The following expression for the free energy of mixing of two idealgases initial

Question

The following expression for the free energy of mixing of two idealgases initially at the same temperature and pressure is given bythe equation:

Gmix = nRT [XAlnXA + XB ln XB]

The same procedure can be used to findGmix when the two gases are originally atdifferent pressures. Consider the system below. 1.000 mol of argon(Ar) is initially confined to the left side of a container, whichhas a volume V = 30.00 L, and 1.000 mol of helium (He) is initiallyconfined to the right side of a container, which has a volume V =10.00 L. The barrier between the two sides of the container isremoved and the gases are allowed to mix until equilibrium isreached. Temperature remains constant at T = 300.0K throughout theprocess. Find Gmix. You may assume that argon andhelium behave ideally. HINT: You cannot simply use the aboveequation to answer this question since it was derived assuming theinitial pressures of the two gases were equal to eachother.

Gmix = nRT [XAlnXA + XB ln XB]

The same procedure can be used to findGmix when the two gases are originally atdifferent pressures. Consider the system below. 1.000 mol of argon(Ar) is initially confined to the left side of a container, whichhas a volume V = 30.00 L, and 1.000 mol of helium (He) is initiallyconfined to the right side of a container, which has a volume V =10.00 L. The barrier between the two sides of the container isremoved and the gases are allowed to mix until equilibrium isreached. Temperature remains constant at T = 300.0K throughout theprocess. Find Gmix. You may assume that argon andhelium behave ideally. HINT: You cannot simply use the aboveequation to answer this question since it was derived assuming theinitial pressures of the two gases were equal to eachother.
The same procedure can be used to findGmix when the two gases are originally atdifferent pressures. Consider the system below. 1.000 mol of argon(Ar) is initially confined to the left side of a container, whichhas a volume V = 30.00 L, and 1.000 mol of helium (He) is initiallyconfined to the right side of a container, which has a volume V =10.00 L. The barrier between the two sides of the container isremoved and the gases are allowed to mix until equilibrium isreached. Temperature remains constant at T = 300.0K throughout theprocess. Find Gmix. You may assume that argon andhelium behave ideally. HINT: You cannot simply use the aboveequation to answer this question since it was derived assuming theinitial pressures of the two gases were equal to eachother.

Explanation / Answer

mixG in terms of partial pressures is mixG = nHeRTln(PHe /P) + nAr RTln (PAr / P) After removing the barrier the total volume would be equal to30L + 10L = 40L toal moles of gas = 1 + 1 = 2mol So pressure, P = nRT / V   =2mol*0.0821Latmmol^-1K^-1*300K / 40L = 1.2315atm Therefore, PHe = mole fraction of He * toalpressure = (1/2)*1.2315atm = 0.61575atm                PAr = mole fraction of Ar * toal pressure =(1/2)*1.2315atm = 0.61575atm mixG = 1mol*8.314Jmol^-1K^-1*300Kln(0.61575atm/1.2315atm) + 1mol*8.314Jmol^-1K^-1*300Kln (0.61575atm /1.2315atm)               = 2* {1mol*8.314Jmol^-1K^-1*300Kln(0.61575atm /1.2315atm)}               = -3457.8J

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