One mole of an ideal gas expands isothermally and reversibly from $10 \mathrm{dm}^{3}$ to $20 \mathrm{dm}^{3}$ at $300 \mathrm{~K}$. $\Delta \mathrm{U}$, q and work done in the process respectively are: Given: $\mathrm{R}=8.3 \mathrm{JK}^{-1}$ and mol$^{-1}$ In $10=2.3$ $\log 2=0.30$ $\log 3=0.48$

(D) $0, 1.718 \mathrm{~kJ}, -1.718 \mathrm{~kJ}$

(A) $0,21.84 \mathrm{~kJ},-1.26 \mathrm{~kJ}$

(B) $0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}$

(C) $0,21.84 \mathrm{~kJ}, 21,84 \mathrm{~kJ}$

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Chemical Thermodynamics

One mole of an ideal gas expands isothermally and reversibly from

10 dm^{3}

20 dm^{3}

300 K

Δ U

, q and work done in the process respectively are:
Given:

R = 8.3 JK^{- 1}

and mol

^{- 1}

10 = 2.3

lo g 2 = 0.30

lo g 3 = 0.48

Answer: (D)

For isothermal expansion of an ideal gas the internal energy is unchanged,

Δ U = 0

. The reversible work is

w = - n R T ln \frac{V _{2}}{V _{1}} = - (1) (8.3) (300) ln 2

; using

ln 2 = 2.3 \times 0.30 = 0.69

w = - (8.3) (300) (0.69) = - 1718 J = - 1.718 kJ

. Because

Δ U = q + w = 0

, the heat absorbed is

q = - w = + 1.718 kJ

. Thus

Δ U = 0

q = + 1.718 kJ

and

w = - 1.718 kJ

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option