Question

What is the entropy change when 1 mole oxygen gas expands isothermally and reversibly from an initial volume of 10 L to 100 L at 300 K?

• A. $19.14J{K}^{-1}$
• B. $109.12J{K}^{-1}$
• C. $29.12J{K}^{-1}$
• D. $10J{K}^{-1}$

Answer 1

The correct option is A $19.14J{K}^{-1}$

The change in entropy for the isothermal reversible process is given as :
$△S=2.303nRlog\frac{V_2}{V_1}$
where $V_2$ is the final volume and $V_1$ is the initial volume
Putting all the values we get :
$△S=2.303\times 1\times 8.314\times log\frac{100}{10}=19.14J{K}^{-1}$


Theory:

Calculation of $\Delta S_{universe}$ for an Isothermal Process:
For an isothermal process $dT=0$
Isothermal processes can happen reversibly and irreversibly. We have to calculate the change in entropy of the system when it is moving from state A to state B.

For a reversible isothermal process change in entropy is obtained as,
Change in entropy for system is given as,

$\Delta S_{sys}=n{c}_{v,m}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$
For an isothermal process $T_1=T_2$ which makes the first term in the equation for change in entropy as zero.

Thus,
$\Delta S_{sys}=nRln\frac{V_2}{V_1}$