Question

The composition of air is approximately 75.52% N_2, 23.15% O_2, 1.28% Ar, and 0.046% CO_2. (a) Ignoring contributions from the other trace components of air, calculate the entropy of mixing of air when it is artificially prepared from its pure gas components. (b) At standard pressure, air liquefies at T = 78.8 K. At this temperature, the Ar and CO_2 have both frozen out of the air solution. The remaining liquid-phase air is stable between 58 K and 78.8 K. What is the entropy of mixing of liquid air (prepared from liquid O_2 and liquid N_2) at 70 K?

Answer 1

(a) dS = - R (x1 ln(x1) + x2 ln(x2) + x3ln(x3) + x4ln(x4)

dS = - 8.314 J/mol K * (0.7552*ln(0.7552) + 0.2315*ln(0.2315) + 0.0128*ln(0.0128) + 0.00046*ln(0.00046))

dS = 5.07 J/mol K

(b) Assume we have a mixture of 100 mol of air. If we Ar and CO2 frozen, we lost:

1.28 + 0.046 moles = 1.326 moles

And we have now:

100 - 1.326 = 98.674 moles of air.

In this case, the new molar compositions are:

X_N2 = 75.52 moles / 98.674 = 0.765

X_O2 = 1- X_N2 = 0.235

And the new entropy is:

dS = - R (x1 ln(x1) + x2 ln(x2)) = -8.314 J/mol L * (0.765*ln(0.765) + 0.235*ln(0.235))

dS = 4.53 J/mol K