What Happens to Enthalpy in Constant Volume and Pressure Processes?

[jason.bourne]

enthalpy is defined as

h = u + pv

so, dh = du + d (pv)

if the process is constant pressure,

dh = du + p dv (change in internal energy + boundary work )

during constant pressure process, if heat is added to the system, it increases its internal energy as well as it does the expansion boundary work. so the net heat added will increase the enthalpy of the system.

if the process is constant volume,

dh = du + v dp (what is this extra term v dp??).

during constant volume process, heat added will just raise the internal energy. so what happens to enthalpy ?

 

[Skrambles]

It will equal the change in internal energy because there is no boundary work performed by the system.

 

[jason.bourne]

thanks for the reply Skrambles !

there won't be any boundary work, agreed.

will internal energy change be equal to enthalpy change?

we know for some change in temperature dT,
du = c_v dT and
dh = c_p dT

and c_p and c_v are never same.

 

[Skrambles]

They are indeed equal for an incompressible substance. (k=1)

dh = c_p dT applies only when pressure is constant.

du = c_v dT applies only when volume is constant.

 

[jason.bourne]

they are equal for incompressible fluids such as water - agreed.

what happens when we consider a gas, assuming compressibility condition holds, in a constant volume process?

"du = cv dT applies only when volume is constant"

so does that mean we can't define change in enthalpy during constant volume process?

this is bit confusing coz the thermodynamics textbooks suggests, property relations can be used irrespective of the processes, reversible or irreversible coz they are state variables and don't depend on the path of integration.

 

[Skrambles]

so does that mean we can't define change in enthalpy during constant volume process?

No, it just means that you can't use dh = c_p dT because of the way c_p is defined.

If you can define the state of the system, then you can define its enthalpy because, like you said, it is a state variable.

The change in enthalpy equals the change in internal energy plus boundary work performed during the process. If there is no boundary work, then dh = du and du = c_v dT.

You can also use the First Law to get ΔQ = ΔU.

Don't feel bad about getting confused, I struggled for a while with the same concept when I took thermo. I went into a test thinking I could use c_p and c_v for every process, so I was very surprised when I saw my grade.

 

[jason.bourne]

hahaa...

alright! got it.

thanks for helping me out !