First slide

Different type of processes

Question

One mole of an ideal gas at an initial temperature of T K does 6 R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $\frac{5}{3}$ ,the final temperature of gas will be :

Easy

A

(T - 2.4)K
B

(T+ 4)K
C

(T - 4)K
D

(T + 2.4)K

Solution

$Q = 0 \Rightarrow ∆ U = - 6 R = \frac{(1) R ∆ T}{γ - 1} = \frac{3}{2} R ∆ T \Rightarrow ∆ T = - 4 \Rightarrow T_{f} = (T - 4) K$

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