Why don't we include VdP in U equation ?

[Joseph95]

Homework Statement

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?

Homework Equations

U=Q-PV
H=U+PV

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

 

[Chestermiller]

Homework Statement

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?

Homework Equations

U=Q-PV
H=U+PV

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

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

 

[Joseph95]

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

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.

 

[Chestermiller]

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.

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

 

[Joseph95]

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

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

 

[Chestermiller]

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

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

 

[Chestermiller]

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:$$\Delta (PV)=\bar{P}\Delta V+\bar{V}\Delta P$$

 

[Joseph95]

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:$$\Delta (PV)=\bar{P}\Delta V+\bar{V}\Delta P$$

Thank you very much for your information and elaboration Sir.