Heat capacity at constant volume and energy

Question

So, my understanding of heat capacity at constant volume is that

$$C_V = \bigg(\frac{\delta Q}{dT}\bigg)_V$$

So far, so good. But according to my professor,

$$dU = C_VdT$$

for an ideal gas, even when work is performed, that is, when $\delta Q\neq dU$. I see that there are a lot of explanations in this forum as to why the above formula is true, so please note that I am NOT asking that again. What I am asking for is some help understanding how can $C_V$ be equal to two different things at the same time... That looks like a contradiction. Any help would be appreciated!

Answer 1

I can feel the confusion, at first it seems contradictory, but you actually need to know about nature of $\Delta U$.

There are two functions in thermodynamics:

1. State Functions: These functions depend only on state variables of system (which describe state of the system like $P,V,T$). $\Delta U$ belongs here.

2. Path Functions: These functions depend on path taken to reach one state of system to other state of the system. Path can be Isobaric, Isothermal, Adiabatic, Isochoric, Polytropic or other complex paths. $\Delta Q,\Delta W$ belong here.

Now as you said: $$C_v=\left(\frac{\Delta Q}{n\Delta T}\right)_v$$

Now from first law of thermodynamics:$\Delta U=\Delta Q-\Delta W$ In isochoric process (volume is constant as described by our first equation), $\Delta W=0$ (Path function). Hence:$\Delta U=\Delta Q$

Now: $$C_v=\left(\frac{\Delta U}{n\Delta T}\right)$$ Hence: $$\Delta U=nC_v\Delta T$$

So while we found this equation in isochoric situation, but $\Delta U$ is state function, it doesn't care which path system is going through! so this established equation is valid for any ideal gas system taking any kind of path.