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Two adiabatic...

Question

Two adiabatic processes $b c$ and $a d$ for the same gas are given to intersect two isotherms at $T_{1}$ and $T_{2}$ (as shown). Then

A

\frac{V_{a}}{V_{b}} = \frac{T_{2}}{T_{1}}

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B

\frac{V_{a}}{V_{d}} = \frac{T_{2}}{T_{1}}

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C

\frac{V_{a}}{V_{b}} = {(\frac{T_{2}}{T_{1}})}^{\frac{1}{γ - 1}}

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D

V_{a} V_{c} = V_{b} V_{d}

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Solution

The correct option is D $V_{a} V_{c} = V_{b} V_{d}$
$\begin{matrix} N o w, T_{A} V_{A}^{Y - 1} = T_{D} V_{D}^{Y - 1} T_{A} V_{A} Y - 1 = T_{2} V_{D} Y - 1 \frac{T_{1}}{T_{2}} = {(\frac{V_{D}}{V_{A}})}^{Y - 1} A l s o, p o i n t s B a n d C l i e o n t h e s a m e a d i a b a t i c, T_{B} V_{B}^{Y - 1} = T_{C} V_{C}^{Y - 1} T_{1} V_{B} Y - 1 = T_{2} V_{C} Y - 1 \frac{T_{1}}{T_{2}} = {(\frac{V_{C}}{V_{B}})}^{Y - 1} F r o m e q u a t i o n (1) a n d (2) {(\frac{V D}{V A})}^{Y - 1} = {(\frac{V C}{V B})}^{Y - 1} \frac{V_{A}}{V_{D}} = \frac{V_{B}}{V_{C}} ∴ \frac{(\frac{V_{A}}{V_{D}})}{(\frac{V_{B}}{V_{C}})} = 1 \end{matrix}$
Hence, Option $D$ is correct .

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