Question

Show that { d(G/T) / d(1/T) }p = H

Show that { d(G/T) / d(1/T) }p = H

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Answer #1

let

dG = -S*dT + VdP

by definition

since this is constant p, then

dG = -S*dT

so...

(dG/dT)´= -S

for Temperature:

(d(G/T) / dT)P = (dG/dT)p / T - G/T^2 =( -S*T - G ) /T^2 = -H/T^2

which is also:

[d(G/T) / d(1/T) ]p = H

and

[d(dG/T) / d(1/T) ] p = d H

so

[d(dG/T) / d(T) ] p = dH^2/(T^2)

some algebra on right side:

d(G/T) dT = -1 / (1/T^2)

[d(dG/T) / d(T) ] p =-T^2 ( 1/T*(dG/dT)p - 1/(T^2) *G)

[d(dG/T) / d(T) ] p =-T*(dG/dT)p + G

[d(dG/T) / d(T) ] p =-T*(-S) + G

[d(dG/T) / d(T) ] p =G + TS

[d(dG/T) / d(T) ] p =H

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