JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Chemical Thermodynamics

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20.0 dm^{3}

of an ideal gas ' X ' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa . Which of the following option is correct?
(Given:

lo g 2 = 0.3010

and

lo g 5 = 0.6989

)

(A) $w = - 9.1 kJ, Δ U = 0, Δ H = 0, q = 9.1 kJ$ ✓ Correct
(B) $w = 9.1 J, Δ U = 9.1 J, Δ H = 0; q = 0$
(C) $w = + 4.1 kJ, Δ U = 0, Δ H = 0; q = - 4.1 kJ$
(D) $w = - 3.9 kJ, Δ U = 0, Δ H = 0; q = 3.9 kJ$

Answer: (A)

Solution

The internal energy and enthalpy of an ideal gas depend only on temperature, so for this isothermal change

Δ U = 0

and

Δ H = 0

. The reversible isothermal work is

w = - n R T ln \frac{P _{1}}{P _{2}} = - P_{1} V_{1} ln \frac{P _{1}}{P _{2}}

. Here

P_{1} V_{1} = (0.5 \times 1 0^{6} Pa) (20 \times 1 0^{- 3} m^{3}) = 10 kJ

and

\frac{P _{1}}{P _{2}} = \frac{0.5}{0.2} = 2.5

, with

ln 2.5 = 2.303 (lo g 5 - lo g 2) = 2.303 (0.3979) = 0.916

. Thus

w = - (10 kJ) (0.916) = - 9.1 kJ

, and since

Δ U = q + w = 0

q = + 9.1 kJ

. For this expansion

w = - 9.1

kJ,

Δ U = 0

Δ H = 0

and

q = 9.1

kJ.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Chemical Thermodynamics

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