JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

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
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

Solution

For an ideal gas undergoing an isothermal process: $Δ U = 0$ and $Δ H = 0$ .

Work done in reversible isothermal expansion:

$w = - p_{1} V_{1} ln \frac{P _{1}}{P _{2}}$

$w = - (0.5 \times 1 0^{6}) (20 \times 1 0^{- 3}) ln \frac{0.5}{0.2}$

$w = - 1 0^{4} \times 2.303 \times (lo g 5 - lo g 2)$

$w = - 1 0^{4} \times 2.303 \times (0.6989 - 0.3010) \approx - 9.16 \times 1 0^{3}$ J $\approx - 9.1$ kJ

By the first law, $q = - w = + 9.1$ kJ.

$∴$ Option (1) is correct.

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