## Enthalpy and PV value Question

## Enthalpy and PV value Question

(OP)

I have a question concerning enthalpy or energy contained in a fluid where dH = dU + d(PV).

In a non flow process with constant pressure dH = dU + Pdv = dU + d(PV) since P = constant. Therefore if dH was caused by the addition of heat (dQ) it is easy to see that:

dQ = dH = dU + Pdv = dU + d(PV) = dU + PdV + VdP (where VdP = 0)in other words the change in energy of the system dH equals the change in enthalpy which is the change internal energy dU plus the boundary work Pdv which is equal to the heat input dQ.

Now for a non flow system that the pressure is not constant - for instance if you heat up an elastic sphere - both the pressure and volume increases so that Pdv is not equal to d(PV) in fact d(PV) is greater than Pdv since d(PV) = PdV + Vdp.

My question is if heat is added to an elastic sphere in this case then the energy input dQ the way I see it can still only go into change in internal energy plus boundary work dU + Pdv so this is the change in energy of the system again as above as I see it.

However, in textbooks the statement is always made that the change in energy of a fluid due to heat input is always the difference in enthalpy = dU + d(PV) but how can this be if the heat added can only go into dU (internal energy change) + Pdv (boundary work)but change in enthalpy dH is always greater than this (= dU + PdV + VdP)in regards to the elasic sphere case where both pressure and volume increases. What work or energy stored in the system does the VdP term represent in regards to the energy (heat) input to the system if enthalpy is the actual total energy change of such a system (since dU + PdV appears to be the only energy being absorbed by the system due to heat transfer dQ)?

In a non flow process with constant pressure dH = dU + Pdv = dU + d(PV) since P = constant. Therefore if dH was caused by the addition of heat (dQ) it is easy to see that:

dQ = dH = dU + Pdv = dU + d(PV) = dU + PdV + VdP (where VdP = 0)in other words the change in energy of the system dH equals the change in enthalpy which is the change internal energy dU plus the boundary work Pdv which is equal to the heat input dQ.

Now for a non flow system that the pressure is not constant - for instance if you heat up an elastic sphere - both the pressure and volume increases so that Pdv is not equal to d(PV) in fact d(PV) is greater than Pdv since d(PV) = PdV + Vdp.

My question is if heat is added to an elastic sphere in this case then the energy input dQ the way I see it can still only go into change in internal energy plus boundary work dU + Pdv so this is the change in energy of the system again as above as I see it.

However, in textbooks the statement is always made that the change in energy of a fluid due to heat input is always the difference in enthalpy = dU + d(PV) but how can this be if the heat added can only go into dU (internal energy change) + Pdv (boundary work)but change in enthalpy dH is always greater than this (= dU + PdV + VdP)in regards to the elasic sphere case where both pressure and volume increases. What work or energy stored in the system does the VdP term represent in regards to the energy (heat) input to the system if enthalpy is the actual total energy change of such a system (since dU + PdV appears to be the only energy being absorbed by the system due to heat transfer dQ)?

## RE: Enthalpy and PV value Question

"since d(PV) = PdV + Vdp"Are you sure?

je suis charlie

## RE: Enthalpy and PV value Question

pis constant and equal to the internal pressure; therefore, under these conditionspVwork,W=pdV.## RE: Enthalpy and PV value Question

so

PE+KE=PE+KE

then insert heat into KE, drop out PE in the left half and in the right half do work by the spring of the elastic balloon so:

dQ=dH + K*dV

which is pretty basic for K since K will actually change depending on how far its streched, but you kind of get the idea

## RE: Enthalpy and PV value Question