Thermodynamics and Thermochemistry - Result Question 98
####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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Solution:
- By first law, $\Delta E=Q+W$
For isothermal expansion, $\Delta E=0$
$$ \begin{gathered} Q=-W \ \quad-Q _{\text {irrev }}=W _{\text {irrev }}=p \Delta V=3(2-1)=3 L atm \end{gathered} $$
Also, $\Delta S _{\text {surr }}=\frac{Q _{\text {irrev }}}{T}=\frac{(-3 \times 101.3) J}{300 K}=-\frac{303.9}{300}=-1.013 JK^{-1}$
