MHT CET · Physics · Thermodynamics
An ideal gas at pressure ' \(p\) ' is adiabatically compressed so that its density becomes twice that of the initial. If \(\gamma=\frac{c_p}{c_v}=\frac{7}{5}\), then final pressure of the gas is
- A \(\mathrm{p}\)
- B \(2 \mathrm{p}\)
- C \(\frac{7}{5} \mathrm{p}\)
- D \(2.63 p\)
Answer & Solution
Correct Answer
(D) \(2.63 p\)
Step-by-step Solution
Detailed explanation
Since density become twice, its volume will become half.
\(\therefore \frac{\mathrm{V}_1}{\mathrm{~V}_2}=2\)
\(\therefore\) For adiabatic process we have
\(
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(2)^{7 / 5}=(2)^{1.4}=2.63
\end{aligned}
\)
\(\therefore \frac{\mathrm{V}_1}{\mathrm{~V}_2}=2\)
\(\therefore\) For adiabatic process we have
\(
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(2)^{7 / 5}=(2)^{1.4}=2.63
\end{aligned}
\)
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