MHT CET · Physics · Thermodynamics
A monoatomic gas is suddenly compressed to \((1 / 8)^{\text {th }}\) of its initial volume adiabatically. The ratio of the final pressure to initial pressure of the gas is ( \(\gamma=5 / 3\) )
- A 32
- B 8
- C \(\frac{40}{3}\)
- D \(\frac{24}{5}\)
Answer & Solution
Correct Answer
(A) 32
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{8}, \gamma=\frac{5}{3}\)
For adiabatic process,
\(\begin{aligned}& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(8)^{5 / 3}=(2)^5=32\end{aligned}\)
For adiabatic process,
\(\begin{aligned}& \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(8)^{5 / 3}=(2)^5=32\end{aligned}\)
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