Kinetic Theory of Gases MCQ Questions & Answers in Heat and Thermodynamics | Physics

For a monatomic gas
$${C_v} = \frac{3}{2}R$$
So correct graph is

Work done is to be done in expanding the gas at constant pressure.

$${v_{rms}} = \sqrt {\frac{{3P}}{\rho }} = \sqrt {\frac{{3PV}}{m}} \Rightarrow P = \frac{{m{{\left( {{v_{rms}}} \right)}^2}}}{{3V}}$$
If $${P'}$$ be the final pressure then
$$\frac{{P'}}{P} = \left[ {\frac{{\frac{{\frac{m}{2}{{\left( {2{v_{rms}}} \right)}^2}}}{{3V}}}}{{\frac{{m{{\left( {{v_{rms}}} \right)}^2}}}{{3V}}}}} \right] = 2 \Rightarrow P' = 2P.$$

From Boyle's law ($$T$$ = constant)
$$\eqalign{ & {P_1}{V_1} = {P_2}{V_2} \cr & \therefore \left( {H{d_{{\text{water}}}} + h{d_{{\text{mercury}}}}} \right)g\left( {\frac{4}{3}\pi {r^3}} \right) = h{d_{{\text{mercury}}}}g\left( {\frac{4}{3}\pi {{\left( {2r} \right)}^3}} \right) \cr & \Rightarrow H{d_{{\text{water}}}} = 8h{d_{{\text{mercury}}}} - h{d_{{\text{mercury}}}} \cr & \Rightarrow H = 7h\frac{{{d_{{\text{mercury}}}}}}{{{d_{{\text{water}}}}}} \cr & \therefore H = 7\;h\rho \cr} $$

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 $$ = 3\left( 3 \right) - 3 = 6$$

$${P_1} > {P_2}$$

$${\text{As}}\,V = {\text{constant}} \Rightarrow P \propto T$$
Hence from $$V-T$$  graph $${P_1} > {P_2}$$

From first law of thermodynamics
$$\eqalign{ & \Delta Q = \Delta U + \Delta W = \frac{3}{2}.\frac{1}{4}R\left( {{T_2} - {T_1}} \right) + 0 \cr & = \frac{3}{8}\;{N_a}{K_B}\left( {{T_2} - {T_1}} \right)\,\,\left[ {\because K = \frac{R}{N}} \right] \cr} $$

The speed of sound in a gas is given by $$v = \sqrt {\frac{{\gamma RT}}{M}} $$
$$\eqalign{ & \therefore \,\,\frac{{{v_{{O_2}}}}}{{{v_{He}}}} = \sqrt {\frac{{{\gamma _{{O_2}}}}}{{{M_{{O_2}}}}} \times \frac{{{M_{He}}}}{{{\gamma _{He}}}}} \cr & = \sqrt {\frac{{1.4}}{{32}} \times \frac{4}{{1.67}}} \cr & = 0.3237 \cr & \therefore \,\,{v_{He}} = \frac{{{v_{{O_2}}}}}{{0.3237}} \cr & = \frac{{460}}{{0.3237}} \cr & = 1421\,\,m/s \cr} $$

$$\eqalign{ & {v_{rms}} = \sqrt {\frac{{8RT}}{{\pi M}}} \cr & {\text{For}}\,\,{v_{rms}}\,\,{\text{to}}\,{\text{be}}\,{\text{equal}}\,\,\frac{{{T_{{H_2}}}}}{{{M_{{H_2}}}}} = \frac{{{T_{{O_2}}}}}{{{M_{{O_2}}}}} \cr & {\text{Here}}\,\,{M_{{H_2}}} = 2;\,\,{M_{{O_2}}} = 32; \cr & {T_{{O_2}}} = 47 + 273 = 320\;K \cr & \therefore \frac{{{T_{{H_2}}}}}{2} = \frac{{320}}{{32}} \Rightarrow {T_{{H_2}}} = 20\,K \cr} $$

$$\eqalign{ & {C_p} = \left( {\frac{f}{2} + 1} \right)R = \frac{9}{2}R \cr & \Rightarrow f = 7 \cr & \frac{{{C_p}}}{{{C_v}}} = 1 + \frac{2}{f} = 1 + \frac{2}{7} = 1.28 \cr} $$