Thermodynamics: T-S slope at constant volume

[2DGamer]

1. Prove that the slope of an isochoric process in a T-S diagram is T/Cv where T is the temperature and Cv is the heat capacity at constant volume



2. dQ = T * dS (I understand this)
dQ = Cv * dT for isochoric processes (I don't understand this)



3. Since the two equations share dQ I set them equal to each other:
Cv * dT = T * dS
Then I just rewrote the equation as:
T/Cv = dT/dS
And dT/dS describes the slope of the line which is T/Cv
My main problem is that I don't understand why dQ = Cv*dT. The book just gave it to us, but I'm thinking that I need to be able to derive it or the problem is just too easy.

 

[Mapes]

dQ = Cv * dT for isochoric processes (I don't understand this)

For isochoric (constant-volume, $dV=0$) processes, no work is done (work being $P\,dV$), so any energy change $dU$ in the system must be in the form of heat transfer $q$. Additionally, we can use the differential energy equation

$$dU(=q)=T\,dS-P\,dV=T\,dS=T\left(\frac{\partial S}{\partial T}\right)_V\,dT=C_V\,dT$$

which uses the definition of $C_V$. Does this make sense?