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  4. 1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27° C. The work done is 3 kJ mol -1. The final temperature of the gas is K (Nearest integer). Given C V =20 J mol -1 K -1

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Q. 1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} C$. The work done is $3\, kJ \,mol ^{-1}$. The final temperature of the gas is _______ $K$ (Nearest integer). Given $C_{ V }=20 \,J\, mol ^{-1} K ^{-1}$

JEE MainJEE Main 2023Thermodynamics

Solution:

$ q =0 $
$\Delta U = w $
$ 1 \times 20 \times\left[ T _2-300\right]=-3000 $
$ T _2-300=-150 $
$ T _2=150 \,K $