Enthalpy change in adiabatic expansion of ideal gas

[ScottHendo]

Homework Statement

Adiabatic expansion of an ideal gas is carried out in a steady flow process. The initial pressure of the gas is 2.5 bar. The volume is expanded from 1.2m³ to 3.8m³. Heat capacity ratio (γ) = 1.42. Calculate enthalpy change of the process.

Homework Equations

PV = nRT

W = ∫ PdV

ΔH = ΔU + Δ(PV)

PVγ = constant

The Attempt at a Solution

Calculate work done by using work done for steady flow process equation:
W = ∫ PdV with PVγ= constant to get W = constant ∫ (dV)/V^γ .
Carry out the integration to get W.
-W = ΔU
Then use ΔH = ΔU + Δ(PV) to find enthalpy change.

I am looking for any help on 1) to make sure I am on the right track and 2) what to do with the Δ(PV) part of the last equation.

Thanks in advance!

 

[dRic2]

I suppose the expansion is a reversible process. In a steady flow process $W = \Delta H = c_p*(T_2-T_1)$. Working with internal energy seems to be correct but it much more complicated.

 

[Chestermiller]

Is this the exact problem statement, or is there something that you left out? The answer is different if the flow is through a porous plug compared to a turbine featuring an adiabatic reversible expansion. Did you leave out the word “reversible” from your description?

 

[ScottHendo]

I missed out that it is a slow steady flow process, would this mean that I could assume it is reversible?