Gibbs Free Energy
Why have I sometimes seen Gibbs Free Energy expressed as: dG=-TdS PdV
Surely this doesn’t equate to the normal:
dG=dH-TdS
Surely this doesn’t equate to the normal:
dG=dH-TdS
(edited 3 years ago)
The correct differential expression for Gibbs Free Energy is dG=Vdp−SdT. The version you quoted looks more like the differential form for internal energy, except that the signs are wrong (dU=TdS−pdV).
To avoid confusing the two, it helps to remember that Gibbs Free Energy is a natural function of pressure and temperature (hence dp and dT), which is why it is normally the most useful state function when these two variables are held constant.
You can show that this follows from definition of G = H - TS => dG = dH - TdS - SdT by substituting for dH = dU - pdV - Vdp, and then using the relation for dU I quoted above. Notice that I took the full derivative of G, using chain rule, whereas the version you quoted above (dG = dH - TdS) is only valid at constant temperature.
To avoid confusing the two, it helps to remember that Gibbs Free Energy is a natural function of pressure and temperature (hence dp and dT), which is why it is normally the most useful state function when these two variables are held constant.
You can show that this follows from definition of G = H - TS => dG = dH - TdS - SdT by substituting for dH = dU - pdV - Vdp, and then using the relation for dU I quoted above. Notice that I took the full derivative of G, using chain rule, whereas the version you quoted above (dG = dH - TdS) is only valid at constant temperature.
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