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

$$\eqalign{ & {P^{1 - \gamma }}{T^\gamma } = {\text{constant}} \cr & {\text{or}}\,\,p \propto {T^{\left[ {\frac{\gamma }{{\left( {\gamma - 1} \right)}}} \right]}} \cr & {\text{Here,}}\,\,p \propto {T^3} \cr & {\text{or}}\,\,\frac{\gamma }{{\gamma - 1}} = 3 \cr & {\text{or}}\,\,\gamma = \frac{3}{2} \cr} $$

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

$$PV = nRT\,\,or{\text{ }}P = \frac{{nRT}}{V}\,\,or{\text{ }}P\,\propto \,T$$
( $$\because $$ $$V$$ and $$n$$ are same.)
Therefore, if $$T$$ is doubled, pressure also becomes two times, i.e, $$2\,P.$$

$$\eqalign{ & V_{rms}^2 = < {V^2} > = \frac{{V_1^2 + V_2^2 + V_3^2 + .......}}{N} \cr & = \frac{{\int {{V^2}} dN}}{{\int d N}}\,\,{\text{here}}\,\,\frac{{dN}}{{dV}} = N\left( V \right) \cr & V_{rms}^2 = \frac{1}{N}\int\limits_0^\infty N \left( V \right){V^2}dV \cr & = \frac{1}{N}\int\limits_0^{{V_0}} {\left[ {\frac{{3N}}{{V_0^3}}{V^2}} \right]} {V^2}dV = \frac{3}{5}V_0^2 \cr & \Rightarrow {V_{rms}} = \sqrt {\frac{3}{5}} {V_0} \cr} $$

$${v_{av}} = \left[ {\frac{{500 + 600 + 700 + 800 + 900}}{5}} \right] = 700\,m/s\,\,{\text{and}}\,\,{v_{rms}} = \sqrt {\frac{{{{500}^2} + {{600}^2} + {{700}^2} + {{800}^2} + {{900}^2}}}{5}} = 714\,m/s$$
Thus $${v_{rms}}$$  is greater than average speed by $$14\,m/s.$$

The heat is supplied at constant pressure. Therefore,
$$\eqalign{ & Q = n{C_p}\Delta t \cr & = 2\left[ {\frac{5}{2}R} \right] \times \Delta t \cr & = 2 \times \frac{5}{2} \times 8.31 \times 5 \cr & = 208\,J\,\left( {\because {C_p} = \frac{5}{2}R\,{\text{for mono - atomic gas}}} \right) \cr} $$

We know that $$\frac{V}{T} = {\text{constant}}$$
$$\therefore \,\,\frac{{V + \Delta T}}{{T + \Delta V}} = \frac{V}{T}\,\,{\text{or }}\frac{{\Delta V}}{{V\,\Delta T}} = \frac{1}{T}$$

$$A$$ is free to move, therefore heat will be supplied at constant pressure
$$\therefore \Delta {Q_A} = n{C_p}\Delta {T_A}\,......\left( {\text{i}} \right)$$
$$B$$ is held fixed, therefore heat will be supplied at constant volume.
$$\eqalign{ & \therefore \Delta {Q_B} = n{C_v}\Delta {T_B}\,......\left( {{\text{ii}}} \right) \cr & {\text{But}}\,\Delta {Q_A} = \Delta {Q_B}\,\left( {{\text{given}}} \right) \cr & \therefore n{C_p}\Delta {T_A} = n{C_v}\Delta {T_B} \cr & \therefore \Delta {T_B} = \left( {\frac{{{C_p}}}{{{C_v}}}} \right)\Delta {T_A} \cr & = \gamma \left( {\Delta {T_A}} \right)\,\,\left[ {\gamma = 1.4\left( {{\text{diatomic}}} \right)} \right] \cr & = \left( {1.4} \right)\left( {30\,K} \right) \cr & \therefore \Delta {T_B} = 42\;K \cr} $$

Concept
Use ideal gas equation to find the ratio between density of a fixed mass of an ideal gas and its pressure.
Ideal gas equation
$$\eqalign{ & pV = nRT \cr & \frac{{pV}}{m} = \frac{1}{M}RT\left( {\because n = \frac{m}{M}} \right) \cr & \frac{p}{\rho } = \frac{{RT}}{M}\,\,\left( {\because \frac{V}{m} = \frac{1}{\rho }} \right) \cr & \therefore \frac{\rho }{p} \propto \frac{1}{T} \cr} $$
Molecular mass $$M$$ and universal gas constant $$R$$ remains same for a gas.
So, for two different situations i.e. at two different temperatures and densities
$$\eqalign{ & \therefore \frac{{\frac{{{\rho _1}}}{{{p_1}}}}}{{\frac{{{\rho _2}}}{{{p_2}}}}} = \frac{{{T_2}}}{{{T_1}}} \Rightarrow \frac{x}{{\left( {\frac{{{\rho _2}}}{{{p_2}}}} \right)}} = \frac{{383\,K}}{{283\,K}} \cr & \Rightarrow \frac{{{\rho _2}}}{{{p_2}}} = \frac{{283}}{{383}}x \cr} $$

Vapour pressure does not depend on the amount of substance. It depends on the temperature alone.