# Enthalpy=Internal Energy at Constant Pressure, and InternalEnergy=q at constant volume. What about..

#### manohman

5+ Year Member
What about Enthalpy at constant volume?

H = U + dPdV

U= I + W

W= -PdV

so 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 wouldnt Enthalpy also = q under constant volume since dPdV = 0 as well if volume isnt changing?

I think i'm confusing the dPdV in the Enthalpy equation!

#### amy_k

You made a slight mistake when you multiplied out d(PV), remember we are dealing with differential change here.

The following equation applies for both isobaric and isochoric processes:

dH = dU + d(PV)
dH = dU + PdV + VdP

***(d(PV) = PdV + VdP, and not dVdP, using the product rule for differentiation)***

Since dU + PdV = dQ, we can simplify it as follows:

dH = dQ + VdP

So for an isobaric process when dP = 0:

dH = dQ

And for isochoric when V is constant, we cannot cancel out the VdP term, so

dH = dQ + VdP

But you are correct about internal energy under isochoric conditions:

dU = dQ - pdV
dU = dQ

I hope that helps you out.

OP
M

#### manohman

5+ Year Member
You made a slight mistake when you multiplied out d(PV), remember we are dealing with differential change here.

The following equation applies for both isobaric and isochoric processes:

dH = dU + d(PV)
dH = dU + PdV + VdP

***(d(PV) = PdV + VdP, and not dVdP, using the product rule for differentiation)***

Since dU + PdV = dQ, we can simplify it as follows:

dH = dQ + VdP

So for an isobaric process when dP = 0:

dH = dQ

And for isochoric when V is constant, we cannot cancel out the VdP term, so

dH = dQ + VdP

But you are correct about internal energy under isochoric conditions:

dU = dQ - pdV
dU = dQ

I hope that helps you out.
Ah Awesome That clears it up entirely and makes complete sense. I was having so much trouble. Thanks!