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One mole of an ideal gas at an initial temperature of T K does 6R joules 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
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- (T – 4) K
- (T + 2.4) K
- (T – 2.4) K
- (T + 4) K
Correct Option: A
| T1 = T , W = 6R joules , γ = | ||
| 3 |
| W = | = | ||
| γ - 1 | γ - 1 |
| = | ||
| γ - 1 |
| n = 1 , T1 = T ⇒ | = 6R | |
| (5 / 3) - 1 |
⇒ T2 = ( T–4)K
