5. [20 (1) The variation of chemical potential, u of a perfect gas with pressure
ID: 1005023 • Letter: 5
Question
5. [20 (1) The variation of chemical potential, u of a perfect gas with pressure can be expressed as follows: Au RT ln Derive the below equation for Gibbs energy of mixing, when two gases of A and B are mixed at constant pressure, p and temperature, T. (2) Calculate the molar Gibbs energy of mixing when the two major components of air (nitrogen and oxygen) are mixed to form air at 298 K and 1 atm. The mole fractions of N2 and O2 in the mixture are 0.78 and 0.22, respectively. Is the mixing spontaneous? (3) Suppose now that argon is added to the mixture above to bring the composition closer to real air, with mole fractions of N2, O2, and Ar in the mixture are 0.78, 0.21, and 0.01, respectively. What is the additional change in molar Gibbs energy at 298 K? Is the mixing spontaneous? T hua Gas constant, R 8.314 J/mol. K: 1 J 1 a m 1 atm 1.01325 x 103 PaExplanation / Answer
1] The total Gibbs free energy of a two-component solution is given by the expression
G = n1G1 + n2G2 -------> 1
where
The molar Gibbs energy of an ideal gas can be found using the equation
G = Go + RT ln P
where dGo is the standard molar Gibbs energy of the gas at 1 bar, and P is the pressure of the system. In a mixture of ideal gases, we find that the system’s partial molar Gibbs energy is equivalent to its chemical potential, or that
G = ui
ui = uio + RTln P -----------------> 2
where
Now pretend we have two gases at the same temperature and pressure, gas 1 and gas 2. The Gibbs energy of the system before the gases are mixed is given by Equation 1 which can be combined with Equation 2 to give the expression
Ginitial = n1[u1+RTlnP] + n2[u2RTlnP]
If gas 1 and gas 2 are then mixed together, they will each exert a partial pressure on the total system, P1 and P2, so that P1+ P2= P. This means that the final Gibbs energy of the final solution can be found using the equation.
Gfinal = n1[u1+RTlnP1] + n2[u2RTlnP2]
delat G = Gfinal - Ginitial
= n1RTln P1/P + n2RTln P2/P
= n1RTln x1 + n2RTln x2
since P1 = x1 * P from Raoults law
This equation can be simplified further by knowing that the mole fraction of a component is equal to the number of moles of that component over the total moles of the system, or x1 = n1/n
delta G = nRT [x1ln x1] + nRT [ x2ln x2 ]
This expression gives us the effect that mixing has on the Gibbs free energy of a solution. Since x1 and x2 are mole fractions, and therefore less than 1, we can conclude that delta G will be a negative number. This is consistent with the idea that gases mix spontaneously at constant pressure and temperature.
2] given ,
Mole fractions of N2 and O2 ===> 0.78 and 0.22
T = 298k and R = 8.314
Total moles take 1 as not given
substituting in obtained formula
delta G = 1*8.314*298 [[0.78ln 0.78] + [0.22ln 0.22]]
delta G = -1305 J
since delta G is negative it is spontaneous process
3] delta G = nRT [[x1ln x1] + [ x2ln x2 ] +[x3lnx3] ]
total moles = x1+x2+x3 = 1
substitute all
delta G = -1406 J
delta G is negative so it is a spontaneous process
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