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

$$\eqalign{ & {\text{Volume}} = \frac{{{\text{mass}}}}{{{\text{density}}}} = \frac{1}{4}{m^3} \cr & K.E = \frac{5}{2}\,PV \cr & = \frac{5}{2} \times 8 \times {10^4} \times \frac{1}{4} \cr & = 5 \times {10^4}\,J \cr} $$

$$\eqalign{ & M = \frac{{{C_V}}}{{{S_V}}} = \frac{{\frac{{3 \times 2}}{2}}}{{0.075}} = 40 \cr & \therefore {C_V} = \frac{3}{2}R \cr} $$

$$\eqalign{ & {C_v} = \frac{{dE}}{{dT}} = \frac{3}{2}R \cr & {C_p} = \frac{3}{2}R + R = \frac{5}{2}R = 2.5\,R \cr} $$

$$\eqalign{ & P = \frac{1}{3}\rho {\upsilon ^{\bar 2}} = \frac{1}{3} \times \left( {6 \times {{10}^{ - 2}}} \right) \times {\left( {500} \right)^2} \cr & = 5 \times {10^3}\;N/{m^2} \cr} $$

Pressure exerted by gas molecules is
$$p = \frac{1}{3}\rho {{\bar v}^2}\,......\left( {\text{i}} \right)$$
where, $$\rho $$ = density of gas
$${\bar v}$$ = average velocity of gas molecules
or $$p = \frac{2}{3}n \cdot \frac{1}{2}m{{\bar v}^2}\,\,\left( {\because \rho = mn} \right)$$
Now, $$\frac{1}{2}m{{\bar v}^2}$$  = average kinetic energy of a gas molecule $$\left( {\overline {KE} } \right)$$
Therefore, $$p = \frac{2}{3}n\,\overline {KE} $$
If $$N$$ is total number of gas molecules in volume $$V,$$ then
No of gas molecules per unit volume
$$\eqalign{ & n = \frac{N}{V} \cr & \therefore p = \frac{2}{3} \cdot \frac{N}{V}\left( {\frac{1}{2}m{{\bar v}^2}} \right) \cr & {\text{or}}\,\,pV = \frac{2}{3}N\left( {\overline {KE} } \right)\,\,\left[ {KE = \frac{1}{2}m{{\bar v}^2}} \right] \cr} $$
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, $$p = \frac{2}{3}E$$

$$\eqalign{ & PV = \frac{m}{M}RT; \cr & \therefore \frac{P}{T} = Cm \cr & {\text{or}}\,\,\frac{{{\text{slope of}}\,B}}{{{\text{slope of}}\,A}} = \frac{{{m_B}}}{{{m_A}}} = \frac{{3m}}{m} = 3 \cr} $$

$$P$$ = rate of change of momentum per unit area.

As the bubble rises the pressure gets reduced for constant temperature, if $$P$$ is the standard atmospheric pressure, then
$$\eqalign{ & \left( {P + \rho gh} \right){V_0} = PV \cr & {\text{or}}\,\,V = {V_0}\left( {1 + \frac{{\rho gh}}{P}} \right) \cr} $$

$$\eqalign{ & \frac{{{M_{{N_2}}}}}{{{M_{{H_e}}}}} = \frac{{14}}{1}{M_{{N_2}}} = 14\;m\,\,{\text{and}}\,\,{M_{{H_e}}} = m \cr & {C_p} = \frac{{{n_1}{C_{{p_1}}} + {n_2}{C_{{p_2}}}}}{{{n_1} + {n_2}}} = \frac{{\frac{{14\;m}}{{28}} \times \frac{7}{2}R + \frac{m}{4} \times \frac{5}{2}R}}{{\frac{{14\;m}}{{28}} + \frac{m}{2}}} \cr & = \frac{{2 \times \frac{7}{2}R + \frac{5}{2}R}}{{2 + 1}} = \frac{{7R + \frac{5}{2}R}}{3} = \frac{{19R}}{6} \cr} $$

$$P{T^2} = {\text{constant }}\left( {{\text{given}}} \right)$$
Also for an ideal gas $$\frac{{PV}}{T} = {\text{constant}}$$
From the above two equations, after eliminating $$P.$$
$$\eqalign{ & \frac{V}{{{T^3}}} = {\text{constant}} \cr & \Rightarrow \,\,V = k{T^3}\,\,{\text{where }}k = {\text{co}}{\text{nstant}} \cr & \Rightarrow \,\,\frac{{dV}}{V} = 3\frac{{dT}}{T} \cr & \Rightarrow \,\,dV = \left( {\frac{3}{T}} \right)VdT\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
We know that change in volume due to thermal expansion is given by $$dV = V\,\gamma\, dT\,\,\,\,.....\left( {{\text{ii}}} \right)$$
where $$\gamma $$ = co-efficient of volume expansion.
From (i) and (ii)
$$\eqalign{ & V\,\gamma \,dT = \left( {\frac{3}{T}} \right)VdT \cr & \Rightarrow \,\,\gamma = \frac{3}{T} \cr} $$