In the P-V diagram shown, there are two adiabatic parts of the same gas intersecting two isothermals at $T_1$ and $T_2$. The ratio $\left(\frac{V_b}{V_a}\right)$ is equal to

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Aicharya · Accepted Answer

For adiabatic process $P V^\gamma=C$ Ideal gas law states $P V=n R T$ Dividing $T V^{\gamma-1}=$ constant above equations. Consider the diagram as follows: For path BC $\begin{aligned} & T_1=V_b^{\gamma-1}=T_2=\underset{c}{V \gamma-1} \ & \Rightarrow\left(\frac{V_b}{V_c}ight)^{Y-1}=\frac{T_2}{T_1}---(1) \end{aligned}$ Now, for path DA. $\begin{aligned} & T_2=\underset{d}{V}-1=T_1=\underset{a}{V \gamma-1} \ & \Rightarrow\left(\frac{V_a}{V_d}ight)^{\gamma-1}=\frac{T_2}{T_1} \quad--(2) \end{aligned}$ $\begin{aligned} & \frac{V_b}{V_c}=\frac{V_a}{V_d} \ & \Rightarrow \frac{V_b}{V_a}=\frac{V_c}{V_d} \end{aligned}$ Option (B) is correct.

In the P-V diagram shown, there are two adiabatic parts of the same gas intersecting two isothermals at…

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