1. Jul 7, 2012

### Carcul

Consider a closed thermodynamic system capable of exchanging energy in the form of work only as PV work. Under these conditions, for an isobaric process we find that the heat exchanged equals the enthalpy change. Now what about the reverse? For such a system, if for some process the heat exchange equals the change in enthalpy can we conclude that the pressure has remained constant? If not can you find a counter example?

2. Jul 10, 2012

### Andrew Mason

Yes, but only if ∂W = PdV. You can prove this from the definition of enthalpy:

H = U + PV
dH = dU + PdV + VdP

If dH = ∂Q = dU + ∂W then ∂W = PdV + VdP

If ∂W = PdV then VdP = 0 which implies that dP = 0 (isobaric)

AM

3. Jul 10, 2012

### Carcul

Thank you very much. But why does ΔH = Q implies dH = δQ?

Last edited: Jul 10, 2012
4. Jul 10, 2012

### Studiot

It doesn't.

Neither the heat nor the work exchanged are true differentials, they are actual values so it is wrong talk of delta (of any sort) q or w. Some people prefer capitals, some prefer lower case some use the (as Andrew has done) Greek delta to show this.

But the bottom line is that the heat exchanged is the heat exchanged it is not a small change in the heat exchanged.