Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The…

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio $\frac{V_a}{V_d}$ and the ratio $\frac{V_b}{V_c}$ is:
  1. $\frac{V_a}{V_d} \neq \frac{V_b}{V_c}$
  2. $\frac{V_a}{V_d}=\frac{V_b}{V_c}$
  3. $\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^{-1}$
  4. $\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^2$

Solution

For adiabatic process $\mathrm{TV}^{\gamma-1}=$ constant $\begin{aligned} & \mathrm{T}_{\mathrm{a}} \cdot \mathrm{V}_{\mathrm{a}}^{\gamma-1}=\mathrm{T}_{\mathrm{d}} \cdot \mathrm{V}_{\mathrm{d}}^{\gamma-1} \\ & \left(\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{d}}}{\mathrm{T}_{\mathrm{a}}} \\ & \mathrm{T}_{\mathrm{b}} \cdot \mathrm{V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_{\mathrm{c}} \cdot \mathrm{V}_{\mathrm{c}}^{\gamma-1} \\ & \left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{b}}} \\ & \frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}=\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}} \quad\left(\begin{array}{c}\because \mathrm{T}_{\mathrm{d}}=\mathrm{T}_{\mathrm{c}} \\ \mathrm{T}_{\mathrm{a}}=\mathrm{T}_{\mathrm{b}}\end{array}\right)\end{aligned}$

Asked in: JEE Main 2024 (08 Apr Shift 1)

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