Thermodynamics - ideal gas properties

[johnsmith456]

Hey guys,

I'm working on calculating the enthalpy, entropy and internal energy of substances. treating them as an ideal gas. I wonder if I could be pointed in the right direction with some calculations.

Enthalpy change:
h₂-h₁=∫c_p dT+ ∫[v-T(∂v/∂T)_P]dP

At ideal gas the pressure integral goes to zero.
So we have just the integral of C_p with respect to T.

All correct so far?

Its with the entropy and internal energy where I get confused.

ds = (C_p)dT - (∂v/∂T)_P dP

is there a way to simplify this further at ideal gas to make it easily calculated ?

I guess internal energy can be calculated from h using h = u + Pv.

Thanks very much

Look forward to any replies.

 

[Andrew Mason]

Enthalpy change:
h₂-h₁=∫c_p dT+ ∫[v-T(∂v/∂T)_P]dP

At ideal gas the pressure integral goes to zero.

I am not sure how you get this. If H = U + PV, then dH = dU + PdV + VdP and ΔH = ΔU + ∫PdV + ∫VdP = ΔQ + ∫VdP

So, if pressure is constant, ∫VdP = 0, so ΔH = ΔQ = ∫C_pdT

So we have just the integral of C_p with respect to T.

All correct so far?

Not in general. This is true only if P is constant.

Its with the entropy and internal energy where I get confused.

ds = (C_p)dT - (∂v/∂T)_P dP

Since U and PV are state functions, ΔU and Δ(PV) are independent of the path between two states so they are the same whether the path is reversible or irreversible. For a reversible path, ΔH = ΔQ + ∫VdP = ∫TdS + ∫VdP. Consequently, this must be true for all paths.

So you can use: dH = TdS + VdP

AM