Consider three gas connected gas containers as shown in the figure below. Consid

ID: 978352 • Letter: C

Question

Consider three gas connected gas containers as shown in the figure below.

CO2 (g) P = 2.13, V = 1.50L

H2(g) P = 0.861 atm, V = 1.00 L

Ar (g) p = 1.15atm, V = 2.00L

T = 298K

Assume, that you open the two stopcock so that the gases can flow freely.

Calculate the change of the chemical potential of the entire system due to mixing the gases.

Using the change of the chemical potential of gases show that the mixing of three gases is spontaneous

Determine the value of DmixH .

mix = ______________________

mixH = _____________________

Explanation / Answer

The chemical potential of a gas is given as

i = i0 + RTlnPi where i is the chemical potential of the i-th component in the system at pressure Pi and i0 is the standard chemical potential of the i-th component at 1 bar pressure.

Let us calculate the number of moles of each component as

Component

Initial pressure (atm)

Initial volume (L)

Number of moles = PV/RT

CO2 = 1

2.13

1.50

(2.13)(1.50)/RT = 3.195/RT

H2 = 2

0.861

1.00

(0.861)(1.00)/RT = 0.861/RT

Ar = 3

1.15

2.00

(1.15)(2.00)/RT = 2.30/RT

Since chemical potential is an additive property, the Gibb’s free energy of the system before mixing the gases is

Gi = n11 + n22 + n33 = [3.195/RT{10 + RTln(2.13)} + 0.861/RT{20 + RTln(0.861)} + 2.30/RT{30 + RTln(1.15)}]

or, Gi = {3.195ln(2.13) + 0.861ln(0.861) + 2.30ln(1.15)} + 1/RT(3.1951 + 0.8612 + 2.303) ……(1)

When the partition is removed, the total volume of the system is 4.5 L and the partial pressures of components 1, 2 and 3 are

P1’= (3.195/RT)(RT/4.00) = 0.79875 atm (since we assume ideal behaviour and P’= nRT/V where V is the total volume and n is the number of mole of the said component)

P2’= (0.861/RT)(RT/4.00) = 0.21525 atm and

P3’= (2.30/RT)(RT/4.00) = 0.575 atm.

As before, the Gibb’s free energy after mixing is

Gf = n11’+ n22’+ n33’

Hence, Gf = [(3.195/RT){10 + RTln(0.79875)} + (0.861/RT){20 + RTln(0.21525)} + (2.30/RT){30 + RTln(0.575)}]

or, Gf = {3.195ln(0.79875) + 0.861ln(0.21525) + 2.30ln(0.575)} + 1/RT(3.19510 + 0.86120 + 2.3030) …….(2)

The Gibb’s free energy of mixing is

Gm = Gf – Gi = {3.195ln(0.79875) + 0.861ln(0.21525) + 2.30ln(0.575)} – {3.195ln(2.13) + 0861ln(0.861) + 2.30ln(1.15)}

or, Gm = 3.195ln(0.79875/2.13) + 0.861ln(0.21525/0.861) + 2.30ln(0.575/1.15) = (- 3.134) + (-1.193) + (-1.594) = -5.921

The free energy change of mixing is -5.92 kJ (Gibb’s free energy change is always expressed in kJ)

Since, Gm < 0, the mixing process is spontaneous.

The entropy of mixing, Sm = -Gm/T where T is the absolute temperature.

The enthalpy of mixing can be calculated using the relation between G, H and S.

Gm = Hm – T.Sm ==è Hm = Gm + T.Sm = Gm + T. (-Gm/T) = 0