1. Use the adiabatic relation

For the same ideal gas undergoing an adiabatic process, $PV^\gamma=\text{constant}$ Also, using $PV=nRT$, $TV^{\gamma-1}=\text{constant}$

So for any adiabatic path between two isotherms at temperatures $T_1$ and $T_2$, $T_1 V_\text{(on upper isotherm)}^{\gamma-1}=T_2 V_\text{(on lower isotherm)}^{\gamma-1}$

2. Apply this to the first adiabatic path

Suppose points $a$ and $b$ lie on the same adiabat, with $a$ on one isotherm and $b$ on the other. Then, $T_1 V_a^{\gamma-1}=T_2 V_b^{\gamma-1}$ Hence, $\left(\frac{V_a}{V_b}\right)^{\gamma-1}=\frac{T_2}{T_1}$

3. Apply this to the second adiabatic path

Similarly, for points $d$ and $c$ on the second adiabat, $T_1 V_d^{\gamma-1}=T_2 V_c^{\gamma-1}$ Hence, $\left(\frac{V_d}{V_c}\right)^{\gamma-1}=\frac{T_2}{T_1}$

4. Compare the two results

Since both right-hand sides are equal, $\left(\frac{V_a}{V_b}\right)^{\gamma-1}=\left(\frac{V_d}{V_c}\right)^{\gamma-1}$ Therefore, $\frac{V_a}{V_b}=\frac{V_d}{V_c}$ Rearranging, $\frac{V_a}{V_d}=\frac{V_b}{V_c}$

5. Check options

• A: $\frac{V_a}{V_d} \ne \frac{V_b}{V_c}$ ❌
• B: $\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^{-1}$ ❌
• C: $\frac{V_a}{V_d}=\frac{V_b}{V_c}$ ✅
• D: $\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^2$ ❌

So the correct answer is C.