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Ideal Gas Equation

Two different...

Question

Two different adiabetic paths for the same gas intersects two isothermal paths $T_{1}$ and $T_{2}$ as shown in the $P - V$ diagram. Then $V_{a} / V_{d}$ is equal to

A

\frac{V_{c}}{V_{b}}

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B

V_{c} V_{b}

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C

\frac{V_{b}}{V_{c}}

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D

\frac{V_{c} V_{b}}{V_{c} + V_{b}}

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Solution

The correct option is C $\frac{V_{b}}{V_{c}}$
$T_{1} (V_{a})^{γ - 1} = T_{2} (V_{d})^{γ - 1}$
$\frac{T_{1}}{T_{2}} = (\frac{V_{d}}{V_{a}})^{γ - 1}$

Also,
$T_{1} (V_{b})^{γ - 1} = T_{2} (V_{c})^{γ - 1}$
$\frac{T_{1}}{T_{2}} = (\frac{V_{c}}{V_{b}})^{γ - 1}$

$\Rightarrow \frac{V_{a}}{V_{d}} = \frac{V_{b}}{V_{c}}$

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