Question

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0L to 2.0L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings $\left(\Delta S\right)in{T}^{-1}$ is (1 L atm = 101.3J)

• A. 5.763
• B. 1.013
• C. -1.013
• D. -5.763

Answer 1

Answer: (C)

-1.013

By first law, $\Delta E=Q+W$
For isothermal expansion, $\Delta E=0$
$\therefore Q=-W$
$-Q_{irrev}=W_{irrev}=p\Delta V=3(2-1)=3Latm$
$Also,\Delta S_{surr}=\frac{Q_{irrev}}{T}=\frac{(-3\times 101.3)J}{300K}=-\frac{303.9}{300}=-1.013J{K}^{-1}$