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In the P-V diagram shown, there are two adiabatic parts of the same gas intersecting two isothermals at $T_1$ and $T_2$. The ratio $\left(\frac{V_b}{V_a}\right)$ is equal to
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Verified Answer
The correct answer is:
$\left(\frac{V_c}{V_d}\right)$
For adiabatic process
$P V Y=C$ Ideal gas law states
$P V=n R T$
Dividing $T V \gamma-1=$ constant above equations.
Consider the diagram as follows:
For path BC
$\begin{aligned}
& T_1=V_b^{\gamma-1}=T_2=\underset{c}{V \gamma-1} \\
& \Rightarrow\left(\frac{V_b}{V_c}\right)^{Y-1}=\frac{T_2}{T_1}---(1)
\end{aligned}$
Now, for path DA.
$\begin{aligned}
& T_2=\underset{d}{V}-1=T_1=\underset{a}{V \gamma-1} \\
& \Rightarrow\left(\frac{V_a}{V_d}\right)^{\gamma-1}=\frac{T_2}{T_1} \quad--(2)
\end{aligned}$
$\begin{aligned}
& \frac{V_b}{V_c}=\frac{V_a}{V_d} \\
& \Rightarrow \frac{V_b}{V_a}=\frac{V_c}{V_d}
\end{aligned}$
Option (B) is correct.
$P V Y=C$ Ideal gas law states
$P V=n R T$
Dividing $T V \gamma-1=$ constant above equations.
Consider the diagram as follows:
For path BC
$\begin{aligned}
& T_1=V_b^{\gamma-1}=T_2=\underset{c}{V \gamma-1} \\
& \Rightarrow\left(\frac{V_b}{V_c}\right)^{Y-1}=\frac{T_2}{T_1}---(1)
\end{aligned}$
Now, for path DA.
$\begin{aligned}
& T_2=\underset{d}{V}-1=T_1=\underset{a}{V \gamma-1} \\
& \Rightarrow\left(\frac{V_a}{V_d}\right)^{\gamma-1}=\frac{T_2}{T_1} \quad--(2)
\end{aligned}$
$\begin{aligned}
& \frac{V_b}{V_c}=\frac{V_a}{V_d} \\
& \Rightarrow \frac{V_b}{V_a}=\frac{V_c}{V_d}
\end{aligned}$
Option (B) is correct.
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