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

1. Jul 4, 2014

### manimaran1605

(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?

2. Jul 4, 2014

### WannabeNewton

Because work and heat individually depend on the path taken or process considered. There are countless examples of this all around you.

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.