Question

2.00mole of nitrogen (N2) gas and 2.00mole of argon (Ar) gas are in separate, equal-sized, insulated containers at the same temperature. The containers are then connected and the gases (assumed ideal) allowed to mix. PART A: What is the change in entropy of the system? PART B: What is the change in entropy of the environment? PART C: Repeat part A but assume one container is twice as large as the other.

Answer 1

n1 = 8 mol
n2 = 8 mol
T1 = T2 = Ti
V1 = V2 = V

A)
Determine initial pressure,
P1V1/n1T1 = P2V2/n2T2
P1 = P2

Final pressure,
P1V1/n1T1 = PfVf/nfTf
P1V1/n = PfVf/nf
P1 = Pf

Use Tds equation
Tds = du - vdp
?S = n*cln(Tf/T) - n*R ln (Pf/P)

Entropy change of Nitrogen
?S_N2 = n*cln(Tf/T1) - n*R ln (Pf/P), T2 = T1
?S_N2 = -8*8.314*ln1
?S_N2 = 0 J/K

Entropy change of Argon
?S_Ar = 0 J/K

Entropy change of system,
?S = ?S_N2+?S_Ar
?S = 0 J/K

B)
?S_environtment = Q/T, Q = 0 (insulated)
?S_environtment = 0

C)
V1 = 2V
V2 = V

Determine initial pressure,
P1V1/n1T1 = P2V2/n2T2
P1*2V/n1T1 = P2V/n2T2
2P1 = P2

Final pressure,
P1V1/n1T1 = PfVf/nfTf
P1V1/n1 = PfVf/nf
2P1/ = 3Pf/2
Pf/P1 =4/3 and Pf/P2 = 2/3

Entropy change of Nitrogen
?S_N2 = - n*R ln (Pf/P1)
?S_N2 = - 8*8.314*ln4/3
?S_N2 = -19.13431 J/K

Entropy change of Argon
?S_Ar = - n*R ln (Pf/P2)
?S_Ar = 26.9683 J/K

Entropy change of system,
?S = ?S_N2 + ?S_Ar
?S = -19.13431 + 26.9683
?S = 7.83399 J/K