Page:Zur Thermodynamik bewegter Systeme.djvu/4

 Let $$U''_0$$ be the value now assumed by $$U_0$$; then it is

D. We bring the body to rest in an adiabatic and isochoric way. There, the work

$U_{0} - \Phi(U_{0},\ v,\ \beta)$

is performed. The state of the body is now given by the variables $$U''_{0},\ v,\ \beta = 0$$. The total work of the external forces is:

$(-p_{0}dv) + (\Phi(U'_{0},\ v',\ \beta) - U'_{0}) + (pdv + \beta d\phi) + (U_{0} - \Phi(U_{0}, v,\ \beta)).$

According to the first thermodynamic main-theorem, the work must be equal to $$U''_0-U_0$$. The second main-theorem additionally requires, that this difference is zero. Otherwise this circular process (or the reverse one) would represent a thermal perpetual motion machine. Thus we put $$U''_0-U_0$$ in the previous expressions; then we notice that

$\begin{array}{lll} d\phi & = & \phi(U_{0},v,\beta)-\phi(U'_{0},v',\beta)=\\ \\ & = & \frac{\partial\phi}{\partial U_{0}}(U_{0}-U'_{0})+\frac{\partial\phi}{\partial v}(v-v')=\\ \\ & = & \frac{\partial\phi}{\partial U_{0}}p_{0}dv-\frac{\partial\phi}{\partial v}dv\end{array}$|undefined

and

$\Phi(U'_{0},v',\beta)-\Phi(U_{0},v,\beta)=\frac{\partial\Phi}{\partial v}dv-\frac{\partial\Phi}{\partial U_{0}}p_{0}dv$|undefined

thus we obtain:

$\frac{\partial\Phi}{\partial v}dv-\frac{\partial\Phi}{\partial U_{0}}p_{0}dv+\beta\frac{\partial\phi}{\partial U_{0}}p_{0}dv-\beta\frac{\partial\phi}{\partial v}dv+pdv=0$|undefined

or according to (2):

Thus we obtain the theorem: If an arbitrary body (in the state of rest) is under the pressure $$p_0$$, then it