Question

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in the $P-V$ diagram. The relation between the ratio $\frac{V_a}{V_d}$ and the ratio $\frac{V_b}{V_c}$ is:

• A. $\frac{V_a}{V_d} = \left(\frac{V_b}{V_c}\right)^{-1}$
• B. $\frac{V_a}{V_d} \neq \frac{V_b}{V_c}$
• C. $\frac{V_a}{V_d} = \frac{V_b}{V_c}$
• D. $\frac{V_a}{V_d} = \left(\frac{V_b}{V_c}\right)^2$

Answer 1

Answer: (C)

$\frac{V_a}{V_d} = \frac{V_b}{V_c}$

Answer 2

For an adiabatic process, the equation $TV^{\gamma-1} = \text{constant}$ holds.

Between points $a$ and $d$:
$T_a \cdot V_a^{\gamma-1} = T_d \cdot V_d^{\gamma-1}.$
$\frac{V_a}{V_d} = \frac{T_d}{T_a}.$

Between points $b$ and $c$:
$T_b \cdot V_b^{\gamma-1} = T_c \cdot V_c^{\gamma-1}.$
$\frac{V_b}{V_c} = \frac{T_c}{T_b}.$

Given $T_d = T_c$ and $T_a = T_b$, we have:
$\frac{V_a}{V_d} = \frac{V_b}{V_c}.$

Final Answer: $\frac{V_a}{V_d} = \frac{V_b}{V_c}$.