MHT CET · Physics · Thermodynamics
A polyatomic gas \((\gamma=4 / 3)\) is compressed to \(\left(\frac{1}{8}\right)^{\text {th }}\) of its volume adiabatically.If its initial pressure is \(\mathrm{P}_0\), its new pressure will be
- A \(2 \mathrm{P}_0\)
- B \(8 \mathrm{P}_0\)
- C \(6 \mathrm{P}_0\)
- D \(16 \mathrm{P}_0\)
Answer & Solution
Correct Answer
(D) \(16 \mathrm{P}_0\)
Step-by-step Solution
Detailed explanation
For adiabatic expression, we have
\(
\begin{aligned}
& \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_1 \mathrm{~V}_1^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(8)^{\frac{4}{3}}=16 \\
& \therefore \mathrm{P}_2=16 \mathrm{P}_1=16 \mathrm{P}_0
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_1 \mathrm{~V}_1^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(8)^{\frac{4}{3}}=16 \\
& \therefore \mathrm{P}_2=16 \mathrm{P}_1=16 \mathrm{P}_0
\end{aligned}
\)
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