###
Entropy Change in Mixing of Two Gases

Home ->
Solved Problems
->
Thermodynamics ->

What is the change in entropy when 0.7 m^{3} of CO_{2} and 0.3 m^{3} of N_{2} each at 1 bar and 25^{o}C blend to form a homogenous mixture at the same conditions. Assume ideal gases.

Calculations:

Since there is no change in temperature after mixing, internal energy change of individual gases is zero.

For a reversible process, dU = TdS - PdV

And for constant internal energy process dU = 0. Therefore TdS = PdV

dS = PdV/T à 1

For an ideal gas, PV = nRT

Therefore, P/T = nR/V

Substituting for P/T in Equn.1,

dS = nR dV/V

Integrating the above equation,

DS = nR ln(V_{2}/V_{1}) à 2

Since mixing is an irreversible process, we can calculate the entropy change of individual gases by assuming the equivalent reversible process given below:

(a) Reversible expansion of gas from its initial volume to the final volume of 1 m^{3}. Entropy change for this process is given by Equn.2

(b) Mixing two gases at identical conditions. (each at a volume of 1 m^{3}). Since there is no change in thermodynamic conditions in this process, entropy change is zero for this step. (Entropy is a state function)

We shall calculate the entropy change of individual gases and total entropy change of the system is the sum of the entropy changes of two gases.

For CO_{2}:

Moles of CO_{2} = 10^{5} x 0.7 / (R x 298) = 234.9 / R

DS = 234.9 ln (1/0.7) = 83.78 J/^{o}C

For N_{2}:

Moles of N_{2} = 10^{5} x 0.3 / (R x 298) = 100.7 / R

DS = 100.7 ln (1/0.3) = 121.24 J/^{o}C

Entropy change of the system = 83.78 +
121.24 = 205.02 J/^{o}C

HOME

Last Modified on: 11-Sep-2014

Chemical Engineering Learning Resources - msubbu

e-mail: msubbu.in[AT]gmail.com

Web: http://www.msubbu.in