One mole of an ideal gas at an initial temperature of T K does 6R joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be
- (T + 2.4) K
- (T - 2.4) K
- (T + 4) K
- (T - 4) K
Answer
(4) (T - 4) K
Solution
For an adiabatic process the work done by the gas is W = nR(T1 - T2) - 1, so
6R = 1 R(T - T2) 53 - 1 = 32R(T - T2)
T - T2 = 4
The gas cools as it does work, hence
T2 = (T - 4) K
Answer: (4)