JEE Advanced 2016 · Paper 1 · Chemistry Q20

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ChemistrySingle correct+3 / -1Question 20 of 54

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings $(\Delta S_{surr})$ in $\mathrm{J\,K^{-1}}$ is

(1 L atm = 101.3 J)

5.763

1.013

−1.013

−5.763

Answer and worked solution

Official answer: C

Work. Irreversible expansion against 3.0 atm:
$w=-p_{ext}\Delta V=-3.0\times1.0=-3.0\text{ L atm}=-303.9\text{ J}.$

Heat. The process is isothermal for an ideal gas, so $\Delta U=0$ and $q_{sys}=+303.9$ J, taken from the surroundings.

Entropy of the surroundings.
$\Delta S_{surr}=\frac{-303.9}{300}=-1.013\text{ J K}^{-1}.$

(C)

Topics:Entropy First law and enthalpy

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