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 $ $JK^{-1}$
BETA
