Question

what is the change in entropy, enthalpy and gibbs free energy when 1 L of ideal gas i, 3 L of ideal gas j and 4 L of ideal gas k, each at 1 atm and room temperature (298.15K) blend to form a gas mixture at the same conditions?

Answer 1

The gibbs free energy of mixing of 3 species is,

ΔG_mix= nRT(x₁lnx₁+x₂lnx₂+x₃lnx₃)

R is the gas constant (8.314 Jmol^-1K^-1), T is the temperature and x₁,x₂, and x₃ are the mole fractions of gases.n is the total moles.

From ideal gas law,

PV =n₁RT

n₁=PV/RT=(1 atm*1 L)/(0.0821 atm L mol^-1K^-1*298.15 K) = 0.04 mol

n₂=PV/RT=(1 atm*3 L)/(0.0821 atm L mol^-1K^-1*298.15 K) = 0.1225 mol

n₃=PV/RT=(1 atm*4 L)/(0.0821 atm L mol^-1K^-1*298.15 K) = 0.1634 mol

Total moles= 0.1634 mol+0.1225 mol+0.04 mol = 0.3259 mol

Mole fraction = moles /total moles

x₁ = 0.04 mol/ 0.3259 mol = 0.1227

x₂ = 0.1225 mol/ 0.3259 mol = 0.3758

x₃= 0.1634 mol/ 0.3259 mol = 0.5013

Calculate the value of ΔG_mix as follows:

ΔG_mix= 0.3259 mol *8.314 Jmol^-1K^-1*298.15 K (0.1227 ln0.1227 + 0.3758 ln 0.3758 + 0.5013 ln 0.5013

=0.3259 mol *8.314 Jmol^-1K^-1*298.15 K (-0.2574 + -0.3677 + -0.3461)

=-784.64 J

Therefore, the gibbs free energy of mixing is -784.64 J.

The entropy of mixing of 3 species is,

ΔS_mix= -nR(x₁lnx₁+x₂lnx₂+x₃lnx₃)

Calculate the value of ΔS_mix as follows:

ΔS_mix= -(0.3259 mol *8.314 Jmol^-1K^-1 (0.1227 ln0.1227 + 0.3758 ln 0.3758 + 0.5013 ln 0.5013))

= -(0.3259 mol *8.314 Jmol^-1K^-1 (-0.2574 + -0.3677 + -0.3461))

= 2.63 J K^-1

Therefore, the entropy of mixing is 2.63 J  K^-1

The enthalpy on mixing of ideal gas is zero.

As the molecules of ideal gas spread out, and there are no interactions between one another on mixing with each other, and therefore no heat is absorbed or released resulting in zero for enthalpy on mixing.

Therefore, ΔH_mix is 0.