Question

Consider the process where nA mol of gas A initially at 1 bar pressure mix with nB mol of gas B also at 1 bar form 1 mol of a uniform mixture of A and B at a final total pressure of 1 bar, and all at constant temperature T. Assume that all gases behave ideally. a. Show that the entropy change, ΔmixSm, for this process is given by ΔmixSm = -xAR ln xA - xBR ln xB, where xA and xB are the mole fractions of A and B, respectively. b. What can you say about the value (especially the sign) of ΔMIXSm, and how does this correlate with the Second Law? c. Derive an expression for ΔmixGm for the conditions of this problem and comment on values (and signs) of its terms.

Answer 1

a) ΔSmix,m = 5.7628 J/mol.K

b) ΔSmix,m > 0:

• since xA <0 and xB < 0
• irreversible process

c) ΔGmix,m = 1717.314 J/mol

Step-by-step explanation:

mixture A and B:

∴ nA + nB = 1 mol = nt

⇒ nA = 0.5 mol

⇒ nB = 0.5 mol

∴ P = 1 bar =

∴ T = 25°C = 298K

assumptions:

• P,T→ constants

• gases beahave ideally

a) ΔSmix,m = - xAR LnxA - xBR LnxB

∴ xA = nA/nt = 0.5

∴ xB = nB/nt = 0.5

∴ R = 8.314 J/mol.K

⇒ ΔSmix,m = - (0.5)(8.314) Ln(0.5) - (0.5)(8.314) Ln(0.5)

⇒ ΔSmix,m = 2.8814 J/mol.K + 2.8814 J/mol.K

⇒ ΔSmix,m = 5.7628 J/mol.K

b) ΔSmix,m > 0:

• since xA <0 and xB < 0
• irreversible process

c) ΔGmix,m = ΔH - TΔS

∴ ΔH = 0.....P constant

⇒ ΔGmix,m = - TΔS = - (298K)(5.7628 J/mol.K)

⇒ ΔGmix,m = 1717.314 J/mol

∴ ΔGmix,m > 0, endothermic reaction