_1\) and \(l_2\) be the lengths of the upper part and the lower part of the cylinder respectively. Clearly, \(l_1=50 \mathrm{~cm}\) and \(l_2=40 \mathrm{~cm}\). Let the pressures in the upper and lower parts be \(p_1\) and \(p_2\) respectively. Let the area of cross-section of the cylinder be \(A\). The temperature in both parts is \(T=300 \mathrm{~K}\).
Consider the equilibrium of the piston. The forces acting on the piston are
(a) its weight \(\mathrm{mg}\)
(b) \(p_1 A\) downward, by the upper part of the gas and (c) \(p_2 A\) upward, by the lower part of the gas.
Thus, \(\quad p_2 A=p_1 A+m g \dots(i)\)
Using \(p V=n R T\) for the upper and the lower parts
\(p_1 l_1 A=n R T \dots(ii)\)
and \(\quad p_2 l_2 A=n R T \dots(iii)\).
Putting \(p_1 A\) and \(p_2 A\) from (ii) and (iii) into (i),
\(
\frac{n R T}{l_2}=\frac{n R T}{l_1}+m g .
\)
Thus,
\(
m=\frac{n R T}{g}\left[\frac{1}{l_2}-\frac{1}{l_1}\right]
\)
\(
\begin{aligned}
& =\frac{(0.1 \mathrm{~mol})\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K})}{9.8 \mathrm{~m} \mathrm{~s}^{-2}}\left[\frac{1}{0.4 \mathrm{~m}}-\frac{1}{0.5 \mathrm{~m}}\right] \\
& =12 \cdot 7 \mathrm{~kg} .
\end{aligned}
\)

Hint

As the separator is conducting, the temperatures in the two parts will be the same. Suppose the common temperature is \(T\) when the separator is in the middle. Let \(n_1\) and \(n_2\) be the number of moles of the gas in the left part and the right part respectively. Using the ideal gas equation,
\(p_1 \frac{V_0}{2}=n_1 R T\)
and \(p_2 \frac{V_0}{2}=n_2 R T\).
Thus, \(
\frac{n_1}{n_2}=\frac{p_1}{p_2} \text {. } \dots(i)
\)
The separator will stay in equilibrium at a position where the pressures on the two sides are equal. Suppose the volume of the left part is \(V_1\) and of the right part is \(V_2\) in this situation. Let the common pressure be \(p^{\prime}\). Also, let the common temperature in this situation be \(T^{\prime}\). Using ideal gas equation,
\(
p^{\prime} V_1=n_1 R T^{\prime}
\)
and
or, \(\quad \frac{V_1}{V_2}=\frac{n_1}{n_2}=\frac{p_1}{p_2}\). [using (i)]
Also, \(\quad V_1+V_2=V_0\).
Thus, \(\quad V_1=\frac{p_1 V_0}{p_1+p_2}\) and \(V_2=\frac{p_2 V_0}{p_1+p_2}\).

Hint

When the tube is kept inclined to the vertical, the length of the upper part is \(l_1=46 \mathrm{~cm}\) and that of the lower part is \(l_2=44 \cdot 5 \mathrm{~cm}\). When the tube lies horizontally, the length on each side is
\(
l_0=\frac{l_1+l_2}{2}=\frac{46 \mathrm{~cm}+44 \cdot 5 \mathrm{~cm}}{2}=45 \cdot 25 \mathrm{~cm} .
\)
Let \(p_1\) and \(p_2\) be the pressures in the upper and the lower parts when the tube is kept inclined. As the temperature is constant throughout, we can apply Boyle’s law. For the upper part,
\(
p_1 l_1 A=p l_0 A
\)
or,
\(
p_1=\frac{p l_0}{l_1} \dots(i)
\)
Similarly, for the lower part,
\(
p_2=\frac{p l_0}{l_2} \dots(ii)
\)

Now consider the equilibrium of the mercury pellet when the tube is kept in inclined position. Let \(m\) be the mass of the mercury. The forces along the length of the tube are
(a) \(p_1 A\) down the tube
(b) \(p_2 A\) up the tube
and (c) \(m g \cos 60^{\circ}\) down the tube.
Thus, \(\quad p_2=p_1+\frac{m g}{A} \cos 60^{\circ}\).
Putting from (i) and (ii),
\(
\frac{p l_0}{l_2}=\frac{p l_0}{l_1}+\frac{m g}{2 A}
\)
or, \(\quad p l_0\left(\frac{1}{l_2}-\frac{1}{l_1}\right)=\frac{m g}{2 A}\)
or,
\(
p=\frac{m g}{2 A l_0\left(\frac{1}{l_2}-\frac{1}{l_1}\right)} .
\)
If the pressure \(p\) is equal to a height \(h\) of mercury,
\(
\begin{aligned}
& p=h \rho g \text {. } \\
& \text { Also, } \quad m=(5 \mathrm{~cm}) A \rho \\
& \text { so that } h \rho g=\frac{(5 \mathrm{~cm}) A \rho g}{2 A l_0\left(\frac{1}{l_2}-\frac{1}{l_1}\right)} \\
& h=\frac{(5 \mathrm{~cm})}{2(45 \cdot 25 \mathrm{~cm})\left(\frac{1}{44 \cdot 5 \mathrm{~cm}}-\frac{1}{46 \mathrm{~cm}}\right)} \\
& =75 \cdot 39 \mathrm{~cm} \text {. } \\
&
\end{aligned}
\)
The pressure \(p\) is equal to \(75.39 \mathrm{~cm}\) of mercury.

Hint

Initially the spring is in its relaxed state. So, the pressure of the gas equals the atmospheric pressure.
Initial pressure \(=p_1=1.0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}\).
Final pressure \(=p_2=p_1+\frac{k x}{A}\)
\(
\begin{aligned}
& =1.0 \times 10^5 \mathrm{Nm}^{-2}+\frac{\left(8000 \mathrm{Nm}^{-1}\right)(0.1 \mathrm{~m})}{8.0 \times 10^{-3} \mathrm{~m}^2} \\
& =2.0 \times 10^5 \mathrm{Nm}^{-2} . \\
\text { Final volume } & =V_2=V_1+A x
\end{aligned}
\)
\(
=2.4 \times 10^{-3} \mathrm{~m}^3+8.0 \times 10^{-3} \mathrm{~m}^2 \times 0.1 \mathrm{~m}=3.2 \times 10^{-3} \mathrm{~m}^3 .
\)
\(
\text { Using } \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \text {, }
\)
\(
T_2=\frac{p_2 V_2}{p_1 V_1} T_1
\)
\(
\begin{aligned}
& =\frac{\left(2 \cdot 0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}\right)\left(3.2 \times 10^{-3} \mathrm{~m}^3\right)}{\left(1 \cdot 0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}\right)\left(2.4 \times 10^{-3} \mathrm{~m}^3\right)} \times 300 \mathrm{~K} \\
& =800 \mathrm{~K} .
\end{aligned}
\)

Hint

Suppose the pressure at height \(h\) is \(p\) and that at \(h+d h\) is \(p+d p\). Then
\(
d p=-\rho g d h \dots(i)
\)
Now considering any small volume \(\Delta V\) of air of mass \(\Delta m\)
or, or,
\(
\begin{aligned}
p \Delta V & =n R T=\frac{\Delta m}{M} R T \\
p & =\frac{\Delta m}{\Delta V} \frac{R T}{M}=\frac{\rho R T}{M} \\
\rho & =\frac{M}{R T} p .
\end{aligned}
\)
Putting in (i),
\(
\begin{array}{r}
d p=-\frac{M}{R T} p g d h \\
\text { or, } \quad \int_{P_0}^p \frac{d p}{p}=\int_0^h-\frac{M}{R T} g d h
\end{array}
\)
or, \(\ln \frac{p}{p_0}=-\frac{M g h}{R T}\) where \(p_0\) is the pressure at \(h=0\).
Thus,
\(
p=p_0 e^{-\frac{M g h}{R T}}
\)

Hint

Consider an element of the gas between the cross sections at distances \(x\) and \(x+d x\) from the fixed end (figure above). If \(p\) be the pressure at \(x\) and \(p+d p\) at \(x+d x\), the force acting on the element towards the centre is \(A d p\), where \(A\) is the cross-sectional area. As this element is going in a circle of radius \(x\), \(
A d p=(d m) \omega^2 x
\)
where \(d m=\) mass of the element. Using \(p V=n R T\) on this element,
or,
\(
\begin{aligned}
p A d x & =\frac{d m}{M} R T \\
d m & =\frac{M p A}{R T} d x .
\end{aligned}
\)
Putting in (i),
\(
\begin{aligned}
A d p & =\frac{M p A}{R T} \omega^2 x d x \\
\int_{p_1}^{p_2} \frac{d p}{p} & =\int_0^l \frac{M \omega^2}{R T} x d x \\
\ln \frac{p_2}{p_1} & =\frac{M \omega^2 l^2}{2 R T} \\
p_2 & =p_1 e^{\frac{M \omega^2 l^2}{2 R T}} .
\end{aligned}
\)

Hint

The pressure due to the air + vapour is \(76 \mathrm{~cm}-70 \mathrm{~cm}=6 \mathrm{~cm}\) of mercury. The vapour is saturated and the pressure due to it is \(1 \mathrm{~cm}\) of mercury. The pressure due to the air is, therefore, \(5 \mathrm{~cm}\) of mercury.
As the tube is lowered and the volume above the mercury is decreased, some of the vapour will condense. The remaining vapour will again exert a pressure of \(1 \mathrm{~cm}\) of mercury. The pressure due to air is doubled as the volume is halved. Thus, \(p_{\text {air }}=2 \times 5 \mathrm{~cm}=10 \mathrm{~cm}\) of mercury. The pressure due to the air + vapour \(=10 \mathrm{~cm}+1 \mathrm{~cm}=11 \mathrm{~cm}\) of mercury. The barometer reading is \(76 \mathrm{~cm}-11 \mathrm{~cm}=65 \mathrm{~cm}\).

Hint

At \(300 \mathrm{~K}\), the saturation vapour pressure \(=3.6 \mathrm{kPa}\). Considering \(1 \mathrm{~m}^3\) of volume,
\(
p V=n R T=\frac{m}{M} R T
\)
where \(m=\) mass of vapour and \(M=\) molecular weight of water.
Thus, \(m=\frac{M p V}{R T}\)
\(
=\frac{\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)\left(3 \cdot 6 \times 10^3 \mathrm{~Pa}\right)\left(1 \mathrm{~m}^3\right)}{\left(8 \cdot 3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K})} \approx 26 \mathrm{~g} .
\)
As the relative humidity is \(50 \%\), the amount of vapour present in \(1 \mathrm{~m}^3\) is \(26 \mathrm{~g} \times 0.50=13 \mathrm{~g}\).

Hint

The relative humidity is
Thus, the vapour pressure at \(20^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& =0.8 \times 17.5 \mathrm{~mm} \text { of } \mathrm{Hg} \\
& =14 \mathrm{~mm} \text { of } \mathrm{Hg} .
\end{aligned}
\)
Consider a volume \(V\) of air. If the vapour pressure is \(p\) and the temperature is \(T\), the mass \(m\) of the vapour present is given by
\(
\begin{aligned}
p V & =\frac{m}{M} R T \\
m & =\frac{M V}{R} \frac{p}{T} \dots(i)
\end{aligned}
\)
The mass present at \(20^{\circ} \mathrm{C}\) is
\(
m_1=\frac{M V}{R} \frac{14 \mathrm{~mm} \text { of } \mathrm{Hg}}{293 \mathrm{~K}} .
\)
When the air is cooled to \(5^{\circ} \mathrm{C}\), some vapour condenses and the air gets saturated with the remaining vapour. The vapour pressure at \(5^{\circ} \mathrm{C}\) is, therefore, \(6.5 \mathrm{~mm}\) of mercury. The mass of vapour present at \(5^{\circ} \mathrm{C}\) is, therefore,
\(
m_2=\frac{M V}{R} \frac{6 \cdot 5 \mathrm{~mm} \text { of } \mathrm{Hg}}{278 \mathrm{~K}} .
\)
The fraction condensed
\(
\begin{aligned}
& =\frac{m_1-m_2}{m_1}=1-\frac{m_2}{m_1} \\
& =1-\frac{6 \cdot 5}{278} \times \frac{293}{14}=0.51 .
\end{aligned}
\)

Hint

Water will be in equilibrium with its vapour when the vapour gets saturated. In this case, the pressure of vapour=saturation vapour pressure \(=15 \mathrm{~mm}\) of mercury
\(
\begin{aligned}
& =\left(15 \times 10^{-3} \mathrm{~m}\right)\left(13600 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right) \\
& =2000 \mathrm{~N} \mathrm{~m}^{-2} .
\end{aligned}
\)
Using gas law, \(p V=\frac{m}{M} R T\)
\(
\begin{aligned}
m & =\frac{M p V}{R T} \\
& =\frac{\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)\left(2000 \mathrm{~N} \mathrm{~m}^{-2}\right)\left(76 \mathrm{~m}^3\right)}{\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(288 \mathrm{~K})} \\
& =1145 \mathrm{~g}=1.14 \mathrm{~kg} .
\end{aligned}
\)
Thus, \(1.14 \mathrm{~kg}\) of water will evaporate.

Hint

At temperature \(T_1\), the total pressure is \(830 \mathrm{~mm}\) of mercury. Out of this, \(30 \mathrm{~mm}\) of mercury is due to the vapour and \(800 \mathrm{~mm}\) of mercury is due to the gas. As the temperature decreases, the pressure due to the gas decreases according to the gas law. Here the volume is constant, so,
or,
\(
\begin{aligned}
& \frac{p_2}{T_2}=\frac{p_1}{T_1} \\
& p_2=\frac{T_2}{T_1} p_1 .
\end{aligned}
\)
As \(T_2\) is \(1 \%\) less than \(T_1\)
\(
T_2=0.99 T_1
\)
and hence,
\(
p_2=0 \cdot 99 p_1
\)
\(=0.99 \times 800 \mathrm{~mm}\) of mercury \(=792 \mathrm{~mm}\) of mercury.
The vapour is still saturated and hence, its pressure is \(25 \mathrm{~mm}\) of mercury. The total pressure at the reduced temperature is
\(
\begin{aligned}
p & =(792+25) \mathrm{mm} \text { of mercury } \\
& =817 \mathrm{~mm} \text { of mercury. }
\end{aligned}
\)

Hint

We have \(p V=\frac{m}{M} R T\)
or,
\(
\rho=\frac{m}{V}=\frac{M p}{R T} \dots(i)
\)
The dew point is \(161^{\circ} \mathrm{C}\) and the saturation vapour pressure is \(13.6 \mathrm{~mm}\) of mercury at the dew point. This means that the present vapour pressure is \(13.6 \mathrm{~mm}\) of mercury.
At this pressure and temperature, the density of vapour will be
\(
\begin{aligned}
& \rho=\frac{M p}{R T} \\
& =\frac{\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)\left(13 \cdot 6 \times 10^{-3} \mathrm{~m}\right)\left(13600 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(9 \cdot 8 \mathrm{~m} \mathrm{~s}^{-2}\right)}{\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K})} \\
& =13.1 \mathrm{~g} \mathrm{~m}^{-3} .
\end{aligned}
\)
Thus, 1 litre of moist air at \(27^{\circ} \mathrm{C}\) contains \(0.0131 \mathrm{~g}\) of vapour.
The pressure of dry air at \(27^{\circ} \mathrm{C}\) is \(753.6 \mathrm{~mm}-13.6 \mathrm{~mm}\) \(=740 \mathrm{~mm}\) of mercury. The density of air at STP is \(0.001293 \mathrm{~g}(\mathrm{cc})^{-1}\). The density at \(27^{\circ} \mathrm{C}\) is given by equation (i),
or,
\(
\begin{aligned}
\frac{\rho_1}{\rho_2} & =\frac{p_1 / T_1}{p_2 / T_2} \\
\rho_2 & =\frac{p_2 T_1}{T_2 p_1} \rho_1 \\
& =\frac{740 \times 273}{300 \times 760} \times 0.001293 \mathrm{~g}(\mathrm{cc})^{-1} . \\
& =.001457 \mathrm{~g}(\mathrm{cc})^{-1} .
\end{aligned}
\)
Thus, 1 litre of moist air contains \(1.145 \mathrm{~g}\) of dry air. The mass of 1 litre of moist air is \(1.1457 \mathrm{~g}+0.0131 \mathrm{~g}\) \(\approx 1 \cdot 159 \mathrm{~g}\)

Hint

Volume of 1 mole gas
\(
\begin{aligned}
& P V=n R T \\
& \Rightarrow V=\frac{R T}{P}=0.082 \times 273(\text { as } P=1) \\
& V=22.386=22.4 \mathrm{~L} \\
& =22.4 \times 10^{-3} \mathrm{~m}^3 \\
& =2.24 \times 10^{-2} \mathrm{~m}^3
\end{aligned}
\)

Hint

Volume of ideal gas at STP \(=22.4 \mathrm{~L}\)
Number of molecules in \(22.4 \mathrm{~L}\) of ideal gas at STP \(=6.022 \times 10^{23}\)
Number of molecules in \(22.4 \times 10^3 \mathrm{~cm}^3\) of ideal gas at STP \(=6.022 \times 10^{23}\)
Now,
Number of molecules in \(1 \mathrm{~cm}^3\) of ideal gas at STP \(=\frac{6.022 \times 10^{23}}{22.4 \times 10^3}=2.688 \times 10^{19}\)

Hint

Given:
Volume of ideal gas, \(V=1 \mathrm{~cm}^3=10^{-6} \mathrm{~m}^3\)
Temperature of ideal gas, \(T=0^{\circ} \mathrm{C}=273 \mathrm{~K}\)
Pressure of mercury, \(P=10^{-8} \mathrm{~m}\) of \(\mathrm{Hg}\)
Density of ideal gas, \(\rho=13600 \mathrm{kgm}^{-3}\)
Pressure \((P)\) is given by
\(P=\rho g h\)
Here,
\(\rho=\) density of ideal gas
\(g=\) acceleration due to gravity,
Using the ideal gas equation, we get
\(
\begin{aligned}
& n=\frac{P V}{R T} \\
& \Rightarrow n=\frac{\rho g h \times V}{R T} \\
& \Rightarrow n=\frac{13600 \times 9.8 \times 10^{-8} \times 10^{-6}}{8.31 \times 273} \\
& \Rightarrow n=5.87 \times 10^{-13}
\end{aligned}
\)
\(
\begin{aligned}
\text { Number of molecules } & =N \times n \\
& =6.023 \times 10^{23} \times 5.874 \times 10^{-13} \\
& =35.384 \times 10^{10} \\
& =3.538 \times 10^{11}
\end{aligned}
\)

Hint

We know that \(22.4 \mathrm{~L}\) of \(\mathrm{O}_2\) contains \(1 \mathrm{~mol} \mathrm{O}_2\) at STP. Thus,
\(22.4 \times 10^3 \mathrm{~cm}^3\) of \(\mathrm{O}_2=1 \mathrm{~mol} \mathrm{O}_2\)
\(1 \mathrm{~cm}^3\) of \(\mathrm{O}_2=\frac{1}{22.4 \times 10^3} \mathrm{~mol} \mathrm{O}_2\)
\(1 \mathrm{~mol}\) of \(\mathrm{O}_2=32 \mathrm{~g}\)
\(\frac{1}{22.4 \times 10^3} \mathrm{~mol}\) of \(\mathrm{O}_2=\frac{32}{22.4 \times 10^3}=1.43 \times 10^{-3} \mathrm{~g}=1.43 \mathrm{mg}\)

Hint

Let the pressure and temperature for the vessels of volume \(V_0\) and \(2 V_0\) be \(P_1, T_1\) and \(P_2, T_2\), respectively. Since the two vessels have the same mass of gas, \(n_1=n_2=n\).
\(
\begin{aligned}
& T_1=300 K \\
& T_2=600 K
\end{aligned}
\)
Using the equation of state for perfect gas, we get \(P V=n R T\)
For the vessel of volume \(\mathrm{V}_o\) :
\(
P_1 V_o=n R T_1 \ldots(1)
\)
For the vessel of volume \(2 \mathrm{~V}_o\) :
\(
P_2\left(2 V_o\right)=n R T_2 \ldots(2)
\)
Dividing eq . (2) by eq . (1), we get
\(
\begin{aligned}
& \frac{2 P_2}{P_1}=\frac{T_2}{T_1}=\frac{600}{300}=2 \\
& \Rightarrow \frac{P_2}{P_1}=1 \\
& \Rightarrow P_2: P_1=1: 1
\end{aligned}
\)

Hint

Given:
Volume of electric bulb, \(V=250 \mathrm{cc}\)
Temperature at which manufacturing takes place, \(T=27+273=300 \mathrm{~K}\)
Height of mercury, \(h=10^{-3} \mathrm{~mm}\)
Density of mercury, \(\rho 13600 \mathrm{kgm}^{-3}\)
Avogadro constant, \(N=6 \times 10^{23} \mathrm{~mol}^{-1}\)
Pressure \((P)\) is given by
\(
P=\rho g h
\)
Using the ideal gas equation, we get
\(
\begin{aligned}
& P V=n R T \\
& P V=n R T \\
& \Rightarrow n=\frac{P V}{R T} \\
& \Rightarrow n=\frac{\rho g h V}{R T} \\
& \Rightarrow n=\frac{10^{-6} \times 13600 \times 10 \times 250 \times 10^{-6}}{8.314 \times 300}
\end{aligned}
\)
Now, number of molecules \(=n N\)
\(
\begin{aligned}
& =\frac{10^{-6} \times 13600 \times 10 \times 250 \times 10^{-6}}{8.314 \times 300} \times 6 \times 10^{23} \\
& =8 \times 10^{15}
\end{aligned}
\)

Hint

Maximum pressure that the cylinder can bear, \(P_{\max }=1.0 \times 10^6 \mathrm{~Pa}\)
Pressure in the gas cylinder, \(P_1=8.0 \times 10^5 \mathrm{~Pa}\)
Temperature in the cylinder, \(T_1=300 \mathrm{~K}\)
Let \(T_2\) be the temperature at which the cylinder will break.
Volume is constant. Thus,
(Given)
\(
V_1=V_2=V
\)
Applying the five variable gas equation, we get
\(
\begin{aligned}
& \frac{P_1 V}{T_1}=\frac{P_2 V}{T_2} \ldots \ldots\left(\because V_1=V_2\right) \\
& \Rightarrow \frac{P_1}{T_1}=\frac{P_2}{T_2} \\
& \Rightarrow T_2=\frac{P_2 \times T_1}{P_1} \\
& \Rightarrow T_2=\frac{1.0 \times 10^6 \times 300}{8.0 \times 10^5}=375 K
\end{aligned}
\)

Hint

Given:
Mass of hydrogen, \(m=2 \mathrm{~g}\)
Volume of the vessel, \(V=0.02 \mathrm{~m}^3\)
Temperature in the vessel, \(T=300 \mathrm{~K}\)
Molecular mass of the hydrogen, \(M=2 \mathrm{u}\)
No of moles, \(n=\frac{m}{M}=\frac{2}{2}=1\) mole
Rydberg’s constant, \(R=8.3 \mathrm{~J} / \mathrm{Kmol}\)
From the ideal gas equation, we get
\(
\begin{aligned}
& \mathrm{PV}=\mathrm{nRT} \\
& \Rightarrow P=\frac{n R T}{V} \\
& \Rightarrow P=\frac{1 \times 8.3 \times 300}{0.02} \\
& \Rightarrow P=1.24 \times 10^5 \mathrm{~Pa}
\end{aligned}
\)

Hint

Let:
\(m\) = Mass of the gas
\(M=\) Molecular mass of the gas
Now,
Density of ideal gas, \(\rho=1.25 \times 10^{-3} \mathrm{gcm}^{-3}=1.25 \mathrm{kgm}^{-3}\)
Pressure, \(P=1.01325 \times 10^5 \mathrm{~Pa} \quad\) (At STP)
Temperature, \(T=273 \mathrm{~K} \quad\) (At STP)
Using the ideal gas equation, we get
\(
\begin{aligned}
& P V=n R T \ldots(1) \\
& n=\frac{m}{M} \ldots(2) \\
& \therefore P V=\frac{m}{M} R T \\
& \Rightarrow M=\frac{m}{V} \frac{R T}{P} \\
& \Rightarrow M=\rho \frac{R T}{P} \\
& \Rightarrow M=1.25 \times \frac{8.31 \times 273}{10^5} \\
& \Rightarrow M=2.83 \times 10^{-2} \\
& =28.3 g \mathrm{mol}^{-1}
\end{aligned}
\)

Hint

Here,
Temperature in Simla, \(T_1=15+273=288 \mathrm{~K}\)
Pressure in Simla, \(\mathrm{P}_1=0.72 \mathrm{~m}\) of \(\mathrm{Hg}\)
Temperature in Kalka, \(\mathrm{T}_2=35+273=308 \mathrm{~K}\)
Pressure in Kalka, \(P_2=0.76 \mathrm{~m}\) of \(\mathrm{Hg}\)
Let the density of air at Simla and Kalka be \(\rho_1\) and \(\rho_2\) respectively. Then,
\(
\begin{aligned}
& P V=\frac{m}{M} R T \\
& \Rightarrow \frac{m}{V}=\frac{P M}{R T} \\
& \Rightarrow \rho=\frac{P M}{R T} \\
& \text { Thus, } \rho_1=\frac{P_1 M}{R T_1} \\
& \rho_2=\frac{P_2 M}{R T_2}
\end{aligned}
\)
Taking ratios, we get
\(
\begin{aligned}
& \frac{\rho_1}{\rho_2}=\frac{P_1}{T_1} \times \frac{T_2}{P_2} \\
& \Rightarrow \frac{\rho_1}{\rho_2}=\frac{0.72}{288} \times \frac{308}{0.76} \\
& \Rightarrow \frac{\rho_2}{\rho_1}=0.987
\end{aligned}
\)

Hint

Since the separator initially divides the cylinder equally, the number of moles of gas are equal in the two parts. Thus,
\(
\mathrm{n}_1=\mathrm{n}_2=\mathrm{n}
\)
Volume of the first part \(=V\)
Volume of the second part \(=3 \mathrm{~V}\)
It is given that the walls are diathermic. So, temperature of the two parts is equal. Thus,
\(
T_1=T_2=T
\)
Let pressure of first and second parts be \(P_1\) and \(P_2\), respectively.
For first part:- Applying equation of state, we get
\(
P_1 V=n R T \dots(1)
\)
For second part:- Applying equation of state, we get
\(
P_2(3 V)=n R T \dots(2)
\)
Dividing eq. (1) by eq. (2), we get
\(
\begin{aligned}
& \frac{P_1 V}{P_2(3 V)}=1 \\
& \Rightarrow \frac{P_1}{P_2}=\frac{3}{1} \\
& \Rightarrow P_1: P_2=3: 1
\end{aligned}
\)

Hint

Here,
Temperature of hydrogen gas, \(T=300 \mathrm{~K}\)
Molar mass of hydrogen, \(M_0=2 \mathrm{~g} / \mathrm{mol}=0.002 \mathrm{~kg} / \mathrm{mol}\)
We know,
\(
\begin{aligned}
& C=\sqrt{\frac{3 R T}{M_0}} \\
& \Rightarrow C=\sqrt{\frac{3 \times 8.3 \times 300}{0.002}} \\
& \Rightarrow C=1932.6 \mathrm{~ms}^{-1}\approx 1930 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
In the second case, let the required temperature be \(T\).
Applying the same formula, we get
\(
\begin{aligned}
& \sqrt{\frac{3 \times 8.3 T}{0.002}}=2 \times 1932.6 \\
& \Rightarrow T=1200 \mathrm{~K}
\end{aligned}
\)

Hint

Here,
\(
\begin{aligned}
& V=10^{-3} \mathrm{~m}^3 \\
& \text { Density }=0.177 \mathrm{kgm}^{-3} \\
& P=10^5 \mathrm{pa} \\
& C=\sqrt{\frac{3 P}{\rho}}=\sqrt{\frac{3 \times 10^5}{0.177}} \\
& =1301.9 \mathrm{~ms}^{-1} \approx 1300 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

We know from kinetic theory of gases that the average translational energy per molecule is \(\frac{3}{2} k T\).
Now,
\(
\begin{aligned}
& \mathrm{E}_{\mathrm{avg}}=0.040 \mathrm{eV}=0.040 \times 1.6 \times 10^{-19}=6.4 \times 10^{-21} J \\
& 6.40 \times 10^{-21}=\frac{3}{2} \times 1.38 \times 10^{-23} \times T \\
& \Rightarrow T=\frac{2}{3} \times \frac{6.40 \times 10^{-21}}{1.38 \times 10^{-23}}=309.2 \mathrm{~K}\approx 310\mathrm{~K}
\end{aligned}
\)

Hint

Here,
\(
\begin{aligned}
& V_{a v g}=\frac{\sqrt{8 R T}}{\sqrt{\pi M}}=\frac{\sqrt{8 \times 8.314 \times 300}}{\sqrt{3.14 \times 0.032}} \\
& =441.44 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
We know
\(
\begin{aligned}
& T=\frac{\text { Distance }}{\text { Speed }}=\frac{6400000 \times 2}{441.44} \\
& =\frac{28996.01 h}{3600}=8.054 h=8 h
\end{aligned}
\)

Hint

Here,
\(
\begin{aligned}
& \mathrm{m}=6.64 \times 10^{-27} \mathrm{~kg} \\
& \mathrm{~T}=273 \mathrm{~K}
\end{aligned}
\)
Average speed of the \(\mathrm{He}\) atom is given by \(V_{a v g}=\sqrt{\frac{8 k T}{\pi m}}\)
\(
\begin{aligned}
& =\sqrt{\frac{8 \times 1.38 \times 10^{-23} \times 273}{3.14 \times 6.64 \times 10^{-27}}} \\
& =1202.31
\end{aligned}
\)
We know,
Momentum \(=m \times V_{\text {avg }}\)
\(
\begin{aligned}
& =6.64 \times 10^{-27} \times 1201.35 \\
& =7.97 \times 10^{-24} \\
& =8 \times 10^{-24} \mathrm{~kg} \mathrm{m} / \mathrm{s}
\end{aligned}
\)

Hint

Mean velocity is given by \(V_{a v g}=\sqrt{\frac{8 R T}{\pi M}}\)
Let temperature for \(\mathrm{H}\) and \(\mathrm{He}\) respectively be \(T_1\) and \(T_2\), respectively.
For hydrogen:
\(
\mathrm{M}_{\mathrm{H}}=2 \mathrm{~g}=2 \times 10^{-3} \mathrm{~kg}
\)
For helium:
\(
\mathrm{M}_{\mathrm{He}}=4 \mathrm{~g}=4 \times 10^{-3} \mathrm{~kg}
\)
Now,
\(
\begin{aligned}
& \mathrm{A} / \mathrm{q} \sqrt{\frac{8 R T_1}{\pi M_H}}=\sqrt{\frac{8 R T_2}{\pi M_{H e}}} \\
& \Rightarrow \sqrt{\frac{8 R T_1}{2 \times 10^{-3} \pi}}=\sqrt{\frac{8 R T_2}{\pi \times 4 \times 10^{-3}}} \\
& \Rightarrow \sqrt{\frac{T_1}{2}}=\sqrt{\frac{T_2}{4}} \\
& \Rightarrow \frac{T_1}{T_2}=\frac{1}{2} \\
& \Rightarrow T_1: T_2=1: 2
\end{aligned}
\)

Hint

Mean speed of the molecule is givne by
\(
\sqrt{\frac{8 R T}{\pi M}}
\)
For \(\mathrm{H}\) molecule, \(M=2 \times 10^{-3} \mathrm{~kg}\)
\(
=\sqrt{\frac{4 R T \times 10^3}{\pi}}
\)
For escape velocity of Earth :
Let \(r\) be the radius of Earth
\(
v=\sqrt{\frac{2 G M}{r}}
\)
Multiplying numerator and denominator by \(R\), we get
\(
\begin{aligned}
& v_c=\sqrt{\frac{G M}{r^2} 2 r} \\
& g=\frac{G M}{r^2} \\
& v_c=\sqrt{2 g r} \\
& \sqrt{\frac{4 R T \times 10^3}{\pi}}=\sqrt{2 g r} \\
& \Rightarrow \frac{2 \times 8.314 \times T \times 10^3}{3.142}=9.8 \times 6.37 \times 10^6 \\
& \Rightarrow T \approx 11800 \mathrm{~K}
\end{aligned}
\)

Hint

we know ,
\(
\begin{aligned}
& V_{a v g}=\sqrt{\frac{8 R T}{\pi M}} \\
& \text { Molar mass of } \mathrm{H}_2=\mathrm{M}_{\mathrm{H}}=2 \times 10^{-3} \mathrm{~kg} \\
& \text { Molar mass of } \mathrm{N}_2=\mathrm{M}_{\mathrm{N}}=28 \times 10^{-3} \mathrm{~kg} \\
& \text { Now, } \\
& <V>_H=\sqrt{\frac{8 R T}{\pi M_H}} \\
& <V>_N=\sqrt{\frac{8 R T}{\pi M_N}} \\
& \frac{<V>_H}{<V>_N}=\sqrt{\frac{M_N}{M_H}}=\sqrt{\frac{28}{2}}=\sqrt{14}=3.74
\end{aligned}
\)

Hint

Given that
The RMS speed of the gas molecule is \(\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}\)
The average velocity of the gas molecule is \(\mathrm{V}_{\text {avg }}=\sqrt{\frac{8 \mathrm{KT}}{\pi \mathrm{m}_2}}\)
Equating the two velocities:
\(
\begin{aligned}
& \mathrm{V}_{\mathrm{rms}}=\mathrm{V}_{\mathrm{avg}} \\
& \Rightarrow \sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}_1}}=\sqrt{\frac{8 \mathrm{KT}}{\pi \mathrm{m}_2}}
\end{aligned}
\)
where \(m_1 \& m_2\) are mass of gas in left and right chamber respectively.
\(
\begin{aligned}
& \Rightarrow \text { So, } \sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}_1}}=\sqrt{\frac{8 \mathrm{KT}}{\pi \mathrm{m}_2}} \\
& \Rightarrow 3 \mathrm{~m}_2=\frac{8 \mathrm{~m}_1}{3.14} \\
& \Rightarrow \frac{\mathrm{m}_1}{\mathrm{~m}_2}=1.18
\end{aligned}
\)
Hence, the answer is 1.18.

Hint

Here,
\(
\begin{aligned}
& \lambda=1.38 \times 10^{-8} \mathrm{~m} \\
& \mathrm{~T}=273 \mathrm{~K} \\
& \mathrm{M}=2 \times 10^{-3} \mathrm{~kg}
\end{aligned}
\)
Average speed of the \(\mathrm{H}\) molecules is given by
\(
\begin{aligned}
& v_{a v g}=\sqrt{\frac{8 R T}{\pi M}} \\
& =\sqrt{\frac{8 \times 8.31 \times 273}{3.14 \times 2 \times 10^{-3}}} \\
& =1700 \mathrm{~ms}^{-1}
\end{aligned}
\)
The time between two collisions is given by
\(
\begin{aligned}
& t=\frac{\lambda}{v_{a v g}} \\
& \Rightarrow t=\frac{1.38 \times 10^{-8}}{1700} \\
& \Rightarrow t=8 \times 10^{-12} s
\end{aligned}
\)
Number of collisions in \(1 \mathrm{~s}=\frac{1}{8.11 \times 10^{-12}}=1.23 \times 10^{11}\)

Hint

\(
\begin{aligned}
& \mathrm{P}=10^5 \mathrm{~Pa} \\
& \mathrm{~T}=300 \mathrm{~K}
\end{aligned}
\)
For \(\mathrm{H}_2\)
\(
M=2 \times 10^{-3} \mathrm{~kg}
\)
(a) Mean speed is given by
\(
\begin{aligned}
& <v>=\sqrt{\frac{8 R T}{\pi M}} \\
& =\sqrt{\frac{8 \times 8.3 \times 300 \times 7}{2 \times 10^{-3} \times 22}} \\
& =1780 \mathrm{~ms}^{-1}
\end{aligned}
\)
Let us consider a cubic volume of \(1 \mathrm{~m}^3\).
\(
V=1 m^3
\)
The momentum of 1 molecule normal to the striking surface before collision \(=\mathrm{mu} \sin 45^{\circ}\)
The momentum of 1 molecule normal to the striking surface after collision \(=-m u \sin 45^{\circ}\)
Change in momentum of the molecule \(=2 \mathrm{mu} \sin 45^{\circ}=\sqrt{2} m u\)
Change in momentum of \(\mathrm{n}\) molecules \(=2 \mathrm{mnu} \sin 45^{\circ}=\sqrt{2} \mathrm{mnu}\)
Let \(\Delta \mathrm{t}\) be the time taken in changing the momentum.
Force per unit area due to one molecule \(=\frac{\sqrt{2} m u}{\Delta t}=\frac{\sqrt{2} m u}{\Delta t}\)
Observed pressure due to collision by \(\mathrm{n}\) molecules \(=\frac{\sqrt{2} m n u}{\Delta t}=10^5\)
\(
\begin{aligned}
& n=\frac{\frac{\sqrt{2} m n u}{\Delta t}}{\frac{\sqrt{2} m u}{\Delta t}}=\frac{10^5}{\sqrt{2} m u} \\
& 6.0 \times 10^{23} \text { molecules }=2 \times 10^{-3} \mathrm{~kg} \\
& 1 \text { molecule }=\frac{2 \times 10^{-} 3}{6 \times 10^{23}}=3.3 \times 10^{-27} \mathrm{~kg} \\
& \Rightarrow n=\frac{10^5}{\sqrt{2} \times 3.3 \times 10^{-27} \times 1780}=1.2 \times 10^{28}
\end{aligned}
\)

Hint

Here,
\(
\begin{aligned}
& P_1=2 \times 10^5 \mathrm{~Pa} \\
& \mathrm{P}_2=? \\
& \mathrm{~T}_1=293 \mathrm{~K} \\
& \mathrm{~T}_2=313 \mathrm{~K} \\
& \mathrm{~V}_2=\mathrm{V}_1+0.02 \mathrm{~V}_1=\mathrm{V}_1(1.02)
\end{aligned}
\)
Now,
\(
\begin{aligned}
& \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \\
& \Rightarrow \frac{2 \times 10^5 V_1}{293}=\frac{P_2 V_1(1.02)}{313} \\
& \Rightarrow P_2=\frac{2 \times 10^5 \times 313}{293 \times 1.02}=209 \mathrm{kPa}
\end{aligned}
\)

Hint

Here,
\(
\begin{aligned}
& \mathrm{V}_1=1.0 \times 10^{-3} \mathrm{~m}^3 \\
& \mathrm{~T}_1=400 \mathrm{~K} \\
& \mathrm{P}_1=1.5 \times 10^5 \mathrm{~Pa} \\
& \mathrm{P}_2=1.0 \times 10^5 \mathrm{~Pa} \\
& \mathrm{~T}_2=300 \\
& M=32 \mathrm{~g}
\end{aligned}
\)
Number of moles in the jar before \(n_1=\frac{P_1 V_1}{R T_1}\)
Volume of the gas when pressure becomes equal to external pressure is given by
\(
\begin{aligned}
& \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \\
& \Rightarrow V_2=\frac{P_1 V_1 T_2}{P_2 T_1} \\
& \Rightarrow V_2=\frac{1.5 \times 10^5 \times 1.0 \times 10^{-3} \times 300}{1.0 \times 10^5 \times 400}=1.125 \times 10^{-3}
\end{aligned}
\)
Net volume of leaked gas \(=V_2-V_1\)
\(
\begin{aligned}
& =1.125 \times 10^{-3}-1.0 \times 10^{-3} \\
& =1.25 \times 10^{-4} \mathrm{~m}^3
\end{aligned}
\)
Let \(n_2\) be the number of moles of leaked gas. Applying the equation of state on this amount of gas, we get
\(
n_2=\frac{P_2 V_2}{R T_2}=\frac{1.0 \times 10^5 \times 1.25 \times 10^{-4}}{8.3 \times 300}=0.005
\)
Mass of leaked gas \(=32 \times 0.005=0.16 \mathrm{~g}\)

Hint

\(
\begin{aligned}
& V_1=\frac{4}{3} \pi\left(2.0 \times 10^{-3}\right)^3 \\
& h=3.3 \mathrm{~m} \\
& P_1=P_0+\rho g h \\
& \Rightarrow P_1=1.0 \times 10^5+1000 \times 9.8 \times 3.3 \\
& \Rightarrow P_1=1.32 \times 10^5 \mathrm{~Pa} \\
& \quad P_2=1.0 \times 10^5 \mathrm{~Pa}
\end{aligned}
\)
Since temperature remains the same, applying Boyle’s law we get
\(
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\
& \Rightarrow \mathrm{V}_2=\frac{P_1 V_1}{P_2} \\
& \Rightarrow \mathrm{V}_2=\frac{1.32 \times 10^5 \times \frac{4}{3} \pi\left(2.0 \times 10^{-3}\right)^3}{1.0 \times 10^5}
\end{aligned}
\)
Let \(R_2\) be the new radius. Then,
\(
\begin{aligned}
& \frac{4}{3} \pi R_2^3=\frac{1.32 \times 10^5 \times \frac{4}{3} \pi\left(2.0 \times 10^{-3}\right)^3}{1.0 \times 10^5} \\
& \Rightarrow R_2^3=\frac{1.32 \times 10^5 \times\left(2.0 \times 10^{-3}\right)^3}{1.0 \times 10^5} \\
& \Rightarrow R_3=\sqrt[3]{\frac{1.32 \times 10^5 \times\left(2.0 \times 10^{-3}\right)^3}{1.0 \times 10^5}} \\
& \Rightarrow R_3=2.2 \times 10^{-3} \mathrm{~m}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \mathrm{P}_1=2 \times 10^5 \mathrm{pa} \\
& \mathrm{V}_1=0.002 \mathrm{~m}^3 \\
& \mathrm{~V}_2=0.0005 \mathrm{~m}^3 \\
& \mathrm{~T}_1=\mathrm{T}_2=300 \mathrm{~K}
\end{aligned}
\)
Number of moles initially , \(\mathrm{n}_1=\frac{P_1 V_1}{R T_1}\)
\(
\begin{aligned}
& \Rightarrow \mathrm{n}_1=\frac{2 \times 10^5 \times 0.002}{8.3 \times 300} \\
& \Rightarrow \mathrm{n}_1=0.16
\end{aligned}
\)
Applying equation of state, we get
\(
P_2 V_2=n_2 R T
\)
Assuming the final pressure becomes equal to the atmospheric pressure, we get
\(
\begin{aligned}
& \mathrm{P}_2=1.0 \times 10^5 \mathrm{pa} \\
& \Rightarrow \mathrm{n}_2=\frac{P_2 V_2}{R T} \\
& \Rightarrow \mathrm{n}_2=\frac{1.0 \times 10^5 \times 0.0005}{8.3 \times 300} \\
& \Rightarrow \mathrm{n}_2=0.02
\end{aligned}
\)
Number of leaked moles \(=n_2-n_1\)
\(
\begin{aligned}
& =0.16-0.02 \\
& =0.14
\end{aligned}
\)

Hint

\(
\begin{aligned}
& m=0.040 \mathrm{~g} \\
& \mathrm{M}=4 \mathrm{~g} \\
& \mathrm{n}=\frac{0.040}{4}=0.01 \\
& T_1=(100+273) \mathrm{K}=373 \mathrm{~K}
\end{aligned}
\)
He is a monoatomic gas. Thus,
\(
\begin{aligned}
& C_v=3 \times\left(\frac{1}{2} R\right) \\
& \Rightarrow C_{\mathrm{v}}=1.5 \times 8.3=12.45
\end{aligned}
\)
Let the initial internal energy be \(U_1\).
Let the final internal energy be \(\mathrm{U}_2\).
\(
\begin{aligned}
& U_2-U_1=n C_V\left(T_2-T_1\right) \\
& \Rightarrow 0.01 \times 12.45\left(T_2-373\right)=12 \\
& \Rightarrow T_2=469 \mathrm{~K}
\end{aligned}
\)
The temperature in \({ }^{\circ} \mathrm{C}\) can be obtained as follows: \(469-273=196^{\circ} \mathrm{C}\)

Hint

\(
\begin{aligned}
& p V=n R T \\
& \Rightarrow p=(n R T / V) \\
& \text { taking }(n R) \text { as a constant } k \\
& \Rightarrow p=k T / V
\end{aligned}
\)
given
\(
\begin{aligned}
& p V^2=\text { constant } \\
& (k T / V) \times V^2=\text { constant } \\
& T \times \mathrm{~V}=\text { constant }
\end{aligned}
\)
As \(V\) becomes \(2 \mathrm{~V}\),
\(\mathrm{T}\) will become \(\mathrm{T} / 2\) to make \(\left(\mathrm{T} \times \mathrm{~V}=\right.\) constant)
temperature \(=\mathrm{T} / 2\)

Hint

\(
\begin{aligned}
& P_{O_2}=\frac{n_{O_2} R T}{V}, \\
& P_{\mathrm{H}_2}=\frac{n_{H_2} R T}{V} \\
& n_{O_2}=\frac{m}{M_{O_2}}=\frac{1.60}{32}=0.05 \\
& \text { Now } P_{\text {mix }}=\left(\frac{n_{O_2}+n_{H_2}}{V}\right) R T \\
& n_{H_2}=\frac{m}{M_{H_2}}=\frac{2.80}{28}=0.1 \\
& P_{\text {mix }}=\frac{(0.05+0.1) \times 8.3 \times 300}{0.166}=2250 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)

Hint

\(
\mathrm{h}=1 \mathrm{~m}
\)
\(
\mathrm{P}_1=0.75 \mathrm{mHg}=0.75 \mathrm{\rho g} \mathrm{Pa}
\)
Let \(\mathrm{h}\) be the height of the mercury above the piston.
\(
P_2=P_1+h \rho g
\)
Let the CSA be A.
\(
\begin{aligned}
& V_1=A h=A \\
& V_2=(1-h) A
\end{aligned}
\)
Applying Boyle’s law, we get
\(
\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& \Rightarrow 0.75 \rho g A=P_2(1-h) A
\end{aligned}
\)
\(
\Rightarrow 0.75 \rho g=(0.75 \rho g+h \rho g)(1-h)
\)
\(
\Rightarrow 0.75=(0.75+\mathrm{h})(1-\mathrm{h})
\)
\(
\begin{aligned}
& \Rightarrow \mathrm{h}=0.25 \mathrm{~m} \\
& \mathrm{~h}=25 \mathrm{~cm}
\end{aligned}
\)

Hint

Let the partial pressure of the gas in chamber \(A\) and \(B\) be \(P_A^{\prime}\) and \(P_B^{\prime}\), respectively.
Applying equation of state for gas \(A\), we get
\(
\begin{aligned}
& \frac{P_A V}{T_A}=\frac{P_A^{\prime} 2 V}{T} \\
& \Rightarrow P_A^{\prime}=\frac{P_A T}{2 T_A}
\end{aligned}
\)
Similarly, For gas B :
\(
P_B^{\prime}=\frac{P_B T}{2 T_B}
\)
Total Pressure is the sum of the partial pressures . It is given by
\(
\begin{aligned}
& \mathrm{P}=\mathrm{P}_{\mathrm{A}}^{\prime}+\mathrm{P}_{\mathrm{B}}^{\prime} \\
& =\frac{P_A T}{2 T_A}+\frac{P_B T}{2 T_B} \\
& \Rightarrow \mathrm{P}=\frac{T}{2}\left(\frac{P_A}{T_A}+\frac{P_B}{T_B}\right) \\
& \Rightarrow \frac{P}{T}=\frac{1}{2}\left(\frac{P_A}{T_A}+\frac{P_B}{T_B}\right)
\end{aligned}
\)

Hint

(a) Here,
\(
\begin{aligned}
& V_1=5 \times 10^{-5} \mathrm{~m}^3 \\
& P_1=10^5 \mathrm{~Pa} \\
& T_1=273 \mathrm{~K} \\
& \mathrm{M}=28.8 \mathrm{~g} \\
& P_1 V_1=n R T_1 \\
& \Rightarrow n=\frac{P_1 V_1}{R T_1} \\
& \Rightarrow \frac{m}{M}=\frac{10^5 \times 5 \times 10^{-5}}{8.3 \times 273} \\
& \Rightarrow m=\frac{10^5 \times 5 \times 10^{-5} \times 28.8}{8.3 \times 273} \\
& \Rightarrow \mathrm{m}=0.0635 \mathrm{~g}
\end{aligned}
\)
(b) Here,
\(
\begin{aligned}
& V_1=5 \times 10^{-5} \mathrm{~m}^3 \\
& P_1=10^5 \mathrm{~Pa}, P_2=10^5 \mathrm{~Pa}, T_1=273 \mathrm{~K}, T_2=373 \mathrm{~K} \\
& \mathrm{M}=28.8 \mathrm{~g} \\
& \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \\
& \Rightarrow \frac{5 \times 10^{-5}}{273}=\frac{V_2}{373} \\
& \Rightarrow V_2=\frac{5 \times 10^{-5} \times 373}{273} \\
& \Rightarrow V_2=6.831 \times 10^{-5}
\end{aligned}
\)
Volume of expelled air \(=6.831 \times 10^{-5}-5 \times 10^{-5}\)
\(
=1.831 \times 10^{-5}
\)
Applying equation of state, we get
\(
\begin{aligned}
& \mathrm{PV}=\mathrm{nRT} \\
& \Rightarrow \frac{m}{M}=\frac{P V}{R T}=\frac{10^5 \times 1.831 \times 10^{-5}}{8.3 \times 373} \\
& \Rightarrow m=\frac{28.8 \times 10^5 \times 1.831 \times 10^{-5}}{8.3 \times 373}=0.017
\end{aligned}
\)
Thus, mass of expelled air \(=0.017 \mathrm{~g}\)
Amount of air in the container \(=0.0635-0.017=0.0465 \mathrm{~g}\)
(c) Here,
\(
\begin{aligned}
& \mathrm{T}=273 \mathrm{~K} \\
& P=10^5 \mathrm{~Pa} \\
& V=5 \times 10^{-5} \mathrm{~m}^3
\end{aligned}
\)
Applying equation of state, we get
\(
\begin{aligned}
& \mathrm{PV}=\mathrm{nRT} \\
& \Rightarrow P=\frac{n R T}{\mathrm{~V}}=\frac{0.0465 \times 8.3 \times 273}{28.8 \times 5 \times 10^{-5}} \\
& P=0.731 \times 10^5 \approx 73 \mathrm{KPa}
\end{aligned}
\)

Hint

Let the CSA of the tube be \(A\).
Initial volume of air, \(\mathrm{V}_1=20 \mathrm{~A} \mathrm{~cm}=0.2 \mathrm{~A}\)
Length of mercury, \(\mathrm{h}=0.1 \mathrm{~m}\)
Let the pressure of the trapped air when the tube is inverted and vertical be \(P_1\).
Now, Pressure of the mercury and trapped air balances the atmospheric pressure. Thus,
\(
\begin{aligned}
& \mathrm{P} 1+0.1 \rho g=0.75 \rho g \\
& \Rightarrow \mathrm{P}_1=0.65 \rho g
\end{aligned}
\)
when the tube is inverted with the closed end down, the pressure acting upon the trapped air is
Atmospheric pressure + Mercury column pressure
Now , Pressure of trapped air \(=\) Atmospheric Pressure + Mercury column Pressure \(\quad\) [In equilibrium]
\(
\mathrm{P}_2=0.75 \rho g+0.1 \rho g=0.85 \rho g
\)
Applying the Boyle’s law when the temperature remains constant, we get
\(
P_1 V_1=P_2 V_2
\)
Let the new height of the trapped air be \(x\).
\(
\begin{aligned}
& \Rightarrow 0.65 \rho g 0.2 \mathrm{~A}=0.85 \rho g \times \mathrm{A} \\
& \Rightarrow \mathrm{x}=0.15 \mathrm{~m}=15 \mathrm{~cm}
\end{aligned}
\)

Hint

Let CSA of the tube be \(A\). on the colder side:
\(
\begin{aligned}
& \mathrm{P}_1=0.76 \mathrm{~m} \mathrm{Hg} \\
& \mathrm{T}_1=300 \mathrm{~K} \\
& \mathrm{~V}_1=\mathrm{V} \\
& \mathrm{T}_2=273 \mathrm{~K} \\
& \mathrm{~V}_2=\mathrm{Ax} \\
& \frac{P_1 V}{T_1}=\frac{P_2 A x}{T_2} \\
& \Rightarrow P_2=\frac{P_1 V T_2}{T_1 A x}
\end{aligned}
\)
on the hotter side :
\(
\begin{aligned}
& \mathrm{P}_1=0.76 \mathrm{~m} \mathrm{Hg} \\
& \mathrm{T}_1=300 \mathrm{~K} \\
& \mathrm{~V}_1^{\prime}=\mathrm{V} \\
& \mathrm{T}_2^{\prime}=400 \mathrm{~K} \\
& \mathrm{~V}_2^{\prime}=\mathrm{Ay} \\
& \frac{{P}_1^{\prime} V}{T_1}=\frac{{P}_2^{\prime} A y}{{T}_1^{\prime}} \\
& \Rightarrow {P}_2^{\prime}=\frac{P_1 V {{T}_1^{\prime}}}{T_1 A y}
\end{aligned}
\)
In equilibrium, the pressures on both side will balance each other.
\(
\Rightarrow P_2^{\prime}=P_2
\)
\(
\begin{aligned}
& \Rightarrow \frac{P_1 V {T}_2^{\prime}}{T_1 A y}=\frac{P_1 V T_2}{T_1 A x} \\
& \Rightarrow \frac{{T}_2^{\prime}}{y}=\frac{T_2}{x}
\end{aligned}
\)
From the length of the tube, we get
\(
\begin{aligned}
& \mathrm{x}+\mathrm{y}+0.1=1 \\
& \Rightarrow \mathrm{y}=0.9-\mathrm{x} \\
& \frac{400}{(0.9-x)}=\frac{273}{x} \\
& \Rightarrow \mathrm{x}=0.365 \mathrm{~m} \\
& \Rightarrow \mathrm{x}=36.5 \mathrm{~cm}
\end{aligned}
\)

Hint

Initial pressure \(=\) Atmospheric pressure + pressure due to mercury
\(
\Rightarrow \mathrm{P}_1=\mathrm{P}_0+\mathrm{P}_{\mathrm{Hg}}
\)
Let the CSA of the tube be \(A\).
\(
\begin{aligned}
& \mathrm{P}_1=0.76+0.2=0.96 \mathrm{~m} \mathrm{Hg} \\
& \mathrm{T}_1=\mathrm{T}_2=\mathrm{T} \\
& \mathrm{V}_1=0.43 \mathrm{~A}
\end{aligned}
\)
If the tube is slanted, then the atmospheric pressure \(\mathrm{P}_0\) remains the same . only the \(\mathrm{P}_{\mathrm{Hg}}\) changes
\(
\begin{aligned}
& P_2=P_0+P_H g \cos 60^{\circ}=0.76+0.2 \times 0.5=0.86 \\
& P_1 V_1=P_2 V_2 \\
& \Rightarrow V_2=\frac{P_1 V_1}{P_2}=\frac{0.96 \times 0.43 A}{0.86}
\end{aligned}
\)
Let the length of the air column be \(l\).
\(
\begin{aligned}
& \Rightarrow \mathrm{Al}=\frac{P_1 V_1}{P_2}=\frac{0.96 \times 0.43 A}{0.86} \\
& \Rightarrow l=0.48 \mathrm{~m} \\
& \Rightarrow l=48 \mathrm{~cm}
\end{aligned}
\)

Hint

The middle wall is weakly conducting. Thus after a long time, the temperature of both parts will equalise. The final position of the separating wall be at distance \(x\) from the left end. So it is at a distance \(30-x\) from the right end Putting combined gas equation of one side of the separating wall,

\(
\begin{aligned}
& \frac{P_1 \times V_1}{T_1}=\frac{P_2 \times V_2}{T_2} \\
& \Rightarrow \frac{P \times 20 A}{400}=\frac{P^{\prime} \times A}{T} \dots(1)\\
& \Rightarrow \frac{P \times 10 A}{100}=\frac{-P^{\prime}(30-x)}{T} \dots(2)
\end{aligned}
\)
Equating (1) and (2)
\(
\Rightarrow \frac{1}{2}=\frac{x}{30-x} \quad \Rightarrow 30-x=2 x \Rightarrow 3 x=30 \Rightarrow x=10 \mathrm{~cm}
\)
The separator will be at a distance \(10 \mathrm{~cm}\) from left end.

Hint

(a) We have,
\(
\begin{aligned}
& d \frac{V}{d t}=r \\
& \Rightarrow d V=r d t
\end{aligned}
\)
Let the pressure pumped out gas \(=d p\) Volume of container \(=V_0\)
At a pump \(d V\) amount of gas has been
pumped out
\(
\begin{aligned}
& P d V=-\left(V_0\right) d P \\
& \Rightarrow \operatorname{Prdt}=-\left(V_0\right) d P \\
& \Rightarrow d \frac{p}{P}=\left(-r \frac{d t}{V_0}\right)
\end{aligned}
\)
On integration, we get
\(
P=\left(e^{-r \frac{t}{V_0}}\right)
\)
(b) Half of the gas has been pumped out, and the pressure will be half.
That is, \(1=(1 / 2) \left(e^{-r^t \frac{t}{V_0}}\right)\)
\(
\begin{aligned}
& \ln 2=r \frac{t}{V_0} \\
& t=\operatorname{In} 2 \times\left(\frac{V_0}{r}\right)
\end{aligned}
\)

Hint

Given:
\(
P=\frac{p_0}{1+\left(\frac{V}{V_0}\right)^2}
\)
Multiplying both sides by \(V\), we get
\(
\begin{aligned}
& p V=\frac{p_0 V}{1+\left(\frac{V}{V_0}\right)^2} \\
& p V=R T
\end{aligned}
\)
Now,
\(
\begin{aligned}
& R T=\frac{p_0 V}{1+\left(\frac{V}{V_0}\right)^2} \\
& T=\frac{1}{R}\left(\frac{p_0 V_0}{1+\left(\frac{V_0}{V_0}\right)^2}\right) \quad\left[V=V_0\right] \\
& \Rightarrow T=\frac{p_0 V_0}{2 R}
\end{aligned}
\)

Hint

Given
\(
\begin{aligned}
& P_1=10^5 \mathrm{~Pa} \\
& A=\pi(0.05)^2 \\
& \mathrm{~L}=0.2 \mathrm{~m} \\
& V=A L=0.0016 \mathrm{~m}^3 \\
& T_1=300 \mathrm{~K} \\
& T_2=600 \mathrm{~K} \\
& \mu=0.20
\end{aligned}
\)
Applying 5 variable equation of state, we get
\(
\begin{aligned}
& \frac{P_1 V}{T_1}=\frac{P_2 V}{T_2} \\
& \Rightarrow \frac{P_1}{T_1}=\frac{P_2}{T_2} \\
& \Rightarrow P_2=\frac{T_2}{T_1} \times P_1=\frac{600}{300} \times 10^5 \\
& \Rightarrow P_2=2 \times 10^5
\end{aligned}
\)
Net pressure, \(P=P_2-P_1=2 \times 10^5-10^5=10^5\)
Total force acting on the stopper \(=P A=10^5 \times \pi \times(0.05)^2\)
Applying law of friction, we get
\(
\begin{aligned}
& F=\mu N=0.2 N \\
& \Rightarrow N=\frac{F}{\mu}=\frac{10^5 \times \pi \times(0.05)^2}{0.2} \\
& \frac{d N}{d l}=\frac{N}{2 \pi r}=\frac{10^5 \times \pi \times(0.05)^2}{0.2 \times 2 \pi \times(0.05)}=0.125 \times 10^5 \\
& \Rightarrow \frac{d N}{d l}=1.25 \times 10^4 \frac{N}{m}
\end{aligned}
\)

Hint

(a)
Since pressure from outside and inside the cylinder is the same, there is no net pressure acting on the pistons. So, tension will be zero.
(b)
\(
\begin{aligned}
& T_1=T_0 \\
& T_2=2 T_0 \\
& P_1=P_0=10^5 \mathrm{~Pa} \\
& \mathrm{CSA}=\mathrm{A}
\end{aligned}
\)
Let the Pistons be \(\mathrm{L}\) distance apart .
\(
V=A L
\)
Applying five variable gas equation, we get
\(
\begin{aligned}
& \frac{P_1 V}{T_1}=\frac{P_2 V}{T_2} \\
& \Rightarrow \frac{10^5}{T_0}=\frac{P_2}{2 T_0} \\
& \Rightarrow P_2=2 \times 10^5=2 P_0
\end{aligned}
\)
Net Force acting outside \(=2 P_0-P_0=P_0\)
Force acting on a piston \(F=P_0 A\)
By the free body diagram, we get
\(
\begin{aligned}
& F-T=0 \\
& T=P_0 A
\end{aligned}
\)

Hint

(a) Pressure of water above the water level of the bigger tank is given by
\(
P=\left(h_2+h_0\right) \rho g
\)
Let the atmospheric pressure above the tube be \(P_0\).
Total pressure above the tube \(=P_0+P\)
\(
=\left(h_2+h_0\right) \rho g+P_0
\)
This pressure initially is balanced by pressure above the tank \(2 P_0\).
\(
\begin{aligned}
& \Rightarrow 2 P_0=\left(h_2+h_0\right) \rho g+P_0 \\
& \Rightarrow h_2=\frac{P_0}{\rho g}-h_0
\end{aligned}
\)
(b) Velocity of the efflux out of the outlet depends upon the total pressure above the outlet.Total pressure above the outlet \(=2 P_0+\left(h_1-h_0\right) \rho g\)
Applying Bernouli’s law, we get
Let the velocity of efflux be \(v_1\) and the velocity with which the level of the tank falls be \(v_2\). pressure above the outlet is \(P_0\). Then,
\(
\frac{2 P_0+\left(h_1-h_0\right) \rho g}{\rho}+\mathrm{gz}+\frac{v_2^2}{2}=\frac{P_0}{\rho}+\mathrm{gz}+\frac{v_2^2}{2}
\)
Now, let the reference point of the liquid be the level of the outlet. Thus, \(z=0\)
\(
\Rightarrow \frac{P_0+\left(h_1-h_0\right) \rho g}{\rho}+\frac{v_2^2}{2}=\frac{v_1^2}{2}
\)
Again, the speed with which the water level of the tank goes down is very less compared to the velocity of the efflux. Thus,
\(
\begin{aligned}
& \mathrm{v}_2=0 \\
& \Rightarrow \frac{P_0+\left(h_1-h_0\right) \rho g}{\rho}=\frac{v_1^2}{2} \\
& \Rightarrow v_1=\left[\frac{2}{\rho}\left(P_0+\left(h_1-h_0\right) \rho g\right)\right]^{\frac{1}{2}}
\end{aligned}
\)
(c) Water maintains its own level, so the height of the water of the tank will be \(h_1\) when water will stop flowing
Thus the height of water in the tube below the tank height will be \(=h_1\)
Hence the height of the water above the tank height will be \(=-h_1\)

Hint

Atmospheric pressure inside the cylinderical vessel, \(P_0=10^5 \mathrm{~Pa}\)
\(
A=10 \mathrm{~cm}^2=10 \times 10^{-4} \mathrm{~m}^2
\)
Pressure due to the weight of the piston \(=\frac{m g}{A}=\frac{1 \times 9.8}{10 \times 10^{-4}}\)
\(
\begin{aligned}
& P_1=10^5+9.8 \times 10^3 \\
& V_1=0.2 \times 10 \times 10^{-4}=2 \times 10^{-4}
\end{aligned}
\)
After evacution, external pressure above the piston \(=0\)
\(
P_2=0+9.8 \times 10^3
\)
Now,
\(
P_1 V_1=P_2 V_2
\)
Let \(\mathrm{L}\) be the final length of the gas column. Then,
\(
\begin{aligned}
& V_2=10 \times 10^{-4} L \\
& \Rightarrow\left(10^5+9.8 \times 10^3\right) \times 0.2 \times 10 \times 10^{-4}=9.8 \times 10^3 \times 10 \times 10^{-4} \mathrm{~L} \\
& L=2.2 \mathrm{~m}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& \Rightarrow\left(\frac{\mathrm{mg}}{\mathrm{A}}+\mathrm{P}_0\right) \mathrm{A}l=\mathrm{P}_0 \mathrm{A}l^{\prime} \\
& \Rightarrow\left(\frac{1 \times 98}{10 \times 10^{-4}}+10^5\right) 0.2=10^5 l^{\prime} \\
& \Rightarrow\left(9.8 \times 10^3+10^5\right) \times 0.2=10^5 l^{\prime} \\
& =109.8 \times 10^3 \times 0.2=10^5 l^{\prime} \\
& \Rightarrow l^{\prime}=\frac{109.8 \times 0.2}{10^2} \\
& =0.2196 \mathrm{~m}=0.22 \mathrm{~m} \approx 22 \mathrm{~cm}
\end{aligned}
\)

Hint

Given,
\(
\begin{aligned}
& P_1=0.76 \mathrm{~m} \mathrm{Hg} \\
& P_2=P \\
& T_1=273 K \\
& T_2=335 K
\end{aligned}
\)
Let each of the bulbs have \(n_1\) moles initially .
Let the number of moles left in second bulb after its pressure reached \(\mathrm{P}\) be \(\mathrm{n}_2\).
Applying equation of state, we get
\(
\begin{aligned}
& \frac{P_1 V}{n_1 T_1}=\frac{P V}{n_2 T_2} \\
& \Rightarrow \frac{0.76}{273 n_1}=\frac{P}{335 n_2} \\
& \Rightarrow n_2=\frac{273 P}{335 \times 0.76} n_1
\end{aligned}
\)
Number of moles left in the second bulb after the temperature rose \(=n_1-n_2\)
\(
=n_1-\frac{273 P}{335 \times 0.76} n_1
\)
Let \(n_3\) moles be left when pressure reached \(P\). Applying equation of state in the first bulb, we get
\(
\begin{aligned}
& \frac{P_1 V}{n_1 T_1}=\frac{P V}{n_3 T_1} \\
& \Rightarrow \frac{0.76}{n_1}=\frac{P}{n_3} \\
& \Rightarrow n_3=\frac{P n_1}{0.76}
\end{aligned}
\)
\(\mathrm{n}_3=\) its own \(n_1\) moles + the it received from the first
\(
\begin{aligned}
& \mathrm{n}_3=n_1+\left(n_1-n_2\right) \\
& \Rightarrow \frac{P n_1}{0.76}=\mathrm{n}_1+n_1-\frac{273 P}{335 \times 0.76} n_1 \\
& \Rightarrow \frac{P}{0.76}=2-\frac{273 P}{335 \times 0.76} \\
& \Rightarrow \mathrm{P}=0.8375 \\
& \Rightarrow \mathrm{P}=84 \mathrm{~cm} \text { of } \mathrm{Hg}
\end{aligned}
\)

Hint

Relative humidity \(=100 \%\)
\(
\mathrm{RH}=\frac{\text { Vapour pressure of air }}{\mathrm{SVP} \text { at the same temperature }}=1
\)
\(\Rightarrow\) Vapour pressure of air \(=\) SVP at the same temperature
So, the air is saturated at \(20^{\circ} \mathrm{C}\). So, the dew point is \(20^{\circ} \mathrm{C}\).

Hint

\(
\begin{aligned}
& \mathrm{T}=298 \mathrm{~K} \\
& \mathrm{RH}=60 \% \\
& \mathrm{P}=1.04 \times 10^5 \mathrm{~Pa} \\
& \mathrm{RH}=\frac{\text { Vapour pressure of water vapour }}{\text { saturated Vapour pressure }} \\
& =0.6 \\
& \text { Saturated vapour pressure }=3.2 \times 10^3 \mathrm{~Pa} \\
& \Rightarrow \text { vapour pressure of water vapour }(\mathrm{VP})=0.6 \times 3.2 \times 10^3=1.92 \times 10^3 \mathrm{~Pa} \\
& \text { If the water vapour is completely removed from the air , then net pressure }=1.04 \times 10^5-1.92 \times 10^3 \\
& =1.02 \times 10^5 \mathrm{~Pa} \\
& =102 \mathrm{kPa}
\end{aligned}
\)

Hint

Temperature \(=20^{\circ} \mathrm{C}\)
Dew Point \(=10^{\circ} \mathrm{C}\)
Air becomes saturated at \(10^{\circ} \mathrm{C}\). But if the room temperature is lowered to \(15^{\circ} \mathrm{C}\), the Dew point will still be at \(10^{\circ} \mathrm{C}\).

Hint

\(
\begin{aligned}
& \mathrm{RH}=40 \% \\
& V_1=10 \times 10^{-6} \mathrm{~m}^3 \\
& \mathrm{RH}=\frac{\mathrm{VP}}{\mathrm{SVP}}=0.4 \\
& \text { Let SVP }=P_0 \\
& \text { condensation occurs when } \mathrm{VP}=P_0 \\
& \Rightarrow P_1=0.4 P_0 \\
& \Rightarrow P_2=P_0
\end{aligned}
\)
Since the process is isothermal, applying Boyle’s law we get
\(
\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& \Rightarrow V_2=\frac{P_1 V_1}{P_2} \\
& \Rightarrow V_2=\frac{0.4 P_0 \times 10 \times 10^{-6}}{P_0} \\
& \Rightarrow V_2=4.0 \times 10^{-6} \\
& \Rightarrow V_2=4.0 \mathrm{~cm}^3
\end{aligned}
\)
Thus water vapour condenses at volume \(4.0 \mathrm{~cm}^3\)

Hint

Atmospheric pressure, \(\mathrm{P}=0.76 \mathrm{~m} \mathrm{Hg}\)
pressure due to water vapour inside, \(\mathrm{P}^{\prime}=0.754 \mathrm{mHg}\)
Vapour pressure \(=P-P^{\prime}=0.76-0.754=0.006 \mathrm{mHg}\)
\(\mathrm{SVH}=0.01 \mathrm{mHg}\)
\(\mathrm{RH}=\frac{\text { Vapour pressure }}{\mathrm{SVH}} \times 100 \%\)
\(=\frac{0.006}{0.01} \times 100 \%=60 \%\)

Hint

Dew point \(=10^{\circ} \mathrm{C} \quad\left[\because\right.\) Dew appears at \(\left.10^{\circ} \mathrm{C}\right]\)
At boiling point, SVP equals atmospheric pressure.
At \(20^{\circ} \mathrm{C}, \mathrm{SVP}=17.5 \mathrm{mmHg}\)
At dew point, \(\mathrm{SVP}=8.9 \mathrm{mmHg}\)
\(
\begin{aligned}
& \text { RH }=\frac{\text { SVP at dew point }}{\text { SVP at air temperature }} \times 100 \% \\
& =\frac{8.9}{17.5} \times 100 \% \\
& =51 \%
\end{aligned}
\)

Hint

We Know that \(1 \mathrm{~m}^3\) of air contains \(30 \mathrm{~g}\) of water vapour at \(30^{\circ} \mathrm{C}\).
So, amount of water vapour in \(50 \mathrm{~m}^3\) of air at \(30^{\circ} \mathrm{C}=(30 \times 50) \mathrm{g}=1500 \mathrm{~g}\)
Also, \(1 \mathrm{~m}^3\) of air contains \(16 \mathrm{~g}\) of water vapour at \(20^{\circ} \mathrm{C}\).
Amount of water vapour in \(50 \mathrm{~m}^3\) of air at \(20^{\circ} \mathrm{C} .=(16 \times 50) \mathrm{g}=800 \mathrm{~g}\)
Amount of water vapour condensed \(=(1500-800) \mathrm{g}=700 \mathrm{~g}\)

Hint

Atmospheric pressure \(=76 \mathrm{~cm} \mathrm{Hg}\)
\(\mathrm{SVP}=0.80 \mathrm{~cm} \mathrm{Hg}\)
When water is introduced into the barometer, water evaporates.
Thus, it exerts its vapour pressure over the mercury meniscus.
As more and more water evaporates, the vapour pressure increases that forces down the mercury level further.
Finally, when the volume is saturated with the vapour at the atmospheric temperature, the highest vapour pressure, i.e. SVP is observed and the fall of mercury level reaches its minimum. Thus,
Net pressure acting on the column \(=76-0.80 \mathrm{cmHg}\)
Net length of \(\mathrm{Hg}\) column at SVG \(=75.2 \mathrm{~cm}\)

Hint

Atmospheric pressure, \(\mathrm{P}_0=99.4 \times 10^3 \mathrm{~Pa}\)
SVP at \(27^{\circ} C,  \mathrm{P}_\omega=3.4 \times 10^3 \mathrm{~Pa}\)
\(\mathrm{T}=300 \mathrm{~K}\)
\(V=50 \times 10^{-6} \mathrm{~m}^3\)
Now, Pressure inside the jar = Pressure outside the jar
\([\because\) Level of water is same inside and outside of the jar]
Pressure outside the jar \(=\) Atmospheric pressure
Pressure inside the jar \(=\) VP of oxygen + SVP of water at \(27^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& \Rightarrow P_0=P+P_\omega \\
& \Rightarrow P=P_0-P_\omega=99.4 \times 10^3-3.4 \times 10^3=96 \times 10^3
\end{aligned}
\)
Applying equation of state, we get
\(
\begin{aligned}
& P V=n R T \\
& \Rightarrow 96 \times 10^3 \times 50 \times 10^{-6}=n \times 8.3 \times 300 \\
& \Rightarrow n=1.9277 \times 10^{-3} \mathrm{~mol} \approx 1.93 \times 10^{-3} \mathrm{~mol}
\end{aligned}
\)

Hint

Let the barometer has a length \(=x\)
Height of air above the air mercury column
\(
=(X-74+1)=(X-73)
\)
Pressure of air \(=(76-74-1=1 \mathrm{~cm})\)
for 2 nd case, height of air above
\(
\Rightarrow(X-72.1+1)=(X-71.1)
\)
pressure of air \(=(74-72.1-1)=0.90\)
\(
\begin{aligned}
& (X-73)(1)=\frac{9}{10}(X-71.1) \\
& \Rightarrow 10(X-73)=9(X-71.1) \\
& \Rightarrow X=10 \times 73-9(71.1) \\
& =730-639.9 \\
& X=90.1
\end{aligned}
\)
Height of air \(=90.1\)
Height of barometer tube above the
mercury column
\(
=90.1+1=91.1 \mathrm{~mm} \text {. }
\)

Hint

Relative humidity \(=40 \%\)
SVP \(=4.6 \mathrm{~mm}\) of \(\mathrm{Hg}\)
\(
\begin{array}{ll}
0.4=\frac{V P}{4.6} & \Rightarrow V P=0.4 \times 4.6=1.84 \\
\frac{P_1 V}{T_1}=\frac{P_2 V}{T_2} & \Rightarrow \frac{1.84}{273}=\frac{P_2}{293} \Rightarrow P_2=\frac{1.84}{273} \times 293
\end{array}
\)
Relative humidity at \(20^{\circ} \mathrm{C}\)
\(
=\frac{V P}{S V P}=\frac{1.84 \times 293}{273 \times 10}=0.109=10.9 \%
\)

Hint

\(
\begin{aligned}
& \mathrm{V}=1 \mathrm{~m}^3 \\
& \mathrm{M}=18 \text { g for water } \\
& R H=50 \% \\
& \Rightarrow \frac{V P}{S V P}=0.5 \\
& \Rightarrow V P=0.5 \times 3600=1800
\end{aligned}
\)
Let \(\mathrm{m}_1\) be the mass of water present in the \(50 \%\) humid air.
\(
\begin{aligned}
& \mathrm{PV}=\mathrm{nRT} \\
& \Rightarrow P V=\frac{m_1}{M} R T \\
& \Rightarrow 1800=\frac{m_1}{18} \times 8.3 \times 300 \\
& \Rightarrow m_1=13 g
\end{aligned}
\)
Required pressure for saturation \(=3600 \mathrm{~Pa}\)
Let \(\mathrm{m}_2\) be the amount of water required for saturation.
\(
\begin{aligned}
& \Rightarrow 3600=\frac{m_2}{M} R T \\
& \Rightarrow m_2=\frac{3600 \times 18}{8.3 \times 300}=26 \mathrm{~g}
\end{aligned}
\)
Total excess water vapour that has to be added \(=\mathrm{m}_2-m_1\) \(=36-13=13 g\)

Hint

\(
\begin{aligned}
& \mathrm{T}=300 \mathrm{~K} \\
& \mathrm{SVP}=3300 \mathrm{~Pa} \text { at } 300 \mathrm{~K} \\
& \mathrm{RH}=20 \% \\
& \Rightarrow \frac{P}{S V P}=0.2 \\
& \Rightarrow P=0.2 \times S V P=0.2 \times 3300=660 \\
& \mathrm{~V}=50 \mathrm{~m}^3 \\
& \mathrm{M}=18 \mathrm{~g}
\end{aligned}
\)
Now ,
\(
\begin{aligned}
& \mathrm{PV}=\mathrm{nRT} \\
& \Rightarrow P V=\frac{m}{M} R T \\
& \Rightarrow 660 \times 50=\frac{m}{18} \times 8.3 \times 300 \\
& \Rightarrow \mathrm{m}=238.55 \mathrm{~g} \approx 238 \mathrm{~g}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \mathrm{M}=18 \mathrm{~g} \text { for water } \\
& \mathrm{m}=500 \mathrm{~g} \\
& \mathrm{~V}=50 \mathrm{~m}^3 \\
& \mathrm{~T}=300 \mathrm{~K} \\
& \mathrm{SVP}=3300 \mathrm{~Pa} \\
& \mathrm{RH}=20 \% \\
& \frac{V P}{S V P}=0.2 \\
& \Rightarrow \mathrm{VP}=P_1=0.2 \times 3300=660 \mathrm{~Pa}
\end{aligned}
\)
Partial pressure \(P_2\) For evaporated water is given by
\(
\begin{aligned}
& P_2 V=\frac{m}{M} R T \\
& \Rightarrow P_2=\frac{500}{18 \times 50} \times 8.31 \times 300 \\
& \Rightarrow P_2=1385 \mathrm{~Pa} \\
& \text { Total pressure }, P=P_1+P_2=1385+660=2045 \mathrm{~Pa} \\
& \mathrm{RH}=\frac{P}{S V P} \times 100=\frac{2045}{3300} \times 100 \%=61.9 \approx 62 \%
\end{aligned}
\)

Hint

(a) Relative humidity is given by
\(
\begin{gathered}
\frac{V P}{\text { SVP at } 15^{\circ} \mathrm{C}} \\
\Rightarrow 0.4=\frac{V P}{1.6 \times 10^3} \\
\Rightarrow V P=0.4 \times 1.6 \times 10^3
\end{gathered}
\)
Evaporation occurs as long as the atmosphere is not saturated.
Net pressure change \(=1.6 \times 10^3-0.4 \times 1.6 \times 10^3\)
\(
\begin{aligned}
& =(1.6-0.4 \times 1.6) 10^3 \\
& =0.96 \times 10^3
\end{aligned}
\)
Let the mass of water evaporated be \(m\). Then,
\(
\begin{aligned}
& \Rightarrow 0.96 \times 10^3 \times 50=\frac{m \times 8.3 \times 288}{18} \\
& \Rightarrow m=\frac{0.96 \times 50 \times 18 \times 10^3}{8.3 \times 288} \\
& =361.45 \approx 361 \mathrm{~g}
\end{aligned}
\)
(b) At \(20^{\circ} \mathrm{C}, \mathrm{SVP}=2.4 \mathrm{KPa}\)
At \(15^{\circ} \mathrm{C}, \mathrm{SVP}=1.6 \mathrm{KPa}\)
Net pressure change \(=(2.4-1.6) \times 10^3 \mathrm{~Pa}\)
\(
=0.8 \times 10^3 \mathrm{~Pa}
\)
Mass of water evaporated is given by
\(
\begin{aligned}
& m=\frac{m^{\prime} \times 8.3 \times 293}{18} \\
& \Rightarrow m^{\prime}=\frac{0.8 \times 50 \times 18 \times 10^3}{8.3 \times 293} \\
& =296.06 \approx 296 \mathrm{~g}
\end{aligned}
\)