Thermo Q:
dU=$$(\frac{dU}{dT})_{V}$$dT+$$(\frac{dU}{dV})_{T}$$dV
First law: dU=dq-PdV
Solve for dq, ($$C_{V}$$=$$(\frac{dU}{dT})_{V}$$.
This gives dq=($$C_{V}$$+[$$(\frac{dU}{dV})_{T}$$+P]dV. I don't understand the steps to get - [$$(\frac{dU}{dV})_{T}$$+P]dV

Question

Thermo Q: 
dU=$$(\frac{dU}{dT})_{V}$$dT+$$(\frac{dU}{dV})_{T}$$dV
First law: dU=dq-PdV
Solve for dq, ($$C_{V}$$=$$(\frac{dU}{dT})_{V}$$.
This gives dq=($$C_{V}$$+[$$(\frac{dU}{dV})_{T}$$+P]dV. I don't understand the steps to get -> [$$(\frac{dU}{dV})_{T}$$+P]dV

anonymous · Answer

It's quite confusing because of the way you formated it. What is the name of this equation and what's it's purpose? I haven't seen this before.

anonymous · Answer

Yeah, i don't know how the format got so messed up. I typed it in normally and it came out that way. Maybe the spaces between terms implies in code that they are separate equations. It is an equation for the infinitesimally small change in internal energy of a system. It is from thermodynamics.

anonymous · Answer

That's alright. Thanks for at least acknowledging my question. It is annoying when no one replies but tons of people flock to answer multiple choice questions on solving for x!

anonymous · Answer

I will work on figuring it out but it might take me a while. Hopefully someone else can help.