Thermodynamics and Thermochemistry - Result Question 23

23. 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 $\left(\Delta S _{\text {surr }}\right)$ in $JK^{-1}$ is $(1 $ $L$ $ atm=101.3$ $ J)$

(2016 Adv.)

(a) $5.763$

(b) $1.013$

(c) $-1.013$

(d) $-5.763$

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Answer:

Correct Answer: 23. (c)

Solution:

By first law, $\Delta E=Q+W$

For isothermal expansion, $\Delta E=0$

$\because \quad \quad Q=-W$

$\quad-Q _{\text {irrev }}=W _{\text {irrev }}=p \Delta V=3(2-1)=3 {~L} {~atm}$

Also, $\Delta S _{\text {surr }}=\dfrac{Q _{\text {irrev }}}{T}=\dfrac{(-3 \times 101.3) J}{300 K}=-\dfrac{303.9}{300}=-1.013 $ $JK^{-1}$