## Proof That the State Space for an Ideal Gas is Not Conservative

The amount of heat delivered by a gas during a change of state is given by
$Q(t_2)-Q_(t_1) = \int^{t_2}_{t_1} ( \frac{dU}{dt} +p \frac{dV}{dt}) dt$
.
We can write
$U$
as a function of
$p$
and
$V$
,
$U=U(p,V)$
then
$dU= \frac{\partial U}{\partial p}_V dp + \frac{\partial U}{\partial V}_p dV$

Hence we can write the integral as
\begin{aligned} Q(t_2)-Q_(t_1) &= \int^{t_2}_{t_1} (\frac{\partial U}{\partial p}_V \frac{dp}{dt} + \frac{\partial U}{\partial V}_p \frac{dV}{dt} +p \frac{dV}{dt}) dt \\ &= \int_C \frac{\partial U}{\partial p}_V dp + (\frac{\partial U}{\partial V}_p +p )dV \\ &= \int_C dU + d dV \end{aligned}
.
where
$C$
is the particular curve between initial and final stats.
For this integral to be independent of
$C$
we must have
$\frac{\partial}{\partial p}(\frac{\partial U}{\partial V}_p +p) =\frac{\partial }{\partial V}(\frac{\partial U}{\partial p}_V)$

Hence
$\frac{\partial^2 }{\partial p}{\partial V} + 1 = \frac{\partial^2 }{\partial p}{\partial V}$
which is impossible if
$U$
has continuous second partial derivatives.
For a simple closed curve the heat received is
$\oint p dV$
.