Question

At a constant pressure, the entropy of mixing of 1 mole of oxygen with 3 moles of nitrogen at 25 $^\circ$C is _ _ _ J K$^{-1}$. (rounded off to two decimal places)

Hint:

For ideal gas mixing, use mole fractions in $\Delta S_{\text{mix}} = -R\sum n_i\ln x_i$

Answer 1

Correct Answer: 18.7

Step 1: Use entropy of mixing formula.
For ideal gases:
\[ \Delta S_{\text{mix}} = -R(n_1\ln x_1+n_2\ln x_2) \]

Step 2: Calculate total moles.
\[ n_{\text{total}} = 1 + 3 = 4 \]

Step 3: Find mole fractions.
\[ x_{\text{O}_2} = \frac{1}{4}=0.25 \]
\[ x_{\text{N}_2} = \frac{3}{4}=0.75 \]

Step 4: Substitute values.
\[ \Delta S_{\text{mix}} = -8.314[1\ln(0.25)+3\ln(0.75)] \]

Step 5: Evaluate logarithmic terms.
\[ \ln(0.25)=-1.3863 \]
\[ 3\ln(0.75)=3(-0.2877)=-0.8631 \]

Step 6: Calculate entropy change.
\[ \Delta S_{\text{mix}}=-8.314[-1.3863-0.8631] \]
\[ \Delta S_{\text{mix}}=18.70 \text{ J K}^{-1} \]

Step 7: Conclusion.
\[ \boxed{18.70 \text{ J K}^{-1}} \]