MHT CET · Physics · Thermodynamics
An ideal gas with pressure \(\mathrm{P}\), volume \(\mathrm{V}\) and temperature \(\mathrm{T}\) is expanded isothermally to a volume \(2 \mathrm{~V}\) and a final pressure \(\mathrm{P}_{\mathrm{i}}\). The same gas is expanded adiabatically to a volume \(2 \mathrm{~V}\), the final pressure is \(\mathrm{P}_{\mathrm{a}}\). In terms of the ratio of the two specific heats for the gas ' \(\gamma\) ', the ratio \(\frac{\mathrm{P}_i}{\mathrm{P}_{\mathrm{a}}}\) is
- A \(2^{\gamma+1}\)
- B \(2^{\gamma-1}\)
- C \(2^{1-\gamma}\)
- D \(2^\gamma\)
Answer & Solution
Correct Answer
(B) \(2^{\gamma-1}\)
Step-by-step Solution
Detailed explanation
For isothermal expansion we have
\(
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\
& \therefore \mathrm{P}_2=\mathrm{P}_1 \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\mathrm{P}_1 \times \frac{1}{2}=\frac{\mathrm{P}}{2} \\
& \therefore \mathrm{P}_{\mathrm{i}}=\frac{\mathrm{P}}{2}
\end{aligned}
\)
For adiabatic process:
\(
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \mathrm{P}_2=\mathrm{P}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=\mathrm{P}_1\left(\frac{1}{2}\right)^\gamma=\frac{\mathrm{P}}{2^\gamma} \\
& \therefore \mathrm{P}_{\mathrm{a}}=\frac{\mathrm{P}}{2^\gamma} \\
& \therefore \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{a}}}=\frac{2^\gamma}{2}=2^{\gamma-1}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\
& \therefore \mathrm{P}_2=\mathrm{P}_1 \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\mathrm{P}_1 \times \frac{1}{2}=\frac{\mathrm{P}}{2} \\
& \therefore \mathrm{P}_{\mathrm{i}}=\frac{\mathrm{P}}{2}
\end{aligned}
\)
For adiabatic process:
\(
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \mathrm{P}_2=\mathrm{P}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=\mathrm{P}_1\left(\frac{1}{2}\right)^\gamma=\frac{\mathrm{P}}{2^\gamma} \\
& \therefore \mathrm{P}_{\mathrm{a}}=\frac{\mathrm{P}}{2^\gamma} \\
& \therefore \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{a}}}=\frac{2^\gamma}{2}=2^{\gamma-1}
\end{aligned}
\)
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