NEET Solved Paper 2014 Question 17
Question: A monoatomic gas at a pressure p, having a volume V expands isothermally to a volume 2 V and then adiabatically to a volume 16 V. The final pressure of the gas is (take $ \gamma =\frac{5}{3} $ ) [AIPMT 2014]
Options:
A) 64p
B) 32p
C) $ \frac{p}{64} $
D) 16p
Show Answer
Answer:
Correct Answer: C
Solution:
For isothermal expansion $ pV=p’\times 2V $
$ [\therefore V’=2V] $ $ p’=\frac{p}{2} $ For adiabatic expansion, $ p{V^{\gamma }} $
= constant $ p’{V^{’\gamma }}=p’‘V’{{’}^{\gamma }} $ $ \frac{p}{2}{{[2V]}^{5/3}}=p’’{{[16V]}^{5/3}} $ $ p’’=\frac{p}{2}{{[ \frac{2V}{16V} ]}^{5/3}}=\frac{p}{2}{{[ \frac{1}{8} ]}^{5/3}} $ $ =\frac{p}{2}[0.03125] $
$ =0.0156p $
$ =p/64 $
BETA
