Question

One mole of an ideal gas at $300K$ in thermal contact with surroundings expands isothermally from $1.0L$ to $2.0L$ against a constant pressure of $3.0atm$.

In this process, the change in entropy of surroundings $\left(\Delta S_{surr}\right)$ in $J{K}^{-1}$ is $(1L$ atm $=103J)$:

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

Answer 1

The correct option is C $-1.013$

$\delta S_{surr}=\frac{\delta q}{T}$
Now $\delta q=\delta u+\delta w$
Since isothermal process
$\delta u=0$
Hence
$\delta q=\delta w$
Now,
Heat gained by the surrounding is heat lost by the system.
Hence
$\delta q_{system}=-\delta q_{surr}$
Therefore,
$\delta S_{surr}=\frac{\delta q_{surr}}{T}$
$=\frac{-\delta q_{sys}}{T}$$=\frac{-\delta w}{T}$
$=\frac{-P\left(dV\right)}{T}$
$=\frac{-3(2-1)}{300}\times 103$ Joules
$=-1.03$ joules.