Q132 / 150 · MHT CET 2024 (09 May Shift 2) · Physics
An ideal gas $(\gamma=1.5)$ is expanded adiabatically. To reduce root mean square velocity of molecules two times, the gas should be expanded
- A20 times
- Correct answerB16 times
- C12 times
- D8 times
Solution
Since r.m.s. velocity $\mathrm{v} \propto \sqrt{\mathrm{T}}$,
$\therefore \quad \frac{\mathrm{v}_2}{\mathrm{v}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}$
Given the r.m.s. velocity is reduced two times. $\Rightarrow \mathrm{v}_2=\frac{\mathrm{v}_1}{2}$ Substituting the above result in (i), $\Rightarrow \frac{1}{2}=\sqrt{\frac{T_2}{T_1}} \Rightarrow \frac{T_1}{T_2}=4...(ii)$
Using Fraction of given heat energy utilised in doing external work is given by the formula, $\left(\frac{\Delta W}{\Delta Q}\right)=\left(1-\frac{1}{\gamma}\right)$ For adiabatic expansion, $T_1 \mathrm{~V}_1^{\gamma-1}=T_2 V_2^{y-1}$ $\begin{array}{ll} \therefore & \left(\frac{V_2}{V_1}\right)^{\gamma-1}=\frac{T_1}{T_2}=4 ...[from (ii)]\\ & \Rightarrow\left(\frac{V_2}{V_1}\right)^{1.5-1}=4 \\ \therefore & \left(\frac{V_2}{V_1}\right)^{0.5}=4 \quad \Rightarrow \quad \frac{V_2}{V_1}=16 \end{array}$
Given the r.m.s. velocity is reduced two times. $\Rightarrow \mathrm{v}_2=\frac{\mathrm{v}_1}{2}$ Substituting the above result in (i), $\Rightarrow \frac{1}{2}=\sqrt{\frac{T_2}{T_1}} \Rightarrow \frac{T_1}{T_2}=4...(ii)$
Using Fraction of given heat energy utilised in doing external work is given by the formula, $\left(\frac{\Delta W}{\Delta Q}\right)=\left(1-\frac{1}{\gamma}\right)$ For adiabatic expansion, $T_1 \mathrm{~V}_1^{\gamma-1}=T_2 V_2^{y-1}$ $\begin{array}{ll} \therefore & \left(\frac{V_2}{V_1}\right)^{\gamma-1}=\frac{T_1}{T_2}=4 ...[from (ii)]\\ & \Rightarrow\left(\frac{V_2}{V_1}\right)^{1.5-1}=4 \\ \therefore & \left(\frac{V_2}{V_1}\right)^{0.5}=4 \quad \Rightarrow \quad \frac{V_2}{V_1}=16 \end{array}$
Hence, the correct answer is option: 'B'.