Question

One mole of an ideal gas at $300K$ in thermal contact with surrounding expands isothermally from $1.0Lto2.0L$ against a constant pressure of $3.0atm.$ In this process, the change in entropy of surroundings $\left(\Delta S_{\text{surr}}\right)$ in $J{K}^{-1}$ is:
$(Given:1Latm=101.3J)$

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

Answer 1

The correct option is C -1.013

We know,
From $1^{st}$ law of thermodynamics
$q_{sys}=\Delta U-w$
As the process is isothermal,so
$q_{sys}=0-[-P_{ext}.\Delta V]$
$=3.0atm\times (2.0L-1.0L)=3.0Latm=3.0\times 101.3J$.

Putting the relevent values in formula of entropy,
$\therefore \Delta S_{\text{surr}}=\frac{q_{surr}}{T}=-\frac{q_{sys}}{T}$

$=\frac{-3.0\times 101.3J}{300K}$

$=-1.013J/K$