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Thermodynamics - ideal gas properties

by johnsmith456
Tags: ideal, properties, thermodynamics
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johnsmith456
#1
Nov20-11, 06:37 AM
P: 5
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=∫cp dT+ ∫[v-T(∂v/∂T)P]dP

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

All correct so far?

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

ds = (Cp)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.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Andrew Mason
#2
Nov20-11, 08:15 AM
Sci Advisor
HW Helper
P: 6,653
Quote Quote by johnsmith456 View Post
Enthalpy change:
h2-h1=∫cp 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 = ∫CpdT

So we have just the integral of Cp 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 = (Cp)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


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