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Thermodynamics

Section: Chemistry

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Question : 25 of 42

Marks: +1, -0

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_{\text{surr}})

K^{-1}

-(1 L atm = 101.3 J)

[JEE Adv 2016 P1]

$5.763$
$1.013$
$-1.013$
$-5.763$

Solution:

From

1^{\text{st}}

law of thermodynamics

q_{\text{sys}} = \Delta U - w = 0 - [-P_{\text{ext}} \cdot \Delta V]

= (3.0 \, \text{atm} \times 2.0 \, \text{L} - 1.0 \, \text{L}) = 3.0 \, \text{L-atm}

\therefore \Delta S_{\text{surr}} = \frac{(q_{\text{rev}})_{\text{surr}}}{T} = \frac{q_{\text{sys}}}{T}

= -\frac{3.0 \times 101.3 \, \text{J}}{300 \, \text{K}}

= -1.013 \, \text{J/K}

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