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

Using Charle's law, we have $$\frac{V}{T} = {\text{constant}}$$
$$ \Rightarrow \frac{{\frac{l}{2} + 5}}{{373}} = \frac{{\frac{l}{2} - 5}}{{273}}$$
As the piston moves $$5\,cm,$$  the length of one side will be $$\left( {\frac{l}{2} + 5} \right)$$  and other side $$\left( {\frac{l}{2} - 5} \right).$$  On solving this equation, we get $$l = 64.6\,cm.$$

Thermal capacity of a body is defined as the amount of heat required to raise the temperature of the (whole) body through $${1^ \circ }C$$ or $$1K.$$
Amount of heat energy required $$\left( {\Delta Q} \right)$$  to raise the temperature of mass $$m$$ of a body through temperature range $$\left( {\Delta T} \right)$$  is $$\Delta Q = sm\left( {\Delta T} \right)$$
where, $$s$$ is specific heat of the body,
when $$\Delta T = 1K,\Delta Q =$$    thermal capacity
∴ Thermal capacity $$ = s \times m \times 1$$
$$ = ms$$
Here, $$m = 40\,g,s = 0.2\,cal/g\,K$$
∴ Thermal capacity $$ = 40 \times 0.2 = 8\,cal{/^ \circ }C$$
$$ = 4.2 \times 8\,J{/^ \circ }C = 33.6\,J{/^ \circ }C$$

Both are diatomic gases and $${C_p} - {C_v} = R$$   for all gases.

$$V = 22.4\,litre = 22.4 \times {10^{ - 3}}{m^3},J = 4200\,J/kcal$$
by ideal gas equation for one mole of a gas,
$$\eqalign{ & R = \frac{{PV}}{T} = \frac{{1.013 \times {{10}^5} \times 22.4 \times {{10}^{ - 3}}}}{{273}} \cr & {C_p} - {C_v} = \frac{R}{J} = \frac{{1.013 \times {{10}^5} \times 22.4}}{{273 \times 4200}} = 1.979\,kcal/kmol\,K \cr} $$

$$\eqalign{ & \frac{{P\left( {{V_1} + {V_2}} \right)}}{T} = \frac{{{P_1}{V_1}}}{{{T_1}}} + \frac{{{P_2}{V_2}}}{{{T_2}}}\,......\left( {\text{i}} \right) \cr & {\text{Also}}\,\,P\left( {{V_1} + {V_2}} \right) = {P_1}{V_1} + {P_2}{V_2}\,......\left( {{\text{ii}}} \right) \cr} $$
After solving above equations, we get
$$T = \left[ {\frac{{\left( {{P_1}{V_1} + {P_2}{V_2}} \right){T_1}{T_2}}}{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}} \right]$$

Volume $$ = \frac{{{\text{mass}}}}{{{\text{density}}}} = \frac{1}{4}{m^3}$$
$$K.E = \frac{5}{2}PV = \frac{5}{2} \times 8 \times {10^4} \times \frac{1}{4} = 5 \times {10^4}\,J$$
Alternatively:
$$\eqalign{ & K.E = \frac{5}{2}nRT = \frac{5}{2}\frac{m}{M}RT = \frac{5}{2}\frac{m}{M} \times \frac{{PM}}{d}\,\,\left[ {\because PM = dRT} \right] \cr & = \frac{5}{2}\frac{{mP}}{d} = \frac{5}{2} \times \frac{{1 \times 8 \times {{10}^4}}}{4} = 5 \times {10^4}\,J \cr} $$

According to kinetic theory of gases, the pressure $$p$$ exerted by one mole of an ideal gas is given by
$$\eqalign{ & p = \frac{1}{3}\frac{M}{V}{c^2}\,\,{\text{or}}\,\,pV = \frac{1}{3}M{c^2} \cr & {\text{or}}\,\,\frac{1}{3}M{c^2} = RT\,......\left( {\text{i}} \right) \cr} $$
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.

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
$$p = {p_1} + {p_2} + {p_3}$$

$$\gamma \,\,{\text{for}}\,\,He = \frac{5}{3};\gamma \,\,{\text{for}}\,\,{O_2} = \frac{7}{5}$$
$$8\,g$$  of $$He$$  is equal to $$2\,moles$$  of $$He$$  and $$16\,g$$  of $${O_2} = \frac{1}{2}$$  $$mole$$  of $${O_2}$$ gas. Total one has $$2\frac{1}{2}\,moles.$$
$$\therefore $$ The weighted average is
$$\eqalign{ & = \left( {2 \times \frac{5}{3} + \frac{7}{5} \times \frac{1}{2}} \right) \times \frac{2}{5} = \left\{ {\frac{{10}}{3} + \frac{7}{{10}}} \right\} \times \frac{2}{5} \cr & = \frac{{121}}{{30}} \times \frac{2}{5} = \frac{{24.2}}{{15}} \cr} $$

A real gas behaves as an ideal gas when the average distance between the gas molecules is large enough so that (i) the force of attraction between the gas molecules becomes almost zero (ii) the actual volume of the gas molecules is negligible as compared to the occupied volume of the gas.
The above conditions are true for low pressure and high temperature.