What is the change in entropy when 0.7 m3 of CO2 and 0.3 m3 of N2 each at 1 bar and 25oC blend to form a homogenous mixture at the same conditions. Assume ideal gases.
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 → 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,
ΔS = nR ln(V2/V1) → 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 m3. Entropy change for this process is given by Equn.2
(b) Mixing two gases at identical conditions. (each at a volume of 1 m3). 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.
Moles of CO2 = 105 x 0.7 / (R x 298) = 234.9 / R
ΔS = 234.9 ln (1/0.7) = 83.78 J/oC
Moles of N2 = 105 x 0.3 / (R x 298) = 100.7 / R
ΔS = 100.7 ln (1/0.3) = 121.24 J/oC
Entropy change of the system= 83.78 + 121.24 = 205.02 J/oC
Last Modified on: 04-Feb-2022
Chemical Engineering Learning Resources - msubbu