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5. [20 (1) The variation of chemical potential, uThis is physical chemistry problem from Atkins

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1] The total Gibbs free energy of a two-component solution is given by the expression

G = n1G1 + n2G2 -------> 1

where

  • G is the total Gibbs energy of the system,
  • n1 and n2 is the number of moles of component 1 and 2
  • G1 is the partial molar Gibbs energy of component 1.

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

  • ui is the chemical potential of the ith component,
  • uio is the standard chemical potential of component i at 1 bar, and
  • Pi is the partial pressure of component i.

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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