Can Enthalpy Be Solved Using Different Methods?

[fubear]

Homework Statement

2 moles of an ideal gas at 300K and 5 bar is expanded adiabatically at a constant pressure of 1 bar till the volume doubles. Cv = 3R/2. Calculate w, q, dE, dH and change in T.

Homework Equations

See below

The Attempt at a Solution

I found that w = dE = -997.78J

I solved dH by:

dH = CpdT = -1.66kJ
(found Cp by Cp=Cv+nR ; found dT by Cv=dE/dT)I was wondering if we could solve dH another way:
dH = dE + d(PV)

However, I get a different answer..
dH = dE + Pext dV (because pressure is constant)
dH = -997.78J +-997.78J (because Pext dV is work)
dH = -2.00 kJ

I also tried another way:
dH = dE + d(nRT)
dH = -977J + (2mol)(8.3145J/K)(-80K)
dH = -1.08kJ

I am not getting the same value for each of the different ways I am using... Am I doing something wrong or am I not allowed to solve dH using the other methods?

Also, at constant pressure, isn't dH = qrev.. and q=0, so shouldn't dH = 0?

 

[rude man]

So the gas starts at 5 bar and then is reversibly changed to1 bar?? Under what conditions?

 

[ehild]

H=E+PV, where P is the pressure of the gas. It is defined for the steady state. During the expansion, the gas in not in equilibrium, its pressure is not defined, you can not use the external pressure.
You can calculate the change of enthalpy from the parameters of the new equilibrium, E, P, T, and it will be the same -1.66 kJ.

ehild