[Physics] How is it possible to equate the internal energy at constant volume with the internal energy of an adiabatic process

Question

I hope my question makes sense. My problem is that, I have read through numerous textbooks that $$nCdT = -PdV$$ (where $NC$ is constant volume) when deriving the relationship between $T$ and $V$ for an adiabatic process, though I can't seem to understand why this relation has been made. Surely, at constant volume, work done by the system should be zero? I am lead to believe that, at constant volume, no work is done by the system, but for an adiabatic process, work is only done by the system, not on it. So why is it possible to equate the two internal energies together when they describe two different systems?

Answer 1

Great question; I myself got confused for a moment there. I'm gonna try to be somewhat thorough, so bear with me. First consider the differential form of the first law of thermodynamics which holds for any quasi-static process. $$ dE = \delta Q - \delta W $$ For an adiabatic process, $\delta Q = 0$ by definition, so one obtains $$ dE = -\delta W $$ On the other hand, suppose that we are considering a system (such as an ideal gas with fixed number of particles) whose thermodynamic state can be specified by $(T,V)$, the pair consisting of its volume and temperature. In this case, one obtains $$ dE = \left(\frac{\partial E}{\partial T}\right)_V dT + \left(\frac{\partial E}{\partial V}\right)_TdV $$ It follows that for $\delta W = PdV$ like for an ideal gas, we get $$ \left(\frac{\partial E}{\partial T}\right)_V dT + \left(\frac{\partial E}{\partial V}\right)_TdV = -PdV $$ Now we simply recall the definition of $C_V$ $$ C_V = \left(\frac{\partial E}{\partial T}\right)_V $$ to obtain $$ C_V dT + \left(\frac{\partial E}{\partial V}\right)_TdV = -PdV $$ We're almost there! Your concern basically boils down to how we can eliminate the $dV$ term on the left hand side seeing as how $dV \neq 0$ for an adiabatic process. Well, that's not the only way for that term to vanish. It also vanishes if the partial derivative of the energy with respect to the volume is zero, namely if the internal energy is volume-independent. For some systems, this is in fact the case (like for an ideal gas), in which case we get the desired relation: $$ C_V dT = -PdV $$