- JEE Mains 2024
- Hard
- MCQ
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 _______.
Answer & Solution
(C) \(\frac{V_a}{V_d}=\frac{V_b}{V_c}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{TV}^{\gamma-1}=\text { constant }\)
\(\mathrm{T}_{\mathrm{a}} \cdot \mathrm{V}_{\mathrm{a}}^{\gamma-1}=\mathrm{T}_{\mathrm{d}} \cdot \mathrm{V}_{\mathrm{d}}^{\gamma-1}\)
\(\left(\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{d}}}{\mathrm{T}_{\mathrm{a}}}\)
\(\mathrm{T}_{\mathrm{b}} \cdot \mathrm{V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_{\mathrm{c}} \cdot \mathrm{V}_{\mathrm{c}}^{\gamma-1}\)
\(\left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{b}}}\)
\(\mathrm{T}_{\mathrm{b}} \cdot \mathrm{V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_{\mathrm{c}} \cdot \mathrm{V}_{\mathrm{c}}^{\gamma-1}\)
\(\left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{b}}}\)
\(\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}=\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}} \quad\left(\begin{array}{r}\because \mathrm{T}_{\mathrm{d}}=\mathrm{T}_{\mathrm{c}} \\ \mathrm{T}_{\mathrm{a}}=\mathrm{T}_{\mathrm{b}}\end{array}\right)\)
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