Two ideal gases A and B of same number of moles expand at constant temperatures $T_1$ and $T_2$ respectively…

Two ideal gases A and B of same number of moles expand at constant temperatures $T_1$ and $T_2$ respectively such that the pressure of gas A decreases by $50 \%$ and the pressure of gas B decreases by $75 \%$. If the work done by both the gases is same, then $T_1: T_2$
  1. $1: 3$
  2. $2: 3$
  3. $3: 4$
  4. $2: 1$

Solution

Work done in isothermal process ($\mathrm{T}=$ constant $)$ is $\begin{aligned} & \mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{P}_1}{\mathrm{P}_2} \\ & \therefore \quad \mathrm{~W}_1=\mathrm{W}_2 \Rightarrow \mathrm{nRT} \mathrm{l}_1 \ln \left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)_{\mathrm{A}}=\mathrm{nRT}_2 \ln \left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)_{\mathrm{B}} \\ & \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\ln \left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)_{\mathrm{B}}}{\ln \left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)_{\mathrm{A}}}=\frac{\ln \left(\frac{\mathrm{P}_1}{0.25 \mathrm{P}_1}\right)}{\ln \left(\frac{\mathrm{P}_1}{0.5 \mathrm{P}_1}\right)}=\frac{\ln 4}{\ln 2} \\ & =\frac{2 \ln 2}{\ln 2}=\frac{2}{1}=2: 1\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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