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: 
- \(\frac{V_a}{V_d} = \left(\frac{V_b}{V_c}\right)^{-1}\)
- \(\frac{V_a}{V_d} \neq \frac{V_b}{V_c}\)
- \(\frac{V_a}{V_d} = \frac{V_b}{V_c}\)
- \(\frac{V_a}{V_d} = \left(\frac{V_b}{V_c}\right)^2\)
The Correct Option is C
Solution and Explanation
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: \(\frac{V_a}{V_d} = \frac{V_b}{V_c}\).
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