JEE Main 2025 Apr 4 Shift 1, Chemistry Q5: Basic Concepts of Thermodynamic

JEE Main2025Apr 4, Shift 1Chemistry

One mole of an ideal gas expands isothermally and reversibly from 10 dm $^{3}$ to 20 dm $^{3}$ at 300 K. $Δ U$ , $q$ and work done in the process respectively are :

Given : $R = 8.3$ JK $^{- 1}$ mol $^{- 1}$

$ln 10 = 2.3$

$lo g 2 = 0.30$

$lo g 3 = 0.48$

A
0, 21.84 kJ, $-$ 1.26 kJ
B
0, $-$ 17.18 kJ, 1.718 J
C
0, 21.84 kJ, 21.84 kJ
D
0, 1.718 kJ, $-$ 1.718 kJ

Solution

For an isothermal process on an ideal gas, internal energy depends only on temperature, so $Δ U = 0$ .

The reversible work done on the gas is $w = - n R T ln \frac{V _{2}}{V _{1}}$ .

Substituting $n = 1$ , $R = 8.3$ J K $^{- 1}$ mol $^{- 1}$ , $T = 300$ K, $V_{2} / V_{1} = 2$ :

$w = - (1) (8.3) (300) ln 2 = - 8.3 \times 300 \times 0.6931 \approx - 1718$ J $= - 1.718$ kJ.

From the first law $Δ U = q + w$ with $Δ U = 0$ , we get $q = - w = + 1.718$ kJ.

$∴ Δ U = 0$ , $q = 1.718$ kJ, $w = - 1.718$ kJ.

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JEE Main 2025 Apr 4 Shift 1, Chemistry Q5: Basic Concepts of Thermodynamic

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