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  
  as a function of  
\[dU= \frac{\partial U}{\partial p}_V dp + \frac{\partial U}{\partial V}_p dV\]

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

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

Add comment

Security code