# Confused about enthalpy change in ideal gas

1. Apr 12, 2013

### susdu

1. The problem statement, all variables and given/known data

The state of 3.75mol ideal gas changed from 0.3atm and 500K to 2.2atm and 330K under constant external pressure (Pex = 2.2atm).

Calculate ΔE, work (w) and heat(Q), ΔH, ΔS.

2. Relevant equations

w= -∫PexdV
ΔE=1.5nRΔT=w+Q
ΔH=ΔE+Δ(PV)
ΔS=dQrev/T

3. The attempt at a solution

I recently started a thermodynamics course and this question seems entry-level but I'm still unable to grasp some fundamental concepts. Calculating the energy change, work and heat was pretty straight forward: I used the appropriate equations to find energy change and work and then used the first law to find the heat.

Problem is when I was asked about the enthalpy change, I thought "ok, this is a constant pressure proccess, so the enthalpy change is equal to the heat". But apparently this is not the case.

What am I missing?

2. Apr 12, 2013

### Staff: Mentor

It's not a constant pressure process. The pressure of the surroundings is constant, but the system pressure is not constant. You already calculated ΔE, and presumably you can calculate the initial and final volumes, so you can get Δ(pV). This will give you ΔH. You can also get ΔH another way, by using nCpΔT, and remembering that Cp=Cv+R for an ideal gas. This should, of course, give you the same answer.

3. Apr 12, 2013

### susdu

So trivial, should have noticed that. Thank you.

4. Apr 15, 2013

### susdu

Can you give an example of a constant pressure process where the enthalpy change equals the heat?
(I can't seem to distinguish between the given scenarios)