Solving Enthaply by other ways?

[fubear]

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.

I found that w = dE = -997.78J

In class we solved dH by:

dH = CpdT = -1.66kJ
(we found Cp by Cp=Cv+nR ; we 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?


Thank you

 

[nasu]

2 moles of an ideal gas at 300K and 5 bar is expanded adiabatically at a constant pressure

It is adiabatic or at constant pressure? How can be both?

 

[fubear]

The external pressure is constant, but the system undergoes an adiabatic process

 

[Philip Wood]

I'm with nasu here. If the gas expands adiabatically pushing out a piston, then it will do work pushing out the piston, and its internal energy will fall. So, therefore, will its temperature, and its pressure (since P = nRT/V and T goes down and V goes up).