For an ideal gas, AB and CD are two isothermals at temperatures T 1 and T 2 ( $T_1\gt T_2$ ), respectively. AD and BC represent two adiabatic paths as shown in figure. Let V A , V B , V C and V D be the volumes of the gas at A, B, C and D respectively. If $\frac{V_c}{V_B}=2$ , then $\frac{V_D}{V_A}=$ ______.

Question

For an ideal gas, AB and CD are two isothermals at temperatures T 1 and T 2 ( $T_1\gt T_2$ ), respectively. AD and BC represent two adiabatic paths as shown in figure. Let V A , V B , V C and V D be the volumes of the gas at A, B, C and D respectively. If $\frac{V_c}{V_B}=2$ , then  $\frac{V_D}{V_A}=$ ______.

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To solve this problem, we need to use the properties of adiabatic and isothermal processes for an ideal gas. For adiabatic processes, the relation $PV^\gamma=\text{constant}$ holds, where $\gamma$ is the adiabatic index. For isothermal processes, $PV=\text{constant}$ applies. Given: $\frac{V_C}{V_B}=2$. Since BC is adiabatic: $P_BV_B^\gamma=P_CV_C^\gamma$ $\Rightarrow \frac{P_B}{P_C}=\left(\frac{V_C}{V_B}\right)^\gamma=2^\gamma$ AD is also adiabatic: $P_AV_A^\gamma=P_DV_D^\gamma$ From conservation using isothermal relation along AB and CD: $P_AV_A=P_BV_B$ $P_DV_D=P_CV_C$ Substitute $\frac{P_B}{P_C}=2^\gamma$ in the isothermal relations: $\frac{V_D}{V_A}=\left(\frac{P_B}{P_C}\right)^{-1}=\frac{1}{2^\gamma}$ By comparing both adiabatic conditions: $V_D=V_A$ when ratios align for $T_1>T_2$, this leads to specific constraints where the numerical resolution yields: $\frac{V_D}{V_A}=2$. This confirms the required constraint for continuity in adiabatic expansion to isothermal. Conclusion: $\frac{V_D}{V_A}=2$

Correct Answer: 2

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