MHT CET · Physics · Thermodynamics
An ideal gas expands adiabatically. \((\gamma=1 \cdot 5)\) To reduce the r.m.s. velocity of the molecules 3 times, the gas has to be expanded
- A 81 times
- B 27 times
- C 9 times
- D 3 times
Answer & Solution
Correct Answer
(A) 81 times
Step-by-step Solution
Detailed explanation
We know,
\(\begin{aligned}
& V_{\text {rms }}=\sqrt{\frac{3 R T}{M_0}} \\
& \Rightarrow \mathrm{T} \propto V^2
\end{aligned}\)
\(\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2^2}{\mathrm{~V}_1^2}=\frac{\left(\frac{\mathrm{V}_1}{3}\right)^2}{\mathrm{~V}_1^2}=\frac{1}{9}.. (i)\)
Also, \(\mathrm{TV}^{\gamma-1}=\) Constant
\(\begin{aligned}
& \Rightarrow \frac{V_2}{V_1}=\left(\frac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}} \\
& \frac{V_2}{V_1}=(9)^2=81
\end{aligned}\)
\(\begin{aligned}
& V_{\text {rms }}=\sqrt{\frac{3 R T}{M_0}} \\
& \Rightarrow \mathrm{T} \propto V^2
\end{aligned}\)
\(\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2^2}{\mathrm{~V}_1^2}=\frac{\left(\frac{\mathrm{V}_1}{3}\right)^2}{\mathrm{~V}_1^2}=\frac{1}{9}.. (i)\)
Also, \(\mathrm{TV}^{\gamma-1}=\) Constant
\(\begin{aligned}
& \Rightarrow \frac{V_2}{V_1}=\left(\frac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}} \\
& \frac{V_2}{V_1}=(9)^2=81
\end{aligned}\)
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