# Why don't we include VdP in U equation ?

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1. Mar 19, 2016

### Joseph95

1. The problem statement, all variables and given/known data
I was trying to obtain Tds=dh-Vdp by differentiating the internal energy equation U=Q-PV and doing some arrangements but at the end I couldn't achieve my goal.The part that I don't understand is when we differentiate H=U+PV we obtain dH=dU + PdV + VdP and then from there we can derive Entropy equation which is TdS=dH-VdP.But when we differentiate U=Q-PV we get dU=dQ-PdV instead of dU=dQ-PdV-VdP.Why do we neglect VdP in internal energy equation?I know the internal pressure is negligible in Ideal gases but so why don't we neglect it in enthalpy equation?
2. Relevant equations
U=Q-PV
H=U+PV

3. The attempt at a solution
U=Q-PV
differentiating
dU=dQ-PdV-VdP
dU+PdV=dQ-VdP (dU+PdV=dH) (dQ=TdS)
dH=TdS-VdP
TdS=dH+VdP but the right equation is:TdS=dH-VdP

2. Mar 19, 2016

### Staff: Mentor

Please provide a reference that gives the equation U = Q - PV. Where did you dream this up from?

3. Mar 19, 2016

### Joseph95

We couldn't calculate the internal energy of a substance directly and we define it as change of state so we differentiate equation U=Q-PV and obtain dU=dQ-PdV and then integrating it as a result we get ΔU=ΔQ-PΔV.

4. Mar 19, 2016

### Staff: Mentor

I still don't see any reference (just a lot of hand waving). The equations you wrote are incorrect.

The correct equations should be $\Delta U=Q-W$, where $W=\int{P_{ext}dV}$, and, for a differential change between two closely neighboring thermodynamic equilibrium states, $dU = TdS-PdV$

Chet

5. Mar 19, 2016

### Joseph95

Okay it is my fault Sir ! Sorry for that.Besides this could we obtain ΔH=ΔU+PΔV+VΔP by differentiating equation H=U+PV ?

6. Mar 19, 2016

### Staff: Mentor

If we differentiate H=U+PV, we obtain dH = dU + PdV + VdP. If we substitute dU = TdS - PdV into this equation, we obtain: $$dH=TdS+VdP$$ or equivalently, $$TdS=dH-VdP$$ This is the equation you started with.

If you write H = U + PV, then, mathematically, with finite $\Delta$'s, $$\Delta H=\Delta U+V\Delta P+P\Delta V+\Delta P \Delta V$$Note the last term in this equation (which is required with finite deltas).

Chet

7. Mar 20, 2016

### Staff: Mentor

I'd like to elaborate a little to what I said in the previous post. If the two equilibrium states of the system are widely separated in terms of the pressures and the volumes, then we can write that:
$$\Delta (PV)=P_fV_f-P_iV_i$$
where the subscript i refers to the initial state and the subscript f refers to the final state. If we write that $P_f=P_i+\Delta P$ and $V_f=V_i+\Delta V$, then we find algebraically that $$\Delta (PV)=P_i\Delta V +V_i\Delta P+ (\Delta P)(\Delta V)$$
If we write that $\bar{P}=(P_i+P_f)/2$ and $\bar{V}=(V_i+V_f)/2$, then we find algebraically that:$$\Delta (PV)=\bar{P}\Delta V+\bar{V}\Delta P$$

8. Mar 20, 2016

### Joseph95

Thank you very much for your information and elaboration Sir.