# NEET Physics Thermodynamics Questions Solved

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 5/3, the final temperature of gas will be -

(1) (T + 2.4)K

(2) (T – 2.4)K

(3) (T + 4)K

(4) (T – 4)K

(4) $W=\frac{R\left({T}_{i}-{T}_{f}\right)}{\gamma -1}$

$6R=\frac{R\left(T-{T}_{f}\right)}{\left(\frac{5}{3}-1\right)}$${T}_{f}=\left(T-4\right)K.$

Difficulty Level:

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