# Prove dQ is an inexact differential

1. Oct 14, 2016

### arpon

1. The problem statement, all variables and given/known data
$dz=Mdx+Ndy$ is an exact differential if $(\frac{\partial M}{\partial y})_x=(\frac{\partial N}{\partial x})_y$.
By invoking the condition for an exact differential, demonstrate that the
reversible heat $Q_R$ is not a thermodynamic property.

2. Relevant equations
$dQ=dU+PdV$
$dU=C_VdT$

3. The attempt at a solution
$dQ=C_VdT+PdV$
So, we have to show that, $(\frac{\partial C_V}{\partial V})_T \neq (\frac{\partial P}{\partial T})_V$
Now,
$LHS=(\frac{\partial C_V}{\partial V})_T$
$=(\frac{\partial}{\partial V}((\frac{\partial U}{\partial T})_V))_T~~~~$[$C_V=(\frac{\partial U}{\partial T})_V$]
$=(\frac{\partial}{\partial T}((\frac{\partial U}{\partial V})_T))_V~~~~$ [Symmetry of second derivatives]
So, we need to show that
$(\frac{\partial U}{\partial V})_T \neq P+F(V)~~~$ [where, $F(V)$ is an arbitary function of $V$]
$(\frac{\partial U}{\partial V})_T - P \neq F(V)$

If the equation of state is given, this can easily be proved. How can I prove this in general? I also do not understand what would be the difference for irreversible heat.

Last edited: Oct 14, 2016
2. Oct 14, 2016

Just an input for you=I don't have the solution, but $TdS=dQ_{reversible}$ so that according to the above, you should be able to show that the functional property holds for $S$ but not for $Q_{reversible}$.

3. Oct 14, 2016

### Staff: Mentor

I think you should work with $$C_V=T\left(\frac{\partial S}{\partial T}\right)_V$$ and determine the partial with respect to V at constant T. This can be expressed in terms of the equation of state parameters by applying a Maxwell relationship.

4. Oct 14, 2016

### arpon

I tried as you mentioned,
$\left(\frac{\partial C_V}{\partial V}\right)_T$
$=\left(\frac{\partial}{\partial V}(T\left(\frac{\partial S}{\partial T}\right)_V)\right)_T$
$=T\left(\frac{\partial}{\partial V}(\left(\frac{\partial S}{\partial T}\right)_V)\right)_T$ ... (i)

Now,
$\left(\frac{\partial S}{\partial T}\right)_V=\left(\frac{\partial S}{\partial P}\right)_V \left(\frac{\partial P}{\partial T}\right)_V=-\left(\frac{\partial V}{\partial T}\right)_S \left(\frac{\partial P}{\partial T}\right)_V$...(ii) [Using Maxwell relationship]
Using (i) and (ii),
$\left(\frac{\partial C_V}{\partial V}\right)_T$
$=-T\left(\frac{\partial}{\partial V}(\left(\frac{\partial V}{\partial T}\right)_S \left(\frac{\partial P}{\partial T}\right)_V)_V)\right)_T$
What should be the next steps?

5. Oct 14, 2016

### Staff: Mentor

I didn't notice before that, in your original post, your equation for dU is incorrect. Do you know what the correct relationship is?

6. Oct 14, 2016

### Staff: Mentor

You are going to kick yourself. This problem is much simpler than what you've been trying to do.

For a reversible process, dQ=TdS

Now, if you express dS in terms of dT, dV, and partials of S with respect to T and V, what do you get? What do you get if you substitute that relationship into dQ=TdS?

7. Oct 14, 2016

@Chestermiller I don't see an additional response from the OP yet, but your input was useful. I see when testing the criteria given by the OP in post #1, there is an extra term so that the two sides are not equal. In the case of $dS=(1/T)(dU+PdV)$ though, the criterion is met, at least in the case of an ideal gas.

8. Oct 14, 2016

### arpon

Thanks for pointing out the mistake. I am used to solve problems for ideal gases, that's why I have made the mistake, I guess.

Thanks for the hints!
$dQ_R=TdS=T\left(\frac{\partial S}{\partial T}\right) _V dT + T\left(\frac{\partial S}{\partial V}\right) _T dV$
So, we have to show that,
$\left(\frac{\partial}{\partial V} \left( T\left(\frac{\partial S}{\partial T}\right) _V \right)\right)_T \neq \left(\frac{\partial}{\partial T} \left(T\left(\frac{\partial S}{\partial V}\right) _T \right)\right )_V$
$RHS=T\left(\frac{\partial}{\partial T} \left(\frac{\partial S}{\partial V}\right) _T \right )_V + \left(\frac{\partial S}{\partial V}\right)_T$
And,
$LHS=T \left(\frac{\partial}{\partial V} \left(\frac{\partial S}{\partial T}\right) _V \right)_T$

But $S(V,T)$ is a state function of $V$ and $T$. So $\left(\frac{\partial S}{\partial V}\right)_T \neq 0$ in general.

Last edited: Oct 14, 2016