# I Adiabatic Process, Internal Energy vs. Enthalpy.

1. Mar 1, 2016

### Jeremy1789

I'm struggling to understand a concept which I assume is basic, but I can't seem to fit the pieces together. When speaking about an ideal gas, I understand that

ΔH = ΔU + Δ(PV) = ΔU + RΔT

So far so good. I also understand the relationship:

ΔU = Q + W... (W here is work being done on the system)

In an adiabatic reversible process, Q = 0, which also makes sense. So,

W = ΔU = nCvΔT

Now, where my confusion lies is in the next part. My book works out a problem, and says:

ΔH = W = nCpΔT

How can both ΔU and ΔH equal W? This doesn't make sense to me, unless Δ(PV) from the first equation was 0. I don't see how this could be 0 unless we were talking about an isothermal case. It also doesn't make sense because W can't equal both nCvΔT and nCpΔT simultaneously, since Cp = Cv + R.

Am I missing something?

2. Mar 1, 2016

### Jeremy1789

Never mind. Just answered my own question. The first instance refers to a closed adiabatic system, and the second refers to an open steady-state adiabatic process.

3. Mar 7, 2016

### akshay86

How nCvΔT will equal to W.It is the heat absorbed/released at constant volume only(no work is done at constant volume,the expression
Cv).i think instead of work it is heat,ie ΔU=ΔH=qv at constant volume and ΔH=qp at constant pressuer
actually it is the proof for Cp-Cv=R