\[Q(t_2)-Q_(t_1) = \int^{t_2}_{t_1} ( \frac{dU}{dt} +p \frac{dV}{dt}) dt\]
.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
\[\begin{equation} \begin{aligned} Q(t_2)-Q_(t_1) &= \int^{t_2}_{t_1} (\frac{\partial U}{\partial p}_V \frac{dp}{dt} + \frac{\partial U}{\partial V}_p \frac{dV}{dt} +p \frac{dV}{dt}) dt \\ &= \int_C \frac{\partial U}{\partial p}_V dp + (\frac{\partial U}{\partial V}_p +p )dV \\ &= \int_C dU + d dV \end{aligned} \end{equation}\]
.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
\[\oint p dV\]
.