Thermodynamics Question 253
Question: The relation between U, P and V for an ideal gas in an adiabatic process is given by relation $ U=a+bPV. $ Find the value of adiabatic exponent $ ( \gamma ) $ of this gas.
Options:
A) $ \frac{b+1}{b} $
B) $ \frac{b+1}{a} $
C) $ \frac{a+1}{b} $
D) $ \frac{a}{a+b} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ U=a+bPV $ ..(i)
In adiabatic change, $ dU=-dW=\frac{nR}{\gamma -1}( T _{2}-T _{1} )=\frac{nR}{\gamma -1}( dT ) $
$ \Rightarrow U=\int _{{}}^{{}}{dU}=\frac{nR}{\gamma -1}\int _{{}}^{{}}{dT} $
$ orU=( \frac{nR}{\gamma -1} )T+a=\frac{PV}{\gamma -1}+a $ .(ii)
Where a is the constant of integration.
Comparing (1) and (2), we get $ b=\frac{1}{\gamma -1}\Rightarrow \gamma =\frac{b+1}{b}. $
BETA
