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Why heat can't be expressed as exact differential function?

by manimaran1605
Tags: thermodynamics
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manimaran1605
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Jul4-14, 08:00 AM
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(1) Why differential of work, heat can't be expressed as exact differential function?
(2) How differential of internal energy is an exact differential function and how it is a function of any
two of thermodynamics coordinates?
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WannabeNewton
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Jul4-14, 08:31 AM
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Quote Quote by manimaran1605 View Post
(1) Why differential of work, heat can't be expressed as exact differential function?
Because work and heat individually depend on the path taken or process considered. There are countless examples of this all around you.

Quote Quote by manimaran1605 View Post
(2) How differential of internal energy is an exact differential function and how it is a function of any
two of thermodynamics coordinates?
Internal energy is defined to be a state function, there's nothing deeper to it. It wouldn't even make sense to call it internal energy if it wasn't a state function. Also it isn't true that the internal energy is necessarily a function of only two thermodynamic coordinates. In general ##U = U(S,x_1,...,x_n)## where ##x_i## are a set of generalized coordinates so one requires ##n+1## coordinates, including ##S## in the usual energy representation, and performing Legendre transforms to go to e.g. the Helmholtz or Gibbs representations will not change this. In the Helmholtz representation for example we would have ##F = F(T,x_1,...,x_n)## where ##F## is as usual the Helmholtz free energy. Of course the choice of ##S## for ##U## and the choice of
##T## for ##F## are just a matter of convenience for that specific thermodynamic potential and its associated fundamental relation (although not necessarily convenient for experiment). Indeed I can just as well write ##U = U(T,x_1,...,x_n)## if I wish by inverting ##S## as a function of ##T## but I still need ##n+1## variables.

Mathematically, ##U## is entirely determined by the equations of state and the heat capacity.


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