[WalkAndJump]

I'm having problems grasping the solution of my professor for a thermodynamic test.

1. Enthalpy & Adiabatic Processes

"A gas undergoes an adiabatic expansion against a constant external pressure of 1.00 bar. During this process, the pressure of the gas changes from 10.00 bar to 1.00 bar and the volume changes from 10.0 L to 60.0 L. Calculate q, w, ΔU, ΔH."

I had no problems with q, w, and ΔU but since it is an adiabatic expansion (q=0) shouldn't ΔH=0 aswell?

My professor's solution was...

ΔH = ΔU + Δ(PV) = ΔU + (PfVf - PiVi) = -5000J - [(1.00bar)(60.0L)-(10.0 bar)(10.0L)] = -9000J

I also thought pressure-volume work used the pressure of the surroundings and not the pressure of the gas as the justification for dH = dqp requires pressure to be of the surroundings.

2. The second problem I thought dealt with Carnot efficiency.

"A solid with a constant volume heat capacity of 515 J K^-1 is initially at a temperature of 385 K. The heat capacity of the solid is independent of temperature. The surroundings are at a constant temperature of 273 K."

then...

"Calculate the maximum work that could have been done on the surroundings while
reversibly producing the same change in state of the solid as in part (a)"

Part (a) would be the cooling of the solid to 273K.

qh = -57.7 kJ

My solution would be -w/qh = (Th - Tc) / Th, solve for w=16.8 kJ

However, the solution was w = qh - qc = 9.4 kJ. qc was calculated through -ΔSc = ΔSh.

I was wondering why using Carnot Efficiency cannot work in this case.

[curiouscat]

 since it is an adiabatic expansion (q=0) shouldn't ΔH=0 as well?

Why? 

[WalkAndJump]

 since it is an adiabatic expansion (q=0) shouldn't ΔH=0 as well?

Why?

ΔH = qp = 0 because of an adiabatic boundary. Without heat flow between the system and the surroundings, then shouldn't it follow that 0 = ΔU + pΔV --> -pΔV = ΔU thus expansion work is the sole factor in determining change in internal energy.

[Jorriss]

 since it is an adiabatic expansion (q=0) shouldn't ΔH=0 as well?

Why?

ΔH = qp = 0 because of an adiabatic boundary.

That is for a constant pressure process.

[WalkAndJump]

 since it is an adiabatic expansion (q=0) shouldn't ΔH=0 as well?

Why?

ΔH = qp = 0 because of an adiabatic boundary.

That is for a constant pressure process.

There is a constant external pressure of 1.00 bar and the proof for ΔH = qp uses the external pressure and not of the system.