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.
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 → 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,
Δ
S = nR ln(V2/V1) → 2Since 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.
For CO2:
Moles of CO2 = 105 x 0.7 / (R x 298) = 234.9 / R
Δ
S = 234.9 ln (1/0.7) = 83.78 J/oCFor N2:
Moles of N2 = 105 x 0.3 / (R x 298) = 100.7 / R
Δ
S = 100.7 ln (1/0.3) = 121.24 J/oCEntropy change of the system
= 83.78 + 121.24 = 205.02 J/oCLast Modified on: 04-Feb-2022
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