Exemplar MCQs

Hint

(b) According to the ideal gas law,
\(\mathrm{P}=\mathrm{nRT} / \mathrm{V}\), here the temperature of the vessel remains unchanged hence, the pressure remains the same from that point of view.
Now, let us discuss the phenomenon inside the vessel. The gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic. The number of collisions per unit volume in a gas remains constant. So, the pressure of the gas inside the vessel remains the same because the motion of the vessel as a whole does not affect the relative motion of the gas molecules with respect to the walls.

Hint

(d) In an ideal gas, the gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic. So, their kinetic energy and momentum remains conserved.

So, the momentum transferred to the face \(A B C D=2 \mathrm{mv}\), And the gas molecule is absorbed by the face EFGH. Hence it does not rebound. So, momentum transferred to the face \(E F G H=m v\).
And the pressure on the faces is due to the total momentum to the faces. So, pressure on EFGH would be half that on \(A B C D\).

Hint

The correct option is b) isothermal process

Explanation: In an isothermal process, when the temperature remains constant, Boyle’s law applies.
Only if the number of molecules (n) and the temperature (T) are both constant can Boyle’s Law be used. Boyle’s Law is used to forecast the outcome of changing the volume and pressure of a constant quantity of gas only, and only to the starting state.

Hint

(c) The situation is shown in the diagram where an ideal gas is contained in a cylinder, having a piston of mass \(\mathrm{M}\).
According to the problem, piston can move up and down without friction. So, the only force present is weight of the piston.
The pressure inside the gas will be
\(
P=P_0+M g / A=\text { constant }
\)
where, \(P_0=\) atmospheric pressure
\(A=\) area of cross-section of the piston.
\(M g=\) weight of piston
Weight of piston and atmospheric pressure are constant, hence pressure remains constant.
According to ideal gas law,
\(
P V=n R T
\)
As the pressure remains constant, so if the temperature is increased, only the volume increases as the piston moves up without friction.

Hint

(a) We have to consider the slope of the \(V-T\) graph. From ideal gas equation, the slope of \(V-T\) curve is \(\frac{V}{T}=\left(\frac{n R}{P}\right)\).
So, if the slope is greater, pressure will be smaller and vice-versa.
We know for an ideal gas,
\(
P V=n R T \Rightarrow V=\left(\frac{n R}{P}\right) T
\)
Slope of the \(V-T\) graph, \(m=\frac{d V}{d T}=\frac{n R}{P}\)
\(
\begin{gathered}
\Rightarrow \quad m \propto \frac{1}{P} \quad[\therefore n R=\text { constant }] \\
\Rightarrow \quad P \propto \frac{1}{m}
\end{gathered}
\)
Hence, \(\frac{P_1}{P_2}=\frac{m_2}{m_1}<1\)
So, \(P_1>P_2\)
where, \(m_1\) is slope of the graph corresponding to \(P_1\) and similarly \(m_2\) is slope corresponding to \(P_2\).
Hence the correct option is (a), i.e. \(P_1>P_2\)

Hint

(d) The situation is shown in the diagram, \(\mathrm{H}_2\) gas is contained in a box is heated, and gets converted to a gas of hydrogen atoms. Then the number of moles would become twice.
According to the gas equation,
\(
\mathrm{PV}=\mathrm{nRT}
\)
\(P=\) Pressure of gas, \(n=\) Number of moles
\(\mathrm{R}=\) Gas constant, \(T=\) Temperature \(P V=n R T\)
As volume \((V)\) of the container is constant. Hence, when temperature \((T)\) becomes 10 times, (from \(300 \mathrm{~K}\) to \(3000 \mathrm{~K}\) ) pressure (P) also becomes 10 times, as P \(\propto T\).
Pressure is due to the bombardment of particles and as gases break, the number of moles becomes twice of initial, so \(n_2=2 n_1\)
So \(\quad P \propto n T\)
\(
\begin{array}{ll}
\Rightarrow & \frac{P_2}{P_1}=\frac{n_2 T_2}{n_1 T_1}=\frac{\left(2 n_1\right)(3000)}{n_1(300)}=20 \\
\Rightarrow \quad & P_2=20 P_1
\end{array}
\)
Hence, the final pressure of the gas would be 20 times the pressure initially.

Hint

Key concept: Maxwell’s Law (or the Distribution of Molecular Speeds):
(1) The \(v_{\text {rms }}\) gives us a general idea of molecular speeds in a gas at a given temperature. This doesn’t mean that the speed of each molecule is \(v_{\text {rms. }}\). Many of the molecules have speeds less than \(v_{\text {rms }}\) and many have speeds greater than \(v_{\text {rms }}\).
(2) Maxwell-derived equation gives the distribution of molecules in different speeds as follows:
\(
d N=4 \pi N\left(\frac{m}{2 \pi k T}\right)^{3 / 2} v^2 e^{-\frac{m v^2}{2 k T}} d v
\)
where \(d N=\) Number of molecules with speeds between \(v\) and \(v+d v\)

The masses of hydrogen and oxygen molecules are different.
For a function \(f(v)\), the number of molecules \(d n=f[v)\), which are having speeds between \(v\) and \(v+d v\). The Maxwell-Boltzmann speed distribution function ( \(N_v=d n / d v\) depends on the mass of the gas molecules. For each function \(f_1(v)\) and \(f_2(v)\), \(n\) will be different, hence each function \(f_1(v)\) and \(f_2(v)\) will obey the Maxwell’s distribution law separately.

Hint

(d) According to the equation of ideal gas, \(P V=n R T\)
\(P=\) pressure
\(V=\) volume
\(n=\) number of moles of gases
\(R=\) gas constant
\(T=\) temperature
Thus we have to rewrite this equation in such a way that no. of moles is given by,
\(
n=\frac{P V}{R T}
\)
As number of moles of the gas remains fixed, hence, we can write
\(
\frac{P_1 V_1}{R T_1}=\frac{P_2 V_2}{R T_2}
\)
\(
\begin{aligned}
& \Rightarrow \quad P_2=\left(P_1 V_1\right)\left(\frac{T_2}{V_2 T_1}\right) \\
& =\frac{(P)(V)(1.1 T)}{(1.05) V(T)} \quad\left[\begin{array}{l}
P_1=P \\
V_2=1.05 V \text { and } T_2=1.1 T
\end{array}\right] \\
& =P \times\left(\frac{1.1}{1.05}\right) \\
& =P(1.0476) \approx 1.05 P \\
&
\end{aligned}
\)
Hence, final pressure \(P_2\) lies between \(P\) and \(1.1 P\).

Hint

(b, d) According to the problem, ionized hydrogen is present inside the cube, they are having charge. Now, due to the presence of positive charge on the surface, A BCD hydrogen ions would experience forces other than the forces due to collision with the walls of the container. So, these forces must be of electrostatic nature. Hence, the Isotropy of the system is lost at only one face \(A B C D\) because of the presence of an external positive charge. The usual expression for pressure on the basis of kinetic theory will be valid.

Hint

(c) According to the kinetic theory equation, \(P V=2 / 3 E\) [where \(P=\) Pressure \(V=\) volume] \(E\) is representing only translational part of energy. Internal energy contains all types of energies like translational, rotational, vibrational, etc. But the molecules of an ideal gas is treated as point masses in kinetic theory, so its kinetic energy is only due to translational motion. Point mass does not have rotational or vibrational motion. Here, we assumed that the walls only exert perpendicular forces on molecules. They do not exert any parallel force, hence there will not be any type of rotation present. The wall produces only change in translational motion.

Hint

Key concept: Kinetic Energy of Ideal Gas:
Molecules of ideal gases possess only translational motion. So they possess only translational kinetic energy.
(i) Kinetic energy of a gas molecule
\(
\left(E_{\text {molecule }}\right)=\frac{1}{2} m v_{\mathrm{rms}}^2=\frac{1}{2} m\left(\frac{3 k T}{m}\right)=\frac{3}{2} k T \quad\left[\text { As } v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}\right]
\)
(ii) Kinetic energy of 1 mole ( \(M\) gram) gas
\(
\left(E_{\mathrm{mole}}\right)=\frac{1}{2} M v_{\mathrm{rms}}^2=\frac{1}{2} M \frac{3 R T}{M}=\frac{3}{2} R T \quad\left[\text { As } v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}\right]
\)
(iii) Kinetic energy of \(1 \mathrm{gm}\) gas
\(
\left(E_{\text {gram }}\right)=\frac{3}{2} \frac{R}{M} T=\frac{3}{2} \frac{k N_A}{m N_A} T=\frac{3}{2} \frac{k}{m} T=\frac{3}{2} r T
\)
Here \(m=\) Mass of each molecule, \(M=\) Molecular weight of gas and \(N_A=\) Avogadro number \(=6.023 \times 10^{23}\)
Molecules of ideal gases possess only translational motion. So, they possess only translational energy. According to the question we have to find out the rotational energy of a diatomic molecule in terms of kinetic energy.
Translational kinetic energy of each molecule, \(\mathrm{K}_{\mathrm{T}}=\frac{3}{2} \mathrm{kT} \dots(1)\), where \(k\) is constant.
Rotational kinetic energy of each molecule, \(\mathrm{K}_{\mathrm{R}}=2\left(\frac{1}{2} \mathrm{kT}\right)\)
\(\Rightarrow \mathrm{kT} \dots(2)\), where \(k\) is constant
Now dividing the equation (1) and (2) we get,
\(
\begin{aligned}
& \frac{\mathrm{K}_{\mathrm{R}}}{\mathrm{K}_{\mathrm{T}}}=\frac{(\mathrm{kT})}{\left(\frac{3}{2} \mathrm{kT}\right)} \\
& \Rightarrow \frac{2}{3}
\end{aligned}
\)
Therefore, \(\mathrm{K}_{\mathrm{R}}=\frac{2}{3} \mathrm{~K}_{\mathrm{T}}\)
Therefore, Rotational kinetic energy at a given temperature is \(\frac{2}{3}\) of the translational kinetic energy of a gas molecule. The translational kinetic energy and rotational kinetic energy both obey Maxwell’s law of distribution independent of each other. Different molecules have different rotational kinetic energy. Ideal gases only possess translational kinetic energy but non-ideal gases possess smaller rotational kinetic energy.

Hint

(a, c) For ideal gas behaviour,
\(
P V=n R T
\)
(a) When pressure, \(P=\) constant
From (i) Volume \(V \propto\) Temperature \(T\)
Graph of \(V\) versus \(T\) will be straight line.
(b) When \(T=\) constant
From (i) \(P V=\) constant
So, graph of \(P\) versus \(V\) will be a rectangular hyperbola. Hence this graph is wrong. The correct graph is shown below:
(c) When \(V=\) constant.
From (i) \(P \propto T\)
So, the graph is a straight line passing through the origin
(d) From (i) \(P V \propto T\)
\(\Rightarrow \quad \frac{P V}{T}=\) constant
So, graph of PV versus T will be a straight line parallel to the temperature axis ( \(x\)-axis). i.e., slope of this graph will be zero.
So, (d) is not correct.

Hint

(a) Since the gas is ideal and the collisions of the molecules are elastic. When the molecules collide
with the moving parts of the wall, its kinetic energy increases. But the total kinetic energy of the system will remain conserved. When the gas is compressed adiabatically, the total work done on the gas increases, its internal energy which in turn increases the KE of gas molecules, and hence, the collisions between molecules also increases.

Hint

Here, we know that,
Molar mass \(=\) Mass of Avogadro’s number of atoms (Molecules) \(=6.023 \times 10^{23}\) atoms.
According to the problem,
Mass of the gold, \(m=39.4 \mathrm{~g}\)
Molar mass of the gold, \(M=197 \mathrm{~g} \mathrm{~mol}^{-1}\)
Now, \(197 \mathrm{~g}\) of gold contains \(6.023 \times 10^{23}\) atoms
So, \(\quad 1 \mathrm{~g}\) of gold contains \(\frac{6.023 \times 10^{23}}{197}\) atoms
\(\therefore \quad 39.4\) g of gold contains \(\frac{6.023 \times 10^{23} \times 39.4}{107}\) atoms \(=1.20 \times 10^{23}\) atoms

Hint

Key concept: Here the temperatures are given in Celsius. To apply ideal gas equation, we must convert the given temperature in kelvin. So, to convert them in kelvin we use the relation
\(
\frac{C-0}{100}=\frac{K-273.15}{373.15-273.15}
\)
Aecording to the problem, we have, \(T_1=27^{\circ} \mathrm{C}\)
Therefore, \(\frac{27}{100}=\frac{T_1(\mathrm{~K})-273.15}{373.15-273.15}\)
\(
T_1 \approx 300 \mathrm{~K}
\)
Similarly, \(T_2=327^{\circ} \mathrm{C} \approx 600 \mathrm{~K}\)
\(
V_1=100 \mathrm{~cm}^3
\)
If the pressure of a given mass of the gas is kept constant, then according to the isobaric process,
\(
V \propto T
\)
\(
\begin{array}{ll}
\Rightarrow \quad \frac{V}{T}=\text { constant } \quad\left[\begin{array}{l}
V=\text { Volume of gas } \\
T=\text { Temperature of gas }
\end{array}\right] \\
\Rightarrow \quad \frac{V_1}{T_1}=\frac{V_2}{T_2} \Rightarrow V_2=V_1\left(\frac{T_2}{T_1}\right) \\
 T_1=273+27=300 \mathrm{~K} \\
T_2=273+327=600 \mathrm{~K} \\
\text { But } V_1=100 \mathrm{cc} \\
 V_2=V_1\left(\frac{600}{300}\right) \\
\therefore \quad V_2=2 V_1 \\
V_2=2 \times 100=200 \mathrm{cc}
\end{array}
\)

Hint

We know that for a given mass of gas
\(
\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}
\)
where \(R\) is a gas constant, \(T\) is temperature in kelvin and \(M\) is molar mass of the gas. Clearly for a given gas, \(v_{r m s} \propto \sqrt{T}\) as \(R, M\) are constants.
\(
\begin{aligned}
& \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}=\sqrt{\frac{T_1}{T_2}} \\
& \text { Given }\left(\mathrm{v}_{\mathrm{rms}}\right)_1=100 \mathrm{~m} / \mathrm{s} \\
& \mathrm{T}_1=27+273=300 \mathrm{~K} \\
& \mathrm{~T}_2=127+273=400 \mathrm{~K} \\
& \frac{100}{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}=\sqrt{\frac{300}{400}=\frac{\sqrt{3}}{2}} \\
& \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\frac{2 \times 100}{\sqrt{3}}=\frac{200}{\sqrt{3}} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

rms speed for w-molecules is defined as:
\(
v_{\mathrm{rms}}=\sqrt{\frac{v_1^2+v_2^2+v_3^2+\cdots+v_n^2}{n}} \quad\left[v_{\mathrm{rms}}=\text { root mean square velocity }\right]
\)
where \(v_1, v_2, v_1, \ldots v_n\) are individual velocities of \(n\)-molecules of the gas. For two molecules,
\(
v_{\mathrm{rms}}=\sqrt{\frac{v_1^2+v_2^2}{2}}
\)
According to the problem, \(v_1=9 \times 10^6 \mathrm{~m} / \mathrm{s}\) and \(v_2=1 \times 10^6 \mathrm{~m} / \mathrm{s}\)
\(
\begin{aligned}
\therefore \quad v_{\mathrm{ms}} & =\sqrt{\frac{\left(9 \times 10^6\right)^2+\left(1 \times 10^6\right)^2}{2}} \\
& =\sqrt{\frac{81 \times 10^{12}+1 \times 10^{12}}{2}} \\
& =10^6 \sqrt{\frac{81+1}{2}}=\sqrt{41} \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)

Hint

We know that oxygen has 5 degrees of freedom. Therefore, energy per mole \(=(5 / 2) \mathrm{RT}\)
Therefore, for 2 moles of oxygen, energy \(=5 R T\)
Neon has 3 degrees of freedom. Therefore, energy per mole for neon \(=(3 / 2) \mathrm{RT}\)
For 4 mole of neon, we can write the expression for energy
\(
\begin{aligned}
& =4 \times(3 / 2) R T \\
& =6 R T
\end{aligned}
\)
Therefore, the total energy \(=6 R T + 5 R T=11 \mathrm{RT}\)

Hint

Mean free path

\(\lambda=\frac{1}{\sqrt{2} \pi n d^2}\)
where \(d=\) Diameter of the molecule, \(n=\) Number of molecules per unit volume
According to the problem, \(d_1=1 Å ; d_2=2 Å\) Since, mean free path,
\(
\begin{aligned}
& \lambda \propto \frac{1}{d^2} \\
\Rightarrow \quad & \text { So, } \frac{\lambda_1}{\lambda_2}=\left(\frac{d_2}{d_1}\right)^2=\left(\frac{2}{1}\right)^2=\frac{4}{1}
\end{aligned}
\)
Hence, \(\lambda_1: \lambda_1=4: 1\)

Hint

As shown in the diagram, container is divided in two chambers.
Volume of first chamber \(V_1=2.0 \mathrm{~L}\),
Volume of second chamber \(V_2=3.0 \mathrm{~L}\)
\(
\begin{aligned}
& \mu_1=4.0 \mathrm{~mol}, \mu_2=5.0 \mathrm{~mol} \\
& P_1=1.00 \mathrm{~atm}, P_2=2.00 \mathrm{~atm}
\end{aligned}
\)
According to the equation of ideal gas,
For chamber \(1 P_1 V_1=\mu_1 R T_1\)
For chamber \(2 P_2 V_2=\mu_2 R T_2\)
When the partition is removed, the gases get mixed without any loss of energy. The mixture now attains a common equilibrium pressure and the total volume of the system is the sum of the volume of individual chambers \(\mathrm{V} 1\) and \(\mathrm{V} 2\). Let \(\mathrm{P}\) be the pressure after the partition is removed.
So, \(\quad \mu=\mu_1+\mu_2, V=V_1+V_2\)
From kinetic theory of gases,
For 1.mole \(P V=\frac{2}{3} E \quad[E=\) translational kinetic energy \(]\)
For \(\mu_1\) moles, \(P_1 V_1=\mu_1\left(\frac{2}{3} E_1\right)\)
For \(\mu_2\) moles, \(P_2 V_2=\mu_2\left(\frac{2}{3} E_2\right)\)
Total energy of the gas in both the chambers \(=\left(\mu_1 E_1+\mu_2 E_2\right)=\frac{3}{2}\left(P_1 V_1+P_2 V_2\right)\)
From the above relation, \(P V=\frac{2}{3} E_{\text {total }}=\frac{2}{3} \mu E_{\text {per mole }}\)
\(
\begin{aligned}
& P\left(V_1+V_2\right)=\frac{2}{3} \times \frac{3}{2}\left(P_1 V_1+P_2 V_2\right) \\
& P=\frac{P_1 V_1+P_2 V_2}{V_1+V_2} \\
&=\left(\frac{1.00 \times 2.0+2.00 \times 3.0}{2.0+3.0}\right) \mathrm{atm} \\
&=\frac{8.0}{5.0}=1.60 \mathrm{~atm}
\end{aligned}
\)

Hint

(a) Average kinetic energy of a molecule
\(
\frac{E}{N}=\frac{3}{2} K_B T
\)
The average kinetic energy of a molecule is proportional to the absolute temperature of the gas, and it is independent of mass of the molecules. The average K.E. will be same for \(A, B\) and \(C\).
Final Answer :
The average K.E will be the same as conditions of temperature and pressure are the same.
(b) Root mean square speed:
\(
\begin{aligned}
& v_{r m s}=\sqrt{\frac{3 K_B T}{m}} \\
& v_{r m s} \propto \frac{1}{\sqrt{m}}
\end{aligned}
\)
Given \(m_A>m_B>m_C\)
Then \(v_C>v_B>v_A\)
Final Answer: \(v_C>v_B>v_A\).

Hint

According to the problem,
\(
\begin{aligned}
& V_i=\left(3 \times 10^{-2}\right)^3=27 \times 10^{-6} \mathrm{~m}^3 \\
& P_i=1 \mathrm{~atm}, P_f=100 \mathrm{~atm}, T=\mathrm{constant}
\end{aligned}
\)
Mass of hydrogen, \(m=0.5 \mathrm{~g}\)
Thus at a constant pressure,
\(
P V=n R T
\)
where, \(n, R\) and \(T\) are constants. So,
\(
\frac{V_f}{V_i}=\frac{P_i}{P_f}
\)
Thus, final volume \(V_f=\frac{P_i V_i}{P_f}=\frac{1 \times 27 \times 10^{-6}}{100}\)
\(
=2.7 \times 10^{-7} \mathrm{~m}^3
\)
Number of molecules in \(0.5 \mathrm{~g}\) of \(\mathrm{H}_2\)
\(
N=\frac{N_A}{M} m
\)
Here, \(\quad N_A=6.022 \times 10^{23} \mathrm{~mol}^{-1}\)
\(
\begin{aligned}
& M=2 \mathrm{~g} \mathrm{~mol}^{-1} \\
\therefore \quad & N=\frac{6.022 \times 10^{23}}{2} \times 0.5=1.5 \times 10^{23}
\end{aligned}
\)
Radius of a hydrogen molecule \(=1 Å=10^{-10} \mathrm{~m}\)
Volume of a hydrogen molecule \(=\left(10^{-10}\right)^3=10^{-30} \mathrm{~m}^3\)
Molecular volume \(=\left(1.5 \times 10^{23}\right) \times 10^{-30}\)
\(=1.5 \times 10^{-7} \mathrm{~m}^3\)
We get \(V_f\) is of same order as the molecular volume. As the molecules lie very close to each other, so intermolecular forces cannot be ignored, i.e. gas cannot be treated as an ideal gas. Therefore, one is not justified in assuming the ideal gas law.

Hint

Here, according to the question, when air is pumped, more molecules are pumped and Boyle’s law is stated for situation where the mass of molecules remains constant.
\(
\begin{aligned}
& P V=P\left(\frac{m}{\rho}\right)=\text { constant } \\
\Rightarrow \quad & \frac{P}{\rho}=\text { constant } \text { or } \frac{P_1}{\rho_1}=\frac{P_2}{\rho_2} \\
& \left(\text { As volume }=\frac{m}{\rho(\text { Density of the gas })} \text { and } m=\text { constant }\right)
\end{aligned}
\)
In this case, when air is pumped into a cycle tyre, the mass of air in it increases as the number of air molecules keeps increasing. Hence, this is a case of variable mass, Boyle’s law (and even Charle’s law) is only applicable in situations, where the mass of gas molecules remains fixed. Hence, Boyle’s law is not applicable in this case.

Hint

Average KE per molecule \(=\frac{3}{2} k T\)
According to the problem, number of moles of helium, \(n=5\) gram mole \(T=7^{\circ} \mathrm{C}=7+273=280 \mathrm{~K}\)
(a) Hence, number of atoms (He is monoatomic)
Number of \(\mathrm{He}\) atom, \(N=n N_A\)
\(
\begin{aligned}
& =5 \times 6.023 \times 10^{23} \\
& =30.015 \times 10^{23}
\end{aligned}
\)
(b) \(\mathrm{As}, \mathrm{KE}=\frac{3}{2} k_B T\)
Here, \(k_B=\) Boltzmann constant.
\(
\begin{aligned}
\therefore \text { Total internal energy } & =\left(\frac{3}{2} k T\right) \times N \\
& =\frac{3}{2} \times\left(1.38 \times 10^{-23}\right) \times 280 \times 30.115 \times 10^{23} \\
& =1.74 \times 10^4 \mathrm{~J}
\end{aligned}
\)

Important point: The above degrees of freedom are shown at room temperature. Further at high temperatures, in the case of diatomic or polyatomic molecules, the atoms within the molecule may also vibrate with respect to each other. In such cases, the molecule will have an additional degree of freedom, due to vibrational motion.

An object which vibrates in one dimension has two additional degrees of freedom. One for the potential energy and one for the kinetic energy of vibration. Helium is a monoatomic gas and It has only 3 degrees of freedom. But after addition, its degree of freedom will be 5.

Hint

Key concept: Total number of degrees of freedom in a thermodynamical system = Number of degrees of
freedom associated per molecule \(\mathrm{x}\) number of molecules.
At NTP, Volume occupied by \(6.023 \times 10^{23}\) molecules of gas \(=22400 \mathrm{cc}\) \(\therefore\) Number of molecules in \(1 \mathrm{cc}\) of hydrogen \(=\frac{6.023 \times 10^{23}}{22400}=2.688 \times 10^{19}\)
\(\mathrm{H}_2\) is a diatomic gas, having a total of 5 degrees of freedom.
\((3\) translational +2 rotational)
\(\therefore\) Total degrees of freedom possessed by all the molecules
\(
=5 \times 2.688 \times 10^{19}=1.344 \times 10^{20}
\)

Hint

Since, the container is suddenly stopped which is initially moving with velocity \(v_0\), there is no time for the exchange of heat in the process. Then total KE of the container is transferred to gas molecules in the form of translational \(\mathrm{KE}\), thereby increasing the absolute temperature.
Let \(\mathrm{n}\) be the no. of moles of the monoatomic gas in the container. Since molar mass of the gas is \(\mathrm{m}\).
Total mass of the container, \(M=m n\)
\(\mathrm{KE}\) of molecules due to velocity \(v_0\),
\(
\mathrm{KE}=1 / 2(\mathrm{mn}) v_0^2
\)
Final KE of gas \(=0\)
Change in kinetic energy, \(\Delta K=\frac{1}{2}(\mathrm{~nm}) v^2 \dots(i)\)
If \(\Delta T=\) change in absolute temperature, Then the internal energy of the gas
\(
\Delta U=n C_v \Delta T=n\left(\frac{3}{2} R\right) \Delta T \dots(ii)
\)
According to the conservation of mechanical energy, we get
\(
\Delta K=\Delta U
\)
By equating Eqs. (i) and (ii), we get
\(
\begin{array}{ll}
\Rightarrow & \frac{1}{2}(m n) v_0^2=n \frac{3}{2} R(\Delta T) \\
& (m n) v_0^2=n 3 R(\Delta T) \\
\Rightarrow & \Delta T=\frac{(m n) v_0^2}{3 n R}=\frac{m v_0^2}{3 R}
\end{array}
\)

Hint

Temperature is defined as the average kinetic energy of the molecules in a gas sample. Average is same for all the molecules of the sample. So, kinetic energy is the same for all.

Hint

At low pressure, the concentration of gas molecules is very low. Hence, the kinetic assumption that the size of the molecules can be neglected compared to the volume of the container applies.

At high temperatures, molecules move very fast. So, they tend to collide elastically and forces of interaction between the molecules minimize. This is the required idea condition.

Hint

According to the kinetic theory, molecules show straight line in motion (translational). So, the kinetic energy is essentially transitional.

Hint

The temperature of a gas is directly proportional to its kinetic energy. Thus, the energy of an ideal gas depends only on its temperature.

Hint

The rms speed of a gas is given by \(\sqrt{\frac{3 R T}{M_o}}\). Since hydrogen has the lowest \(\mathrm{M}_0\) compared to other molecules, it will have the highest rms speed.

Hint

The straight line T1 has a greater slope than T2. This means \(\frac{P}{\rho}\) ratio is greater for T1 than T2. Now, rms velocity of a gas is given by \(\sqrt{\frac{3 P}{\rho}}\). This means rms velocity of gas with \(\mathrm{T} 1\) molecules is greater than \(\mathrm{T} 2\) molecules. Again, gas with higher temperature has higher rms velocity.
So, \(T 1>T 2\).

Hint

Root mean squared velocity is given by \(v_{r m s}=\sqrt{\frac{3 R T}{M}}\)
\(
\begin{aligned}
& \Rightarrow\left(v_{r m s}\right)^2=\frac{3 R T}{M} \\
& \Rightarrow\left(v_{r m s}\right)^2 \alpha T
\end{aligned}
\)

Hint

\(
\text { Speed is constant and same for a single molecule. Thus, rms speed will be equal to its average speed. }
\)

Hint

Given,
Molecular mass of hydrogen, \(M_H=2\)
Molecular mass of oxygen, \(M_0=32\)
RMS speed is given by,
\(
\begin{aligned}
& v_{r m s=} \sqrt{\frac{3 R T}{M}} \\
& \Rightarrow \sqrt{\frac{3 R T}{M_O}}=500
\end{aligned}
\)
Now ,
\(
\begin{aligned}
& \Rightarrow \frac{v_{O r m s}}{v_{H r m s}}=\frac{\sqrt{\frac{3 R T}{M_O}}}{\sqrt{\frac{3 R T}{M_H}}} \\
& \Rightarrow \frac{v_O r m s}{v_{H r m s}}=\frac{\sqrt{\frac{3 R T}{32}}}{\sqrt{\frac{3 R T}{2}}} \\
& \Rightarrow \frac{v_{O r m s}}{v_{H r m s}}=\frac{1}{4} \\
& \Rightarrow \frac{500}{v_{H r m s}}=\frac{1}{4} \\
& \Rightarrow v_{H r m s}=4 \times 500=2000 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

Let the number of moles in the gas be \(n\).
Applying equation of state, we get
\(
\begin{aligned}
& P V=n R T \\
& \Rightarrow P=\frac{n R T}{V} \\
& \Rightarrow 2 \times 10^5=\frac{n R T}{V} \ldots(1)
\end{aligned}
\)
When half of the gas is removed, the number of moles left behind \(=\frac{n}{2}\) Let the pressure be \(\mathrm{P}^{\prime}\).
\(
P^{\prime}=\frac{n}{2} \frac{R T}{V}
\)
Now
\(
\begin{aligned}
& P^{\prime}=\frac{1}{2} \times 2 \times 10^5=10^5[\text { From eq . (1)] } \\
& =100 \mathrm{kPa}
\end{aligned}
\)

Hint

Given \(: v=\sqrt{\frac{3 R T}{32}}\)
Let the new rms speed be \(v^{\prime}\).
Molecule dissociate, \(M=16 v^{\prime}=\sqrt{\frac{3 R(2 T)}{16}}\)
\(
\begin{aligned}
& =\sqrt{\frac{3 R(4 T)}{32}} \\
& =2 \sqrt{\frac{3 R T}{32}}=2 v
\end{aligned}
\)

Hint

number of molecules in the gas.
Here,
\(
P V=n R T \ldots(1)
\)
Also,
\(
\begin{aligned}
& k=\frac{R}{N} \\
& \Rightarrow R=k N \ldots(2) \text { Now }, P V=n k N T[\text { From eq } \cdot(1) \text { and } \text { eq } \cdot(2)] \\
& \Rightarrow n N=\frac{P V}{k T} \\
& n N=\text { Number of molecules } \\
& \frac{P V}{k T}=\text { Number of molecules }
\end{aligned}
\)

Hint

According to the graph, \(\mathrm{P}\) is directly proportional to \(\mathrm{T}\).
Applying the equation of state, we get
\(
\begin{aligned}
& \mathrm{PV}=\mathrm{nRT} \\
& \Rightarrow P=\frac{n R}{V} T
\end{aligned}
\)
Given : \(P \alpha T\)
This means \(\frac{n R}{V}\) is a constant. So, \(\mathrm{V}\) is also a constant.
Constant \(\mathrm{V}\) implies the process is isochoric.

Hint

As the liquid is decreasing, the liquid is vapourised. We know that vapourisation cannot occur in saturated air and there cannot be any liquid with no vapour at all. So, the vapour in the remaining part is unsaturated.

Hint

Since the amount of liquid is constant, there is no vapourisation of the liquid inside the bottle. Also, since there cannot be a liquid with no vapours at all and vapourisation cannot take place in the remaining saturated part, the remaining part must be saturated with the vapours of the liquid.

Hint

As the vapour is injected, the pressure of the chamber increases. But when the pressure becomes equal to the saturated vapour pressure, it condenses. So, if more vapour is injected beyond the saturated vapour pressure, the vapour will condense and thus the vapour pressure will be constant.

Hint

The maximum pressure attainable above the water will be saturated vapour pressure at that temperature. Since saturated vapour pressure does not depend upon volume, both the vessels will have same pressure.

Hint

According to Kinetic theory, postulates collision between molecules are elastic. This means that kinetic energy after any collision is conserved because while one gains kinetic energy, another loses it. Both options, (c) and (d) consider the conservation of kinetic energy in the collision.

Hint

The average speed of molecules is given by \(\sqrt{\frac{8 k T}{\pi m}}\).
We observe that the greater the mass, the lesser is the average speed of the molecule. Since an oxygen molecule is heavier than a hydrogen molecule, the oxygen molecule will hit the wall with a smaller average speed.

Hint

The molecules move in all possible directions in an ideal gas at equilibrium. Since momentum is a vector quantity for every direction of motion of the molecules, there exists an opposite direction of motion of the other. Hence, the average momentum is zero for an ideal gas at equilibrium.

Hint

Pressure of an ideal gas is given by PV \(=\frac{1}{3} m n u^2\).
We know that pressure depends on volume, number of molecules and root mean square velocity. Also, root mean square velocity depends on the temperature of the gas. Since the number of molecules, volume and temperature are constant, pressure of the gas will not change.

Hint

\(
\text { Average momentum of a gas sample is zero, so it does not depend upon any of these parameters. }
\)

Hint

Kinetic energy per mole of an ideal gas is directly proportional to T. So, it will be the same for all ideal gases.
The number of molecules in 1 mole of an ideal is the same for all ideal gases because ideal gases obey
Avogadro’s law.
Thus, (a) and (c) are correct answers.

Hint

nature of the gas.
In an ideal gas, the equation of state is given by \(P V=n R T \Rightarrow P V=n N_A \frac{R}{N_A} T\)
\(
\begin{aligned}
& \Rightarrow P V=n N_A k T \\
& \Rightarrow \frac{1}{n N_A}=\frac{k T}{P V}
\end{aligned}
\)
Multiplying both sides by mass of the gas \(\mathrm{M}\), we get \(\frac{M}{n N_A}=\frac{M k T}{P V}\) Now, \(n N_A\) gives the total number of molecules of the gas.
Also, \(\frac{M}{n N_A}\) gives the mass of a single molecule.
Hence, \(\frac{M k T}{P V}\) is the mass of a single molecule of the gas,
Molecular mass is a property of the gas.