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

$$\Delta Q = n{C_p}\Delta T$$    at constant pressure
$$\Delta U = n{C_v}\Delta T$$
Fraction $$ = \frac{{\Delta U}}{{\Delta Q}} = \frac{{{C_V}}}{{{C_P}}} = \frac{1}{\gamma } = \frac{5}{7}$$     for diatomic gas.

Pressure of the gas will not be affected by motion of the system, hence by
$${v_{rms}} = \sqrt {\frac{{3P}}{\rho }} \Rightarrow {{\bar c}^2} = \frac{{3P}}{\rho } = \frac{1}{3}p{{\bar c}^2}$$

Molar mass of the gas $$= 4g/mol$$
Speed of sound
$$\eqalign{ & V = \sqrt {\frac{{\gamma RT}}{m}} \Rightarrow 952 = \sqrt {\frac{{\gamma \times 3.3 \times 273}}{{4 \times {{10}^{ - 3}}}}} \cr & \Rightarrow \gamma = 1.6 = \frac{{16}}{{10}} = \frac{8}{5} \cr & {\text{Also,}}\,\,\gamma = \frac{{{C_P}}}{{{C_V}}} = \frac{8}{5} \cr & {\text{So,}}\,\,{C_P} = \frac{{8 \times 5}}{5} = 8J{K^{ - 1}}mo{l^{ - 1}}\,\left[ {{C_V} = 5.0J{K^{ - 1}}{\text{given}}} \right] \cr} $$

$$\eqalign{ & \because C = \sqrt {\frac{{3RT}}{M}} \cr & {\left( {1930} \right)^2} = \frac{{3 \times 8.314 \times 300}}{M} \cr & M = \frac{{3 \times 8.314 \times 300}}{{1930 \times 1930}} \approx 2 \times {10^{ - 3}}kg \cr} $$
The gas is $${H_2}.$$

$$\eqalign{ & \left( {P + \frac{{a{T^2}}}{V}} \right){V^c} = \left( {RT + b} \right) \cr & \Rightarrow P = \left( {RT + b} \right){V^{ - c}} - \left( {a{T^2}} \right){V^{ - 1}} \cr} $$
On comparing with equation, $$P = A{V^m} - B{V^n},m = - c\,\,{\text{and}}\,\,n = - 1.$$

Higher is the temperature greater is the most probable velocity.

Let the mass of the gas be $$m.$$
At a fixed temperature and pressure, volume is fixed.
Density of the gas, $$\rho = \frac{m}{V}$$
$$\eqalign{ & {\text{Now}}\,\,\frac{\rho }{P} = \frac{m}{{PV}} = \frac{m}{{nRT}} \cr & \Rightarrow \frac{m}{{nRT}} = x\,\,\left( {{\text{By}}\,{\text{question}}} \right) \cr & \Rightarrow xT = {\text{constant}} \Rightarrow {x_1}{T_1} = {x_2}{T_2} \cr} $$
\[ \Rightarrow {x_2} \Rightarrow \frac{{{x_l}{T_1}}}{{\;{T_2}}} = \frac{{283}}{{383}}x\left[ {\begin{array}{*{20}{l}} \therefore \\ {{T_1} = 283\,K}\\ {\;{T_2} = 383\,K} \end{array}} \right]\]

$$C = \frac{Q}{{m\Delta T}};\,{\text{If}}\,\Delta T = 0,C = \infty \,{\text{and}}\,{\text{if}}\,Q = 0,\,{\text{then}}\,C = 0$$

$${v_{rms}} = \sqrt {\frac{{3RT}}{M}} $$
Same $${v_{rms}},$$  implies $$\frac{T}{M}$$ same.
Hence, $$T \propto M$$
$$M$$ for oxygen is more than that of hydrogen, therefore, oxygen is at higher temperature than hydrogen.

$${v_{rms}} = \sqrt {\frac{{3RT}}{M}} = \sqrt {\frac{{3 \times 8.31 \times 273}}{{28 \times {{10}^{ - 3}}}}} = 493\,m/s$$
Thus $$\frac{1}{2}mv_{rms}^2 = \frac{{mgh}}{{1 + \frac{h}{R}}}.$$     After substituting the values and simplifying, we get $$h = 12.4\,km$$