Question
If the ratio of specific heat of a gas at constant pressure to that at constant volume is $$\gamma ,$$ the change in internal energy of a mass of gas when the volume changes from $$V$$ to $$2V$$ at constant pressure $$p$$ is
A.
$$\frac{R}{{\left( {\gamma - 1} \right)}}$$
B.
$$pV$$
C.
$$\frac{{pV}}{{\left( {\gamma - 1} \right)}}$$
D.
$$\frac{{\gamma pV}}{{\left( {\gamma - 1} \right)}}$$
Answer :
$$\frac{{pV}}{{\left( {\gamma - 1} \right)}}$$
Solution :
Change in internal energy of a gas having atomicity $$\gamma $$ is given by
$$\Delta U = \frac{1}{{\left( {\gamma - 1} \right)}}\left( {{p_2}{V_2} - {p_1}{V_1}} \right)$$
Given, $${V_1} = V,{V_2} = 2V$$
So, $$\Delta U = \frac{1}{{\gamma - 1}}\left[ {p \times 2V - p \times V} \right]$$
$$\eqalign{
& = \frac{1}{{\gamma - 1}} \times pV \cr
& = \frac{{pV}}{{\gamma - 1}} \cr} $$
NOTE
The internal energy of an ideal gas depends only on its absolute temperature $$\left( T \right)$$ and is directly proportional to $$T.$$