Work done in adiabatic compression

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  • #36
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This problem is really straightforward if one chooses as the "system" the combination of gas plus piston. Then, from the first law of thermodynamics, we have: $$dU=(nC_v+C)dT=-PdV=-\frac{nRT}{V}dV$$where n is the number of moles of gas, Cv is the molar heat capacity of the gas, and C is the heat capacity of the piston.
 
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  • #37
Delta2
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This problem is really straightforward if one chooses as the "system" the combination of gas plus piston. Then, from the first law of thermodynamics, we have: $$dU=(nC_v+C)dT=-PdV=-\frac{nRT}{V}dV$$where n is the number of moles of gas, Cv is the molar heat capacity of the gas, and C is the heat capacity of the piston.
Ehm, is it ok to say that the ##dU## of piston is just ##CdT##?

That leads to a nice and neat separable ODE btw, we can find T(V) relatively easily...
 
  • #38
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Ehm, is it ok to say that the ##dU## of piston is just ##CdT##?
Sure. It is considered an incompressible solid.
 
  • #39
TSny
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we have: $$dU=(nC_v+C)dT=-PdV=-\frac{nRT}{V}dV$$
Yes. This leads to a very simple result for the work done on the gas during the process in terms of the overall temperature change ##\Delta T## for the process. Maybe this is what they want. So, perhaps ##dT## in the statement of the problem is meant to be ##\Delta T##. I don't know.

You can also use Chestermiller's equations to derive the relation between ##T## and ##V## during the process. Then you can express the work done on the gas in terms of the initial temperature and the initial and final volumes of the gas. This is what I had in mind in post #17.
 
  • #40
so_gr_lo
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Yeah maybe it was that simple, I just wasn’t sure if they wanted me to use nm as it wasn’t in the variable list
 

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