Evolution of temperature (adiabatic procsses)

ted1986
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Hello,

I'm trying to derive a differential equation as requested in the attached exercise (thermal1.jpg).
I'm not quite sure my solution is the right answer (my_solution1.jpg).
How do I get rid of the energy dU ?

Thnks


Ted
 

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well in an adiabatic processes dQ = 0 so ds = 0 giving you quite a simple derivation of the differential equation
 
sgd37 said:
well in an adiabatic processes dQ = 0 so ds = 0 giving you quite a simple derivation of the differential equation


OK, It sure gives a simple derivation of the differential equation, but what is the physical explanation for that?

Thanks
 
it most generally means that it is a reversible process
 
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