Thermodynamics and Thermochemistry 1 Question 20
23. One mole of an ideal gas at $300 \mathrm{~K}$ in thermal contact with surroundings expands isothermally from $1.0 \mathrm{~L}$ to $2.0 \mathrm{~L}$ against a constant pressure of $3.0 \mathrm{~atm}$.
In this process, the change in entropy of surroundings $\left(\Delta S_{\text {surr }}\right)$ in $\mathrm{JK}^{-1}$ is $(1 \mathrm{~L} \mathrm{~atm}=101.3 \mathrm{~J})$
(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, $\Delta E=0$
$$ \begin{gathered} Q=-W \ \quad-Q_{\text {irrev }}=W_{\text {irrev }}=p \Delta V=3(2-1)=3 \mathrm{~L} \mathrm{~atm} \end{gathered} $$
Also, $\Delta S_{\text {surr }}=\frac{Q_{\text {irrev }}}{T}=\frac{(-3 \times 101.3) \mathrm{J}}{300 \mathrm{~K}}=-\frac{303.9}{300}=-1.013 \mathrm{JK}^{-1}$
