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

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Chemical Thermodynamics

1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of

2 7^{\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}

Answer: 150

The expansion is adiabatic, so no heat crosses the boundary and the whole internal-energy change equals the work. The gas does

3 kJ

of work in expanding, so its internal energy falls by that amount,

Δ U = - 3000 J

. For an ideal gas this change is

n C_{v} Δ T

, hence

1 \times 20 \times (T_{2} - 300) = - 3000,

T_{2} - 300 = - 150.

Starting from

2 7^{\circ} C = 300 K

, the final temperature of the gas is

150 K

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer