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