Ideal Gases: Isochoric Process

An isochoric process occurs where the volume is constant, $V_1=V_2$. This means that for the ideal gas law, $PV=nRT$, $T \propto P$.

 Since $\Delta{V}=0$, then the relationship $w=-P\Delta{V}=0$ as well.

Examining the heat at constant volume, we find
$q_v=C_v\int_{T_1}^{T_2}dT=C_v(T_2-T_1)$

 Now, when we look at the internal energy
 $\Delta{U}=q_v+w=q_v=C_v(T_2-T_1)$ 

And finally, the entropy for when $n=1$, with
 $$\Delta{H}=\Delta{U}+\Delta(PV)$$ $$\Delta{H}=q_v+\Delta{RT}$$ $$\Delta{H}=C_v(T_2-T_1)+R(T_2-T_1)$$ $$\Delta{H}=(C_v+R)(T_2-T_1)$$

So, in summary, for an isochoric process $$w_{rev}=0$$ $$q_v=\delta{U}=C_v(T_2-T_1)$$ $$\Delta{H}=(C_v+R)(T_2-T_1)$$

Note, that like for isobaric processes, isochoric processes occur on two different isotherms, as the temperature is not constant. These two isotherms of the isochoric process can be the same isotherms as for an isobaric process, however the work and heat supplied vary.