Question

Two different adiabatic paths for the same gas intersects two isothermals at $T_1$ and $T_2$ as shown in the $P-V$ diagram. Then $\frac{V_a}{V_d}$ is equal to

• A. $\frac{V_c}{V_b}$
• B. $V_c{V}_b$
• C. $\frac{V_b}{V_c}$
• D. $\frac{V_c{V}_b}{V_c+V_b}$

Answer 1

The correct option is C $\frac{V_b}{V_c}$

For curve $bc$,
$T{V}^{\gamma -1}=constant{T}_1{V_b}^{\gamma -1}=T_2{V_c}^{\gamma -1}--\left(1\right)$
For curve $ad$
$T_1{V_a}^{\gamma -1}=T_2{V_d}^{\gamma -1}--\left(2\right)\left(1\right)÷\left(2\right)\Rightarrow {\left(\frac{V_b}{V_a}\right)}^{\gamma -1}={\left(\frac{V_c}{V_d}\right)}^{\gamma -1}\Rightarrow \frac{V_b}{V_a}=\frac{V_c}{V_d}\Rightarrow \frac{V_a}{V_d}=\frac{V_b}{V_c}$