Question

A rigid container is divided into two compartments of equal volume by a partition. One compartment contains 1 mole of ideal gas A at 1 atm, and the other contains 1 mole of ideal gas B at 1 atm. Calculate the increase in entropy which occurs when the partition between the two compartments is removed. If the frst compartment had contained 2 moles of ideal gas A, what would have been the increase in entropy when the partition was removed? Calculate the corresponding increases in entropy in each of the preceding two situations if both compartments had contained ideal gas A.

Answer 1

Answer:

The entropy is $R\,ln\frac{32}{27}$.

Explanation:

When the partition was removed,

The volume occupied by one mole of "A" doubles.

$\Delta S_{A}=R\,ln2$

The volume occupied by one mole of "B" doubles.

$\Delta S_{B}=R\,ln2$

$\Delta S_{total}=\Delta S_{A}+\Delta S_{B}=R\,ln2+R\,ln2=R\,ln4$

The volume occupied by two moles of "A" doubles.

$\Delta S_{A}=2R\,ln2$

The volume occupied by one mole of "B" doubles.

$\Delta S_{B}=R\,ln2$

$\Delta S_{total}=\Delta S_{A}+\Delta S_{B}=2R\,ln2+R\,ln2=R\,ln8$

At no change,

$\Delta S_{total}=0$

Double the volume of the 2 moles of "A"

$\Delta S_{total}=2R\,ln2$

Remove the partition

$\Delta S_{total}=0$

Decrease the volume of the 3 moles of "A" by the factor $\frac{2}{3}$

$\Delta\,S=3R\,ln\frac{2}{3}$

$\Delta\,S_{total}=2Rln\,2+3R\,ln\frac{2}{3}=R\,ln \frac{32}{27}$

Therefore, The entropy is $R\,ln\frac{32}{27}$.