Question

Suppose now that argon is added to the mixture in Exercise 6 to bring the composition closer to real air, with mole fractions 0.780, 0.210, and 0.0096, respectively. What is the additional change in molar Gibbs energy and entropy? Is the mixing spontaneous?

Answer 1

Solution:

Calculate the additional change in molar Gibbs energy and entropy

• Find the molar Gibbs energy

By the formula ΔGmix=RT(xAln⁡xA+xBln⁡xB+xCln⁡xC)

We have:

R=8.31J/mol⋅K
T=298K
xA=xN2=0.78
xB=xO2=0.2
xC=xAr=0.0096

Therefore, ΔGmix=8.31×298(0.78ln⁡0.78+0.2ln⁡0.2+0.0096ln⁡0.0096)=−1401.631J/mol

Therefore, the additional change in molar Gibbs energy is -1401.631 J/mol

• Find the molar entropy

From the formula ΔSmix=−ΔGmixT

Since ΔGmix=−1401.631J/mol and T=298K

Then ΔSmix=−(−1401.631)298=4.70J/mol

According to ΔG<0 and ΔS>0, it is spontaneous.

Therefore, the molar entropy is 4.70 J/mol and it is spontaneous.

a. Therefore, the additional change in molar Gibbs energy is -1401.631 J/mol

b. Therefore, the molar entropy is 4.70 J/mol and it is spontaneous.