Question

Calculate the change in entropy of 2 moles of an ideal gas upon isothermal expansion at 243.6 K from 20 litre until the pressure becomes 1 atm.

• A. $-4.5cal/K$
• B. $-1.25cal/K$
• C. $1.25cal/K$
• D. $2.77cal/K$

Answer 1

The correct option is D $2.77cal/K$

Given: $P_1=1atm$
Number of moles = 2 $mol$
T = 243.6 $K$
$V_1$ = 20 $L$
We know, PV = nRT
Or, $P=\frac{nRT}{V}$
$P_1=\frac{2\times 0.0821\times 243.6}{20}$
$P_1=2atm$

Change in entropy during isothermal process is:
$△S=nRln\frac{P_1}{P_2}$
where $P_2$ = final pressure
$P_1$ = Initial pressure
R = 2 $cal/(K.mol)$
$△S=2\times 2\times ln\frac{2}{1}$
$\Delta S=2.77cal/K$


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 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,
$\Delta S_{sys}=nRln\frac{V_2}{V_1}$
as change is isothermal,
$\Delta S_{sys}=nRln\frac{P_1}{P_2}$