Past NEET Papers

Hint

We know that, the rms speed of the gas molecules,
\(
v_{\mathrm{rms}}=\sqrt{3 R T / M}
\)
Here, \(T\) is the temperature of the gas, \(R\) is the universal gas constant, \(M\) is the molar mass of the gas. Pressure exerted by an ideal gas,
\(
p=\frac{1}{3} n m \bar{v}^2
\)
Here, \(n\) is the number of moles, \(m\) is the mass of the gas, \(\bar{v}\) is the average speed of the gas molecules.
The average kinetic energy of a molecule,
\(
K E_{a v}=\frac{3}{2} k_B T
\)
Here, \(k_B\) is the Boltzmann constant, \(T\) is the temperature of the gas.
For diatomic gas, degree of freedom, \(f=5\)
As, total internal energy of 1 mole of diatomic gas,
\(
\begin{aligned}
\Delta U & =\frac{n f R T}{2} \\
\Rightarrow \Delta U & =\frac{1(5) R T}{2} \Rightarrow \Delta U=\frac{5 R T}{2}
\end{aligned}
\)
The correct match is \(A \rightarrow 2, B \rightarrow 1, C \rightarrow 4\) and \(D \rightarrow 3\).

Hint

The mean free path \(l\) for a gas molecule is given as
\(
l=\frac{1}{\sqrt{2} \pi n d^2} \Rightarrow l \propto \frac{1}{d^2}
\)
where, \(d=\) diameter of molecule of gas.

Hint

Ideal gas equation is given as
\(
\begin{aligned}
\qquad p & =\frac{\rho R T}{M_0} \Rightarrow p \cdot \frac{M_0}{\rho}=R T \\
\Rightarrow \quad p V & =R T \\
\text { where, } V & =\frac{M_0}{\rho}
\end{aligned}
\)
Hence, \(\rho\) and \(M_0\) are mass density and mass of gas, respectively.

Hint

The mean free path \(\lambda\) for a gas, with molecular diameter d and number density \(n\) is given by the relation
\(
\lambda=\frac{1}{\sqrt{2} n \pi d^2}
\)
Hence, the correct option is (a).

Hint

Given,
pressure \(p=249 \mathrm{kPa}=249 \times 10^3 \mathrm{~Pa}\)
Temperature, \(T=27^{\circ} \mathrm{C}\)
\(
=273+27 \mathrm{~K}=300 \mathrm{~K}
\)
Density, \(\rho=\) ?
As, from ideal gas equation,
\(
p V=n R T
\)
\(
\begin{aligned}
& \Rightarrow \quad p V=\frac{m}{M} R T \quad\left[\text { as } n=\frac{m}{M}\right] \\
& \Rightarrow \quad p V M=m R T \\
& \Rightarrow \quad p M=\frac{m}{V} R T=\rho R T \quad\left[\text { as } \frac{m}{V}=\rho\right] \\
&
\end{aligned}
\)
\(
\begin{aligned}
& \begin{aligned}
& \Rightarrow \quad \rho=\frac{p M}{R T} \\
&=\frac{249 \times 10^3 \times 2 \times 10^{-3}}{8.3 \times 300} \\
& {\left[\because \text { for hydrogen gas, } M=2 \mathrm{~g}=2 \times 10^{-3} \mathrm{~kg}\right] }
\end{aligned} \\
& \Rightarrow \rho=0.2 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}
\)
Hence, the correct option is (a).

Hint

The average thermal energy of a system with degree of freedom \(f\) is equals to its average energy, which is given as
\(
=\frac{f}{2} k_B \cdot T
\)
For monoatomic gas, \(f=3\)
\(\therefore\) Average thermal energy \(=\frac{3}{2} k_B \cdot T\)
Hence, the correct option is (a).

Hint

As the temperature of the gas in the container is increased, the kinetic energy also increases. This is because the average kinetic energy of a gas is given by
\(
\mathrm{KE}=\frac{f}{2} n R T \dots(i)
\)
where, \(f=\) degree of freedom,
\(n=\) number of moles of gas molecules,
\(R=\) universal gas constant, and
\(T=\) absolute temperature of the gas.
From Eq. (i),
\(
\mathrm{KE} \propto T
\)
option (b) is incorrect as the increase in temperature will lead to an increase in pressure as \(p \propto T\). Other options (c) and (d) are also incorrect as molecular distance increases while mass remains the same for the increase in temperature.

Hint

The poisson’s ratio,
\(
\gamma=\frac{C_P}{C_V} \dots(i)
\)
where, \(C_p=\) molar heat capacity constant pressure
and \(C_V=\) molar heat capacity at constant volume

Also, \(C_p=C_v+R\) (from Mayer’s relation)
\(
\begin{aligned}
& C_v=\frac{f}{2} R \quad \text { (where, } f=\text { degree of freedom) } \\
& \Rightarrow C_p=\left(\frac{f}{2}+1\right) R \\
& \text { So, Eq. (i) becomes, } \\
& \Rightarrow \gamma=1+\frac{2}{f}
\end{aligned}
\)
For hydrogen gas, which is diatomic, the degree of freedom is 5 (3 translational, 2 rotational).
\(
\therefore \quad \gamma=1+\frac{2}{5}=\frac{7}{5}
\)
For helium gas, which is monoatomic, the degree of freedom is 3 (3 translational only).
\(
\therefore \quad \gamma=1+\frac{2}{3}=\frac{5}{3}
\)
The diatomic gas \(X\) also have vibrational motion, so degree of freedom is 7 (3 translational, 2 rotational and 2 vibrational).
\(
\therefore \quad \gamma=1+\frac{2}{7}=\frac{9}{7}
\)

Hint

Key Concept The minimum velocity with which the body must be projected vertically upwards, so that it could escape from the Earth’s atmosphere, is its escape velocity \(\left(v_e\right)\).
As, \(v_e=\sqrt{2 g R}\)
Substituting the value of \(g\left(9.8 \mathrm{~ms}^{-2}\right)\) and radius of Earth \(\left(R=6.4 \times 10^6 \mathrm{~m}\right)\), we get
\(
\begin{aligned}
v_e & =\sqrt{2 \times 9.8 \times 6.4 \times 10^6} \\
& \cong 11.2 \mathrm{~km} \mathrm{~s}^{-1}=11200 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)
Let the temperature of molecule beT when it attains \(v_e\).
According to the question,
\(
v_{\mathrm{rms}}=v_e
\)
where, \(v_{\mathrm{rms}}\) is the rms speed of the oxygen molecule.
\(
\begin{array}{ll}
\Rightarrow \quad \sqrt{\frac{3 k_B T}{\mathrm{~m}_{\mathrm{O}_2}}}=11.2 \times 10^3 \\
\text { or } & T=\frac{\left(11.2 \times 10^3\right)^2\left(\mathrm{~m}_{\mathrm{O}_2}\right)}{\left(3 k_B\right)}
\end{array}
\)
Substituting the given values, i.e.,
\(
\begin{aligned}
& R_B=1.38 \times 10^{-23} \mathrm{JK}^{-1} \text { and } \\
& m_{\mathrm{O}_2}=m=2.76 \times 10^{-26} \mathrm{~kg}
\end{aligned}
\)
We get,
\(
\begin{aligned}
T & =\frac{\left(11.2 \times 10^3\right)^2\left(2.76 \times 10^{-26}\right)}{\left(3 \times 1.38 \times 10^{-23}\right)} \\
& =8.3626 \times 10^4 \mathrm{~K}
\end{aligned}
\)

Hint

(d)
The total internal energy of system \(=\) Internal energy of oxygen molecules + Internal energy of argon molecules
\(
\begin{aligned}
& =\frac{f_1}{2} n_1 R T+\frac{f_2}{2} n_2 R T=\frac{5}{2} \times 2 R T+\frac{3}{2} \times 4 R T \\
& =11 R T
\end{aligned}
\)

Hint

Here \(\mathrm{v}_1=200 \mathrm{~m} / \mathrm{s}\);
temperature \(\mathrm{T}_1=27^{\circ} \mathrm{C}=27+273=300 \mathrm{k}\)
temperature \(\mathrm{T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{k}, \mathrm{v}_2=\) ?
R.M.S. Velocity, \(\mathrm{v} \propto \sqrt{T}\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{\mathrm{v}_2}{200}=\sqrt{\frac{400}{300}} \\
& \Rightarrow \quad \mathrm{v}_2=\frac{200 \times 2}{\sqrt{3}} \mathrm{~m} / \mathrm{s} \Rightarrow \mathrm{v}_2=\frac{400}{\sqrt{3}} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

As we know that
Pressure, \(p=\frac{1}{3} \cdot \frac{n m}{V} v_{\mathrm{rms}}^2\)
\(\because n m=\) mass of the gas, \(V=\) volume of the gas
\(\therefore \frac{m n}{V}=\) density of the gas. Thus,
\(p=\frac{1}{2} p v_{\mathrm{rms}}^2=\frac{1}{3} \rho \frac{3 R T}{M_0}=\frac{\rho R T}{M_0}\)
\(
\left(\because v_{\mathrm{mms}}=\sqrt{\frac{3 R T}{M_0}}\right)
\)
\(
\begin{aligned}
& \rho=\frac{p M_0}{R T}=\frac{p m N_A}{k N_A T} \\
& {\left[\because R=N_A k \text { and } M_0=m N_A\right]} \\
& \rho=\frac{p m}{k T} \\
&
\end{aligned}
\)

Hint

As we know that for polytropic process of index \(\alpha\) specific heat capacity
\(
\begin{aligned}
& \qquad=C_v+\frac{R}{1-\alpha} \\
& \because \text { Process, } \rho V^3=\text { constant } \Rightarrow \alpha=3 \\
& \therefore \quad C=C_v+\frac{R}{1-\alpha}=\frac{f R}{2}+\frac{R}{1-3} \\
& \text { where, } C_v=\frac{f R}{2}=\frac{3 R}{2}
\end{aligned}
\)
For monatomic gas, \(f=3=\frac{3 R}{2}\)
\(
\Rightarrow \quad C=\frac{3 R}{2}-\frac{R}{2}=R
\)

Hint

Molar mass of the gas \(=4 \mathrm{~g} / \mathrm{mol}\) Speed of sound
\(
\begin{aligned}
& \mathrm{v}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{m}}} \Rightarrow 952=\sqrt{\frac{\gamma \times 3.3 \times 273}{4 \times 10^{-3}}} \\
& \Rightarrow \gamma=1.6=\frac{16}{10}=\frac{8}{5} \\
& \text { Also, } \gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{8}{5}
\end{aligned}
\)
So, \(\mathrm{C}_{\mathrm{P}}=\frac{8 \times 5}{5}=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)
\(
\left[\mathrm{C}_{\mathrm{V}}=5.0 \mathrm{JK}^{-1} \text { given }\right]
\)

Hint

From \(\mathrm{PV}=\mathrm{nRT}\)
\(
P_A-\frac{\rho_A M_A}{R T} \text { and } P_B=\frac{\rho_B M_B}{R T}
\)
From question,
\(
\frac{P_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}} \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}}=2 \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}}=\frac{3}{2}
\)
So, \(\frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}}=\frac{3}{4}\)

Hint

The specific heat of gas at constant volume in terms of degree of freedom \(n\) is
\(
C_v=\frac{n}{2} R
\)
Also  C \(_p-C_v=R\)
\(
\begin{aligned}
& C_p=\frac{n}{2} R+R=R\left(1+\frac{n}{2}\right) \\
& \gamma=\frac{C_p}{C_v}=\frac{R\left(1+\frac{n}{2}\right)}{\frac{n}{2} R}=\frac{2}{n}+1
\end{aligned}
\)

Hint

Change in internal energy from \(\mathrm{A} \rightarrow \mathrm{B}[latex]
[latex]
\begin{aligned}
& \Delta \mathrm{U}=\frac{\mathrm{f}}{2} \mathrm{nR} \Delta \mathrm{T}=\frac{\mathrm{f}}{2} \mathrm{nR}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right) \\
& =\frac{5}{2}\left\{\mathrm{P}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}\right\} \\
& =\frac{5}{2}\left\{2 \times 10^3 \times 6-5 \times 10^3 \times 4\right\} \\
& =\frac{5}{2}\{12-20\} \times 10^3 \mathrm{~J}=5 \times(-4) \times 10^3 \mathrm{~J} \\
& \Delta \mathrm{U}=-20 \mathrm{KJ}
\end{aligned}
\)
(As gas is diatomic \(\, therefore, f=5\) )

Hint

Mean free path \(\lambda_{\mathrm{m}}=\frac{1}{\sqrt{2} \pi \mathrm{d}^2 \mathrm{n}}\) where \(\mathrm{d}=\) diameter of molecule and \(\mathrm{d}=2 \mathrm{r}\)
\(
\therefore \quad \lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{r}^2}
\)

Hint

\(
\begin{aligned}
& \text { (a) } \mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R} \Rightarrow \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+\mathrm{R} \\
& \because \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{C}_{\mathrm{v}}+\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{v}}}+\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}} \\
& \Rightarrow \gamma=1+\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}} \Rightarrow \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=\gamma-1 \\
& \Rightarrow \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}
\end{aligned}
\)

Hint

According to the question,
Slope of the graph \(\propto \frac{1}{\text { Pressure } p}\)
So, \(p_2<p_1\)

Note:

As \(V=\) constant \(\Rightarrow P \propto T\)
Hence from \(\mathrm{V}-\mathrm{T}\) graph \(\mathrm{P}_1>\mathrm{P}_2\)
From the Gay – Lussac’s law, volume remain constant, the pressure of a given mass of a gas is directly proportional to its absolute temperature.
\(
\begin{aligned}
& \therefore \mathrm{P} \propto \mathrm{T} \\
& \frac{\mathrm{P}}{\mathrm{T}}=\text { constant } \Rightarrow \frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2}
\end{aligned}
\)

Hint

According to question,
\(
p \propto T^3 \dots(i)
\)
\(
\left(\begin{array}{l}
p=\text { pressure } \\
T=\text { temperature }
\end{array}\right)
\)
and we know that
\(
p V=n R T \text { and } p V \propto T \dots(ii)
\)
So, putting Eq. (ii) in (i),
\(
\begin{array}{lc}
& p \propto(p V)^3 \\
\Rightarrow & p^2 V^3=\text { constant } \\
\Rightarrow & p V^{3 / 2}=\text { constant } \dots(iii)
\end{array}
\)
\(\Rightarrow\) Comparing Eq. (iii) with
pV \(\gamma=\) constant.
We have \(\gamma=3 / 2\)

Hint

From first law of thermodynamics
\(
\begin{aligned}
& \Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}=\frac{3}{2} \cdot \frac{1}{4} \mathrm{R}\left(\mathrm{T}_2-\mathrm{T}_1\right)+0 \\
& =\frac{3}{8} \mathrm{~N}_{\mathrm{a}} \mathrm{K}_{\mathrm{B}}\left(\mathrm{T}_2-\mathrm{T}_1\right)\left[\because \mathrm{K}=\frac{\mathrm{R}}{\mathrm{N}}\right]
\end{aligned}
\)

Hint

\(P=\frac{1}{3} \frac{m N}{V} v_{\mathrm{rms}}^2\)
where \(\mathrm{m}=\) mass of each molecules
\(\mathrm{N}=\) total number of molecules
\(\mathrm{V}=\) volume of the gas
When the mass is halved and the speed is doubled then
\(
\text { Resultant pressure, } P^{\prime}=\frac{1}{3} \times\left(\frac{m}{2}\right) \times \frac{\mathrm{N}}{\mathrm{V}}\left(2 v_{\mathrm{ms}}\right)^2=2P
\)

Hint

Let the mass of the gas be \(\mathrm{m}\). At a fixed temperature and pressure, volume is fixed.
Density of the gas, \(\rho=\frac{\mathrm{m}}{\mathrm{V}}\)
\(
\begin{aligned}
& \text { Now } \frac{\rho}{\mathrm{P}}=\frac{\mathrm{m}}{\mathrm{PV}}=\frac{\mathrm{m}}{\mathrm{nRT}} \\
& \Rightarrow \frac{\mathrm{m}}{\mathrm{nRT}}=x \text { (By question) } \\
& \Rightarrow \mathrm{xT}=\text { constant } \Rightarrow \mathrm{x}_1 \mathrm{~T}_1=\mathrm{x}_2 \mathrm{~T}_2 \\
& \Rightarrow \mathrm{x}_2 \Rightarrow \frac{\mathrm{x}_1 \mathrm{~T}_1}{\mathrm{~T}_2}=\frac{283}{383} \mathrm{x}\left[\begin{array}{c}
\text { Given } \\
T_1=283 K \\
T_2=383 K
\end{array}\right]
\end{aligned}
\)

Hint

We have given molar specificheat at instant pressure
\(
C_p=\frac{7}{2} R
\)
Mayer’s relation can be written as :
Molar specificheat at constant pressure
– Molar specificheat at constant volume
\(=\) Gas constant,
i.e. \(C_p-C_v=R \Rightarrow C_v=C_p-R\)
\(
=\frac{7}{2} R-R=\frac{5}{2} R\left[\because C_p=\frac{7}{2} R\right]
\)
Hence, the required ratio is
\(
\gamma=\frac{C_p}{C_V}=\frac{\left(\frac{7}{2}\right) R}{\left(\frac{5}{2}\right) R}=\frac{7}{5}
\)

Hint

Number of moles,
\(
n=\frac{m}{\text { molecular weight }}=\frac{5}{32}
\)
As from ideal gas equation
\(
p V=n R T \Rightarrow p V=\frac{5}{32} R T
\)

Hint

Given: Initial temperature of gas
\(\left(T_1\right)=27^{\circ} \mathrm{C}=300 \mathrm{~K}\)
Initial pressure \(\left(P_1\right)=30 \mathrm{~atm}\)
Initial volume \(\left(V_1\right)=V\)
Final pressure \(\left(P_2\right)=1 \mathrm{~atm}\)
Final volume \(\left(V_2\right)=10 \mathrm{~V}\).
We know from the general gas equation that
\(
\begin{aligned}
& \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \quad \text { or, } \frac{30 \times V}{300}=\frac{1 \times 10 \mathrm{~V}}{T_2} \\
& \text { or, } T_2=100 \mathrm{~K}=-173^{\circ} \mathrm{C} .
\end{aligned}
\)

Hint

We know that ratio of specific heat,
\(
\gamma=1+\frac{2}{n} \text { or } n=\frac{2}{\gamma-1}
\)
[where \(\mathrm{n}=\) Degree of freedom]

Hint

The gases carbon-monoxide \((\mathrm{CO})\) and nitrogen \(\left(\mathrm{N}_2\right)\) are diatomic, so both have equal kinetic energy \(\frac{5}{2} k T\), i.e. \(E_1=E_2\).

Hint

In an adiabatic process
\(
\begin{aligned}
& p=\text { pressure } \\
& V=\text { volume }
\end{aligned}
\)
\(
\begin{aligned}
\gamma & =\text { atomicity of gas } \\
p V^\gamma & =\text { constant } \dots(i)
\end{aligned}
\)
Now from the ideal gas equation,
\(
p V=R T \quad \text { (for one mole) }
\)
or \(\quad p=\frac{R T}{V} \dots(ii)\)
\(
(R=\text { gas constant })
\)
From Eqs. (i) and (ii), we have
\(
\begin{aligned}
& \left(\frac{R T}{V}\right) V^\gamma=\text { constant } \\
& T V^{\gamma-1}=\text { constant } \dots(iii)
\end{aligned}
\)
So for two different cases of temperature and volume
So, \(T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}\)
or \(\quad \frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma-1} \dots(iv)\)
Given, \(\quad T_1=27^{\circ} \mathrm{C}\)
\(
=27+273=300 \mathrm{~K}
\)
Given, \(\quad \frac{V_2}{V_1}=\frac{8}{27}, \gamma=\frac{5}{3}\)
Substituting in Eq. (i), we get
\(
\frac{T_2}{300}=\left(\frac{27}{8}\right)^{5 / 3-1}
\)
or
\(
\frac{T_2}{300}=\left[\left(\frac{3}{2}\right)^3\right]^{2 / 3}
\)
or \(\quad \frac{T_2}{300}=\left(\frac{3}{2}\right)^2=\frac{9}{4}\)
\(
\therefore T_2=\frac{9}{4} \times 300=675 \mathrm{~K}=402^{\circ} \mathrm{C}
\)
Thus, rise in temperature
\(
=T_2-T_1=402-27=375^{\circ} \mathrm{C}
\)

Hint

No. of degree of freedom \(=3 K-N\) where \(K\) is no. of atom and \(N\) is the number of relations between atoms. For triatomic gas, \(K=3, N={ }^3 C_2=3\)
No. of degree of freedom \(=3(3)-3=6\)

Hint

At \(0 \mathrm{~K}\), molecular motion stops. Hence, the kinetic energy of molecules becomes zero.

Hint

According to adiabatic process the relation between temperature and volume is given by
\(
T V^{\gamma-1}=\text { constant }
\)
So, for two different cases
\(
\therefore \quad T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \dots(i)
\)
Given, initial temperature
\(
T_1=18^{\circ} \mathrm{C}=291 \mathrm{~K}
\)
Let initial volume \(V_1=V\)
and final volume \(V_2=\left(\frac{1}{8}\right) V\)
Putting these values in Eq. (i)
\(
\begin{aligned}
T_2 & =291\left(\frac{V_1}{V_2}\right)^{\gamma-1}=291\left(\frac{V}{\left(\frac{1}{8}\right) \mathrm{V}}\right)^{7 / 5-1} \\
& \left(\gamma=\frac{7}{5} \text { for diatomic gas }\right) \\
& =291 \times 2.297=668.4 \mathrm{~K}
\end{aligned}
\)

Hint

As per ideal gas equation \(\mathrm{PV}=\mathrm{nRT}\) where \(\mathrm{n}=\) number of moles
Also number of moles \((\mathrm{n})=\frac{\text { Total mass }}{\text { Molar mass }}\)
Total mass of oxygen \(=8 \mathrm{~g}\) and molar mass of \(\mathrm{O}_2=16 \times\) \(2=32\)
Therefore number of moles \((\mathrm{n})=\quad \frac{\text { Total mass }}{\text { Molar mass }}=\frac{8}{32}=\frac{1}{4}\)
Thus \(\mathrm{PV}=\mathrm{nRT}\)
\(
\mathrm{PV}=\frac{\mathrm{RT}}{4}
\)

Hint

\(v_{\mathrm{rms}} \propto \sqrt{T}\)
As temperature increases from \(300 \mathrm{~K}\) to \(1200 \mathrm{~K}\) that is four times, so, \(v_{\mathrm{rms}}\) will be doubled.

Hint

Velocity of sound \(\left(C_S\right)=\sqrt{\frac{\gamma P}{\rho}}\)
R.M.S. velocity of gas molecules \((C)=\sqrt{\frac{3 P}{\rho}}\)
\(
\frac{C_s}{C}=\sqrt{\frac{\gamma P}{\rho} \times \frac{\rho}{3 P}}=\sqrt{\frac{\gamma}{3}} \Rightarrow C_s=C \times \sqrt{\frac{\gamma}{3}}
\)

Hint

Number of degree of freedom of a dynamical system is obtained by subtracting the number of independent relations from the total number of coordinates required to specify the positions of constituent particles of the system.
If \(A=\) number of particlesin the system, \(R=\) number of independent relations among the particles,
\(N=\) number of degree of freedom of the system, then
\(
N=3 A-R
\)
Each monoatomic, diatomic, and triatomic gas has three translatory degree of freedom.

Hint

\(
C_V=\frac{R}{0.67}=1.5 R=\frac{3}{2} R
\)
This is the case of monoatomic gases. when \(\quad C_v=\frac{3}{2} R\)

Note:

For a gas having \(f\) degree of freedom specific heat at constant volume, \(C_v=\frac{1}{2} f R\) specific heat at constant pressure,
\(
C_p=\left(\frac{f}{2}+1\right) R
\)
where, \(R=\) universal gas constant.

Since \(\frac{R}{C_V}=0.67 \Rightarrow C_V=\frac{3}{2} R\), hence gas is monoatomic.

Hint

Pressure exerted by gas molecules is
\(
p=\frac{1}{3} \rho \bar{v}^2 \dots(i)
\)
where, \(\rho=\) density of gas
\(\bar{v}=\) average velocity of gas molecules or \(p=\frac{2}{3} n \cdot \frac{1}{2} m \bar{v}^2 \quad(\because \rho=m n)\) Now,\(\frac{1}{2} m \bar{v}^2=\) average kinetic energy of a gas molecule \((\overline{\mathrm{KE}})\) Therefore, \(p=\frac{2}{3} n \overline{\mathrm{KE}}\) If \(N\) is total number of gas molecules in volume \(V\), then
No of gas molecules per unit volume
\(
\begin{aligned}
n & =\frac{N}{V} \\
\therefore \quad p & =\frac{2}{3} \cdot \frac{N}{V}\left(\frac{1}{2} m \bar{v}^2\right) \\
\text { or } \quad p V & =\frac{2}{3} N(\overline{K E}) \quad\left[K E=\frac{1}{2} m \bar{v}^2\right]
\end{aligned}
\)
Also, from Eq. (i),
\(
p=\frac{2}{3} \cdot \frac{1}{2} \rho \bar{v}^2
\)
Now, \(\frac{1}{2} \rho \bar{v}^2=\) average kinetic energy of the gas per unit volume.
Therefore, \(p=\frac{2}{3} E\)

Hint

According to Dalton’s law of partial pressure, the total pressure exerted by a mixture of gases, which do not interact with each other, is equal to sum of the partial pressures which each would exert, if alone occupied the same volume at the given temperature. When gases are put in one container, then pressure \(p\) of the mixture will be
\(
p=p_1+p_2+p_3
\)

Hint

The mean square velocity of A type molecules \(=\omega^2+\omega^2+\omega^2=3 \omega^2\)
Therefore, \(\frac{1}{2} m\left(3 \omega^2\right)=\frac{1}{2}(2 m) v^2\)
This gives \(\omega^2 / v^2=2 / 3\)
Mean kinetic energy of the two types of molecules should be equal.

Hint

Both hydrogen and oxygen are diatomic gases and \(C_p-C_v=R\) is the same for all gases, hence \(a=b\).

Hint

The average kinetic energy of gas molecules is directly proportional to absolute temperature only; this implies that all molecular motion ceases if the temperature is reduced to absolute zero.

According to kinetic theory of gases, the pressure pexerted by one mole of an ideal gas is given by
\(
p=\frac{1}{3} \frac{M}{V} c^2 \text { or } p V=\frac{1}{3} M c^2
\)
or \(\quad \frac{1}{3} M c^2=R T \dots(i)\)
where \(c\) is root mean square velocity of gas.
From Eq. (i), when \(T=0, c=0\)
Hence, absolute zero of temperature may be defined as that temperature at which root mean square velocity of the gas molecules reduces to zero. It means molecular motion ceases at absolute zero.

Hint

\(
\begin{aligned}
& \gamma=\frac{C_P}{C_V}=\frac{15}{10}=\frac{3}{2} \Rightarrow C_V=\frac{2}{3} C_P \\
& C_P-C_V=\frac{R}{J} \Rightarrow C_P-\frac{2}{3} C_P=\frac{R}{J} \\
& \Rightarrow \frac{C_P}{3}=\frac{R}{J} \Rightarrow C_P=\frac{3 R}{J}
\end{aligned}
\)

Hint

\(C_p=\frac{5}{2} R\) and \(C_v=\frac{3}{2} R\)
We know that \(Q_v=n C_v \Delta T\) and \(Q_p=n C_p \Delta T\)
\(
\Rightarrow \frac{Q_v}{Q_p}=\frac{3}{5}
\)
Given \(Q_p=207 \mathrm{~J} \Rightarrow Q_v \equiv 124 \mathrm{~J}\)

Hint

As the temperature increases, the average velocity increases. So, the collisions are faster.

Hint

According to the law of equipartition of energy, the energy per degree of freedom is \(\frac{1}{2} k T\).
For a polyatomic gas with \(\mathrm{n}\) degrees of freedom, the mean energy per molecule \(=\frac{1}{2} n k T\)

Alternate:

Concept If there is sudden compression without exchange of heat the process will be adiabatic.
According to law of equipartition of energy for any dynamical system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom and the energy associated with each molecule per degree of freedom is \(\frac{1}{2} k T\). For a polyatomic gas with \(n\) degrees of freedom the mean energy per molecule \(=\frac{1}{2} n k T\).
\(K=\) Boltzmann constant
\(n=\) degree of freedom
\(T=\) Temperature

Hint

The adiabatic relation between \(p\) and \(V\) for a perfect gas is
\(
\left.p V^\gamma=k \text { (a constant }\right) \dots(i)
\)
Again from standard gas equation
\(
p V=n R T \Rightarrow V=\frac{R T}{p}
\)
Putting in Eq. (i), we get
\(
p \frac{R^\gamma T^\gamma}{p^\gamma}=k
\)
or \(\quad p^{1-r} T^\gamma=\frac{k}{R^\gamma}=\) ano ther constant
i.e. \(p^{1-\gamma} T^\gamma=\) constant
Comparing two different situations,
\(
\begin{aligned}
& p_1^{1-\gamma} T_1^\gamma=p_2^{1-\gamma} T_2^\gamma \\
& \text { Here, } \quad p_2=\left(\frac{1}{8}\right) p_1 \\
& T_1=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K} \\
& T_2=?, \gamma=\frac{5}{3} \\
& \therefore\left(\frac{T_2}{T_1}\right)^\gamma=\left(\frac{p_1}{p_2}\right)^{1-\gamma} \\
& \text { or }\left(\frac{T_2}{300}\right)^{5 / 3}=(8)^{1-5 / 3}=(8)^{-2 / 3} \\
& \Rightarrow T_2=130.6 \mathrm{~K} \\
& \therefore \quad T_2=-142^{\circ} \mathrm{C} \\
&
\end{aligned}
\)

Hint

Vapour pressure does not depend on the amount of substance. It depends on the temperature alone. As both containers are at same temperature so vapour pressure of both A and B will be same hence ratio of vapour pressure is 1:1