Enthelpy of Reaction under Constant Volume?

[Hereformore]

Homework Statement

So we know Enthalpy under constant Pressure and Internal Energy under Constant Volume. By

H = U + dPdV

U= I + W

W= -PdV

under Isobaric conditions

H = I -PdV + dPdV
= U

Enthalpy = Internal Energy
_________________________________________
Under Isochoric Conditions
W= O because change in V =o

so U = q + 0
Internal Energy = q

But wouldn't Enthalpy also = q under constant volume since dPdV = 0 as well if volume isn't changing?

Homework Equations

H = U + dPdV

U= I + W

W= -PdV

The Attempt at a Solution

( Outlined above)

Am i confusing the dPdV specific to enthalpy and PdV in work?

 

[rude man]

Looks confusing. what's the question?

H = U + PV so dH = dU + pdV + Vdp
dH = dU + Vdp under isochoric
but dU = dQ - pdV = dQ under isochoric
so dH = dQ + Vdp under isochoric.

Is that something like what you're looking for?

 

[Hereformore]

Looks confusing. what's the question?

H = U + PV so dH = dU + pdV + Vdp
dH = dU + Vdp under isochoric
but dU = dQ - pdV = dQ under isochoric
so dH = dQ + Vdp under isochoric.

Is that something like what you're looking for?

Yeah. I see. So Enthalpy = Internal Energy + Change in Pressure times volume.
While change in internal energy (dU) = q + Change in Volume times pressure.

is this correct?

 

[rude man]

Yeah. I see. So Enthalpy = Internal Energy + Change in Pressure times volume.
While change in internal energy (dU) = q + Change in Volume times pressure.

is this correct?

enthalpy = internal energy plus pressure times volume ... basic statement, always true
change in internal energy = heat added minus pressure times change in volume. "Change" should read "differential change". To get change you integrate differential changes.