JEE MAIN - Physics (2024 - 8th April Morning Shift - No. 20)

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:

JEE Main 2024 (Online) 8th April Morning Shift Physics - Heat and Thermodynamics Question 36 English

$$\frac{V_a}{V_d} \neq \frac{V_b}{V_c}$$
$$\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^{-1}$$
$$\frac{V_a}{V_d}=\frac{V_b}{V_c}$$
$$\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^2$$

Explanation

$$\begin{aligned} \text { (1) } P_a V_a & =P_b V_b \quad \text{.... (i)}\\ P_c V_c & =P_d V_d \quad \text{.... (ii)} \end{aligned}$$

$$\begin{aligned} \text { (2) } P_a V_a^{\gamma-1} & =P_d V_d^{\gamma-1} \quad \text{.... (iii)}\\ P_b V_b^{\gamma-1} & =P_c V_c^{\gamma-1} \quad \text{.... (iv)} \end{aligned}$$

$$\begin{aligned} & \text { (i) } \div \text { (iii) } \Rightarrow \frac{V_a}{V_a^{\gamma-1}}=\frac{P_b V_b}{P_d V_d{ }^{\gamma-1}} \text { (v) } \\ & \text { (ii) } \div \text { (iv) } \Rightarrow \frac{V_c}{V_c^{\gamma-1}}=\frac{P_d V_d}{P_b V_b{ }^{\gamma-1}} \text { (vi) } \\ & \text { (v) } \times \text { (vi) } \Rightarrow \frac{V_a}{V_a^{\gamma-1}} \frac{V_c}{V_c{ }^{\gamma-1}}=\frac{P_b V_b}{P_d V_d^{\gamma-1}} \times \frac{P_d V_d}{P_b V_b^{\gamma-1}} \\ & \Rightarrow V_a^{\gamma-2} V_c{ }^{\gamma-2}=V_d^{\gamma-1} V_b{ }^{\gamma-2} \\ & \Rightarrow \frac{V_a}{V_d}=\frac{V_b}{V_c} \end{aligned}$$

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