# Heat integral and molar heat capacity?

1. Aug 24, 2011

### Inertigratus

dQ = nCvdT if volume is constant.
However, n = pV/RT.
What I don't understand is, why are we thinking n as constant when doing the integral?
I had two problems that involved this on a test I had today. At first I kept it constant and then changed n. But then I thought, wait... isn't there a T in n? then that T should be in the integral.
I understand the point, heat capacity per mole. But mathematically, the T that is in the equation for n should matter, right?
dS = dQ / T, if we substitute dQ in that equation we should get 1 / T2 in the integral also.
I know I'm wrong however, so if someone could tell me what's wrong?

2. Aug 24, 2011

### Mapes

The equation dQ = nCvdT also assumes a closed system (i.e., constant n). Otherwise you could effect a temperature change by simply removing gas molecules at constant volume without heating or cooling the system, and this would violate the equation.

3. Aug 24, 2011

### Inertigratus

That's true... higher temperature with lower amount of moles doesn't sound right. When you say that it assumes a closed system, is that a result of the second law?
Or do we always speak of closed systems when talking about heat capacity?

4. Aug 24, 2011

### Mapes

It's not a result of the second law. It's a typical assumption of the definition of heat capacity; that is, we mean

$$C_{V,N}=T\left(\frac{\partial S}{\partial T}\right)_{V,N}$$

but we generally just write $C_V$.

(In some esoteric circumstances, we want to work with systems at constant chemical potential rather than constant matter, but that's an advanced topic.)