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Topic: Thermodynamics - heat capacity  (Read 429 times)

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Offline ZHR

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Thermodynamics - heat capacity
« on: December 03, 2020, 04:55:52 AM »
1))Why DU=CvdT for ideal gases even if the volume isnt constant?
2))Why DH=CpdT for ideal gases even if the pressure isnt constant?

3))Joule–Thomson effect I didnt manage to understand properly the mathmathics behind those equasions if someone can help me with that I'd really apprieciate, thanks.
« Last Edit: December 03, 2020, 05:20:34 AM by ZHR »

Offline mjc123

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Re: Thermodynamics about ideal gases
« Reply #1 on: December 03, 2020, 05:28:44 AM »
Because you are dealing with state functions, the path doesn't matter. You can always get from initial (P1, V1, T1) to final (P2, V2, T2) state by a combination of a constant volume (or pressure) step and an isothermal step. For ideal gases, ΔU = 0 for the isothermal step. So for example
Constant vol: P1, V1, T1  :rarrow: P3, V1, T2   ΔU = CvΔT
Isothermal: P3, V1, T2  :rarrow: P2, V2, T2   ΔU = 0

Constant pressure: P1, V1, T1  :rarrow: P1, V3, T2   ΔH = CpΔT
Isothermal: P1, V3, T2  :rarrow: P2, V2, T2   ΔH = ΔU + Δ(PV) = 0

Offline mjc123

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Re: Thermodynamics - heat capacity
« Reply #2 on: December 03, 2020, 05:35:09 AM »
Don't edit your post to ask a fresh question. Make a new post.

What is it you don't understand about the maths? It looks pretty straightforward. Are you familiar with algebraic rearrangement of equations? (We do get people sometimes who don't realise that if a = b/c, then c = b/a.) Otherwise, what's the problem?

Offline ZHR

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Re: Thermodynamics - heat capacity
« Reply #3 on: December 03, 2020, 05:41:38 AM »
Thanks for the help.
and my fault about the equasion i was confused  ;)

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