Problem involving an adiabatic process

[Another]

Homework Statement

adiabatic process equation derivation

Relevant Equations

dU = U_V dT + U_T dV ; U_V is derivative U by T (volume constant) and U_T is derivative U by V (temperature constant)
dQ = dU + p dV

in this textbook : http://www.fulviofrisone.com/attach...tatistical Mechanics 2Ed (Wiley)(T)(506S).pdf ;page 20

I don't understand about Eq 1.11 come to 1.12 ? I know

dU = U_V dT + U_T dV

dQ = dU + p dV

put dU into dQ. So dQ = U_V dT + (U_T +p) dV

and i know that c_v = U_V = dU/dT when volume constant. So

dQ = c_v dT + (U_T +p) dV

and dS = dQ/T .

dS = c_v/T dT + 1/T (U_T +p) dV and ds is exact differential

d/dV ( c_v / T) = d/dT ((1/T)(U_T +p)))

i think derivative of c_v/ T by dV when T (Temperature constant) Equal to 0 . but I not sure

I need someone to explain to me. why Eq 1.11 come to Eq 1.12

 

[etotheipi]

Here $U = U(T,V)$ and $P = P(T,V)$. Write:$$\begin{align*}\left( \frac{\partial }{\partial V} \right)_{T} \frac{C_v}{T} &= \left( \frac{\partial }{\partial T} \right)_{V} \left[ \frac{1}{T} \left( \frac{\partial U}{\partial V} \right)_{T} + \frac{P}{T} \right] \\ \\

\left( \frac{\partial }{\partial V} \right)_{T} \left[ \frac{1}{T} \left( \frac{\partial U}{\partial T} \right)_{V} \right]&=

-\frac{1}{T^2} \left( \frac{\partial U}{\partial V} \right)_{T} + \frac{1}{T} \left( \frac{\partial }{\partial T} \right)_{V} \left( \frac{\partial U}{\partial V} \right)_{T} + \left( \frac{\partial }{\partial T} \right)_{V} \frac{P}{T}\end{align*}$$where we used the commutativity of mixed partial derivatives. Rearrange:$$\frac{1}{T^2} \left( \frac{\partial U}{\partial V} \right)_{T}= \left( \frac{\partial }{\partial T} \right)_{V} \frac{P}{T} = - \frac{P}{T^2} + \frac{1}{T} \left( \frac{\partial P}{\partial T} \right)_{V}$$Simplify:$$\left( \frac{\partial U}{\partial V} \right)_{T} = -P + T \left( \frac{\partial P}{\partial T} \right)_{V}$$