MHT CET · Physics · Thermodynamics
' \(\mathrm{V}\) ' cc volume of a gas having \(\mathrm{Y}=\frac{5}{2}\) is suddenly compressed to \(\left(\frac{V}{4}\right)\) c.c. The initial pressure of the gas is P. The final pressure of the gas will be
- A \(\frac{\mathrm{P}}{32}\)
- B \(16 \mathrm{P}\)
- C \(\frac{\mathrm{P}}{16}\)
- D \(32 \mathrm{P}\)
Answer & Solution
Correct Answer
(D) \(32 \mathrm{P}\)
Step-by-step Solution
Detailed explanation
Here, \(\mathrm{V}_1=\mathrm{V}\) cc, \(\mathrm{V}_2=\frac{\mathrm{v}}{4} \mathrm{cc}, \mathrm{Y}=\frac{5}{2}, \mathrm{P}_1=\mathrm{P}\)
for adiabatic change,
\(\begin{aligned} & \mathrm{P}_2 \mathrm{~V}_2=\mathrm{P}_1 \mathrm{~V}_1^{\mathrm{Y}} \\ & \therefore \mathrm{P}_1=\mathrm{P}\left\{\frac{\mathrm{V}}{\left(\frac{\mathrm{V}}{4}\right)}\right\}^Y \\ & \mathrm{P}_2=\mathrm{P}(4)^{5 / 2}=\mathrm{P}\left(2^5\right)=32 \mathrm{P}\end{aligned}\)
for adiabatic change,
\(\begin{aligned} & \mathrm{P}_2 \mathrm{~V}_2=\mathrm{P}_1 \mathrm{~V}_1^{\mathrm{Y}} \\ & \therefore \mathrm{P}_1=\mathrm{P}\left\{\frac{\mathrm{V}}{\left(\frac{\mathrm{V}}{4}\right)}\right\}^Y \\ & \mathrm{P}_2=\mathrm{P}(4)^{5 / 2}=\mathrm{P}\left(2^5\right)=32 \mathrm{P}\end{aligned}\)
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