\[Q(t_2)-Q_(t_1) = \int^{t_2}_{t_1} ( \frac{dU}{dt} +p \frac{dV}{dt}) dt\]

.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

\[\oint p dV\]

.