Question

Hi guys please show how to derive the following with apropriate steps thanx!

(a) Derive an equation for the change in free energy, AGne when ideal gasses with the same temperature and pressure, are mixed. (b) Also derive and expression for Hmivrin ASmning and AV, et en mixing

Answer 1

a)))

Dependence of Gibbs energy on mixture composition is and at constant T and P, systems tend towards a lower Gibbs energy The simplest example of mixing: Consider IDEAL gas A and gas B, both in separate containers at pressure p at temperature T. The chemical potentials are at their “pure values” at this point. Gibbs energy is

G = n A µ A + n B µ B

G = n A (µ0 A + RT ln(P/P0)) + n B (µ0B + RT ln(P/P0))

We can simplify things by letting p denote the pressure relative to po, writing

G = n A (µ0 A + RT lnp) + n B (µ0B + RT lnp)

After mixing, the partial pressures of the gases are pA and pB, where the total pressure is p = pA + pB. The total Gibbs energy is then

G = n A( µ0A + RT ln p ( A )) + n B (µ0B + RT ln p ( B ))

The differ ence i n Gibbs energies, Gf - Gi, is the Gibbs energy of mixing

∆ mix G = n A RT ln (pA/ p) + n B RT ln (p B/ p )

We use mole fractions, replacing nJ with xJ n

∆ mix G = nRT (x A ln x A + x B ln x B )

Since the mole fractions are never greater than 1, the ln terms are negative, and ∆mixG < 0 This allows is to conclude that mixing processes are spontaneous, and gases mix spontaneously in all proportions

b)))

dG/dT)P=−S

This means that differentiating Equation of DelGmix at constant pressure with respect to temperature will give an expression for the effect that mixing has on the entropy of a solution. We see that

(dGmix / dT)P=nR(x A ln x A + x B ln x B ) = −ΔSmix

ΔG=ΔH−TΔS

ΔHmix=ΔGmix+TΔSmix

Plugging in our expressions for ΔmixG and ΔmixS , we get

ΔmixH=nRT(xAlnxA+xBlnxB)+T[−nR(xAlnxA+xBlnxB)]=0