Why no Vdp?

[pivoxa15]

1. The problem statement, all variables and given/known data
The books often always present pdV but never Vdp. Why is that? surely it is present in expressions like dE or dG where G is the grand potential. Is it because we don't treat p as a variable? If so why not? For each of these function we can only have 3 variables? So having V and p both as variables would be a bit redundant? If so why?

[pivoxa15]

In a grand canonical ensemble is pressure constant for any system? Hence dp=0? The grand potential dosen't appear to have dp in it.

[pivoxa15]

Is it also because P is a state function so dP can always be expressed as other variables just like dU can. However temperture is also a state function? but it is treated as a variable.

[Andrew Mason]

1. The problem statement, all variables and given/known data
The books often always present pdV but never Vdp. Why is that? They mention VdP. VdP is related to the change in internal energy of the gas at constant volume. PdV is the work done by/on the gas. Together, VdP + PdV = nRdT. Since nCvdT is the heat flow at constant volume and nCpdT is the heat flow at constant pressure, and Cp-Cv = R, VdP + PdV = nRdT = n(Cp-Cv)dT

AM

[pivoxa15]

Why don't they have Vdp in the expression of the grand potential?

G = grand potential = -pV

So dG = -pdV - Vdp = -SdT -pdV - Ndu

So -Vdp = -SdT - Ndu?

or Vdp = SdT + Ndu which must be true by defintion would it? It's more convineint to use the right hand side instead of the left hand side?

[Andrew Mason]

Why don't they have Vdp in the expression of the grand potential?

G = grand potential = -pV

So dG = -pdV - Vdp = -SdT -pdV - Ndu

So -Vdp = -SdT - Ndu?

or Vdp = SdT + Ndu which must be true by defintion would it? It's more convineint to use the right hand side instead of the left hand side?Careful.

dQ = TdS
dU = nCvdT
dW = PdV

So the first law dQ = dU + dW can be written:

(1) TdS = nCvdT + PdV

Since d(PV) = VdP + PdV = d(nRT) = nRdT,

PdV = nRdT - VdP

So substituting this for PdV in (1):

(2) TdS = nCvdT + nRdT - VdP

(3) VdP = nCvdT + nRdT - TdS = nCpdT - TdS = nCpdT - dQ


So the first law can be written:

(4) dQ = nCpdT - VdP

AM

[pivoxa15]

You haven't disproved Vdp = SdT + Ndu have you?

[Andrew Mason]

You haven't disproved Vdp = SdT + Ndu have you?I am not sure what you mean by Ndu. I think it refers to added molecules.

From the Euler equation:

U = TS - PV

dU = d(TS) - d(PV) = (TdS + SdT) - (VdP + Pdv)

But the first law states that dU = dQ - PdV. Since dQ = TdS:

dU = TdS - PdV so

SdT - VdP = 0

AM

[pivoxa15]

u stands for the chemical potential constant. If you have dU=dQ-Pdv you have not accounted for the chemical potential of particles entering and leaving the system. If you did then dU=TdS-pdV+udN. So

We have
U = TS - PV + uN

dU = d(TS) - d(PV) + d(uN)= (TdS + SdT) - (VdP + Pdv) + (Ndu + udN)

We have
dU=TdS-pdV+udN

So SdT - Vdp + Ndu = 0
hence Vdp = SdT + Ndu

Although why is U = TS - PV + uN?

Or even why is U = TS - PV when not accounting for chemical potential?

Is it more the fact that we first know the expressions for dU then work out U from those expressions?