Thermodynamics, Exact differentials

[manimaran1605]

There exist exact differential for P (Pressure) ,V (volume) ,T (temperature), U (Internal Energy) but not for W(work), Q (heat) . Why?

 

[Jano L.]

That refers to "differential" as in "total differential of multi-variable function". Work and heat are said to not have such total differential because neither work nor heat are usually considered to be such functions. That is because they do not describe one state of a physical system (like internal energy, which is multi-variable function - of volume and temperature, for example), but a whole process such system may undergo.

 

[stevendaryl]

Here's a long-winded description of the difference between exact differentials and inexact differentials:

A generalized differential can be written in the form:

$\sum_j Q_j dx_j$

where the $x_j$ are independent state variables, and $Q_j = Q_j(x_1, x_2, ...)$ are functions of those state variables. In contrast, an exact differential is the special case where there is a single state function $F(x_1, x_2, ...)$ and

$Q_j = \dfrac{\partial F}{\partial x_j}$

In that special case, $\sum_j Q_j dx_j = dF$, an exact differential.

How do you know whether there is such an $F$? Well, you can figure it out by using a special property of partial derivatives, which is that the order of differentiation doesn't matter:

$\dfrac{\partial}{\partial x} \dfrac{\partial}{\partial y} F = \dfrac{\partial}{\partial y} \dfrac{\partial}{\partial x} F$

In terms of the $Q_j$, this means that:

$\dfrac{\partial Q_j}{\partial x_k} = \dfrac{\partial Q_k}{\partial x_j}$

Now, relating all this back to the question about work:

$dW = -P dV$

A complete set of independent state variables for a monoatomic gas is volume, temperature and number of particles: $V, T, N$. So to make an exact differential out of $-P dV$, you would have to add other terms, to get something like:

$d ? = -P dV + Q_1 dT + Q_2 dN$

where the mixed derivatives work out:

• $\dfrac{\partial Q_1}{\partial V} = - \dfrac{\partial P}{\partial T}$
  
• $\dfrac{\partial Q_2}{\partial V} = -\dfrac{\partial P}{\partial N}$
  
• $\dfrac{\partial Q_1}{\partial N} = \dfrac{\partial Q_2}{\partial T}$

One set of choices that work out are:

$Q_1 = S$ (entropy)
$Q_2 = \mu$ (chemical potential)

With these choices, our inexact differential $dW$ is turned into the exact differential, $dU = =Pdv + S dT + \mu dN$