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

For equilibrium in case 1 at $${0^ \circ }C$$
Upthrust = Wt. of body
$$\eqalign{ & \therefore \,\,{K_1}V{d_2}g = V{d_1}g \cr & \Rightarrow \,\,{K_1} = \frac{{{d_1}}}{{{d_2}}}\,\,\,\,\,.....\left( {\text{i}} \right) \cr} $$

For equilibrium in case 2 at $${60^ \circ }C$$
Note : When the temperature is increased the density will decrease.
$$\eqalign{ & \therefore \,\,{d_1}' = {d_1}\left( {1 + {\gamma _{Fe}} \times 60} \right) \cr & {\text{and }}{d_2}' = {d_2}\left( {1 + {\gamma _{Hg}} \times 60} \right) \cr} $$
Again upthrust = Wt. of body
$$\eqalign{ & \therefore \,\,{K_2}V'{d_2}'g = V'{d_1}'g \cr & \therefore \,\,{K_2}\left[ {\frac{{{d_2}}}{{1 + {\gamma _{Hg}} \times 60}}} \right] = \frac{{{d_1}}}{{1 + {\gamma _{Fe}} \times 60}} \cr & \therefore \,\,{K_2}\left[ {\frac{{1 + {\gamma _{Fe}} \times 60}}{{1 + {\gamma _{Hg}} \times 60}}} \right] = \frac{{{d_1}}}{{{d_2}}} \cr & \Rightarrow \,\,\frac{{{K_1}}}{{{K_2}}} = \frac{{1 + {\gamma _{Fe}} \times 60}}{{1 + {\gamma _{Hg}} \times 60}} \cr} $$

Average $$K.E. = \frac{1}{2}m{c^2} = \frac{3}{2}kT \propto T$$
$$\therefore \frac{{{{\left( {Av.K.E.} \right)}_{600\;K}}}}{{{{\left( {Av.K.E.} \right)}_{300\;K}}}} = \frac{{600}}{{300}} = 2$$
$$\therefore $$ At $$600\,K$$  it will be $$2$$ times than that at $$300\,K.$$

As temperature increases, velocity of molecules increases and hence the pressure.

$$\eqalign{ & \frac{{{v_{rms}}\left( {{\text{helium}}} \right)}}{{{v_{rms}}\left( {{\text{argon}}} \right)}} = \sqrt {\frac{{{M_{{\text{argon}}}}}}{{{M_{{\text{helium}}}}}}} \cr & = \sqrt {\frac{{40}}{4}} \cr & = \sqrt {10} \cr & \approx 3.16 \cr} $$

According to Dalton’s law of partial pressures, we have $$P = {P_1} + {P_2} + {P_3}$$

$$\eqalign{ & \frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}} \cr & {\text{Here,}}\,{P_1} = 200\,kPa \cr & {T_1} = {22^ \circ }C = 295\,K \cr & {T_2} = {42^ \circ }C = 315\,K \cr & {V_2} = {V_1} + \frac{2}{{100}}{V_1} = 1.02{V_1} \cr & \therefore {P_2} = \frac{{200 \times 315{V_1}}}{{295 \times 1.02{V_1}}} = 209.37\,kPa \cr} $$

As we know that,
$${\text{Pressure,}}\,\,p = \frac{1}{3} \cdot \frac{{nm}}{V}v_{rms}^2$$
$$\because nm = $$   mass of the gas, $$V =$$  volume of the gas
$$\therefore \frac{{mn}}{V} = $$   density of the gas. Thus,
$$\eqalign{ & p = \frac{1}{2}pv_{rms}^2 = \frac{1}{3}\rho \frac{{3RT}}{{{M_0}}} = \frac{{\rho RT}}{{{M_0}}}\,\,\,\,\left( {\because {v_{rms}} = \sqrt {\frac{{3RT}}{{{M_0}}}} } \right) \cr & \rho = \frac{{p{M_0}}}{{RT}} = \frac{{pm{N_A}}}{{k{N_A}T}}\,\,\,\,\left[ {\because R = {N_A}k\,\,{\text{and}}\,\,{M_0} = m{N_A}} \right] \cr & \rho = \frac{{pm}}{{kT}} \cr} $$

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 amongest all the degrees of freedom and the energy associated with each molecule per degree of freedom is $$\frac{1}{2}kT.$$  For a polyatomic gas with $$n$$ degrees of freedom the mean energy per molecule $$ = \frac{1}{2}nkT.$$
$$K =$$  Boltzmann constant
$$n =$$  degree of freedom
$$T =$$  Temperature

$$\eqalign{ & {C_p} - {C_v} = R \Rightarrow {C_p} = {C_v} + R \cr & \because \gamma = \frac{{{C_p}}}{{{C_v}}} = \frac{{{C_v} + R}}{{{C_v}}} = \frac{{{C_v}}}{{{C_v}}} + \frac{R}{{{C_v}}} \cr & \Rightarrow \gamma = 1 + \frac{R}{{{C_v}}} \Rightarrow \frac{R}{{{C_v}}} = \gamma - 1 \Rightarrow {C_v} \cr & = \frac{R}{{\gamma - 1}} \cr} $$

$$\eqalign{ & C = \frac{R}{{\gamma - 1}} + \frac{R}{{1 - n}}\,\,{\text{for}}\,{\text{a}}\,{\text{polytropic}}\,{\text{process}} \cr & \Rightarrow C = \frac{{5R}}{2} + \frac{R}{{1 - n}}\,\,\left( {\because r = \frac{7}{5}\,{\text{for}}\,{\text{a}}\,{\text{diatomic}}\,{\text{gas}}} \right) \cr} $$
For $$C$$ to be negative, the possible value of $$n$$ will be $$\frac{9}{7}.$$