Heat capacity at constant volume

Homework Statement

The question asks whether it is always true that
$$dU=C_VdT$$

and their answer is no, because from the 1st law of TD we derive that:
$$dU=C_VdT+\left(\frac{dU}{dV}\right)_TdV$$

However, if we hold the volume constant, dV=0 and therefore the second term disappears, is this right?

The Attempt at a Solution

$$dU=C_VdT+\left(\frac{dU}{dV}\right)_TdV$$

However, if we hold the volume constant, dV=0 and therefore the second term disappears, is this right?

From Wikipedia:
Wikipedia said:
If the process is performed at constant volume, then the second term of this relation vanishes and one readily obtains,

$$\left(\frac{\partial U}{\partial T}\right)_V=\left(\frac{\partial Q}{\partial T}\right)_V=C_V$$

This defines the heat capacity at constant volume, CV.
http://en.wikipedia.org/wiki/Heat_capacity

This is consistent with your assumption.

Andrew Mason
Homework Helper

Homework Statement

The question asks whether it is always true that
$$dU=C_VdT$$

and their answer is no, because from the 1st law of TD we derive that:
$$dU=C_VdT+\left(\frac{dU}{dV}\right)_TdV$$

However, if we hold the volume constant, dV=0 and therefore the second term disappears, is this right?
Are you speaking about an ideal gas? If so, $dU=C_VdT$ is always true, whether volume is constant or not. [itex]\left(\frac{\partial U}{\partial V}\right)_T = 0[/tex] for an ideal gas. This is not so for gases that obey the Van der Waals equation, for example.

AM

But if the gas isn't ideal, then again dV=0 for constant volume and the second term vanishes.

Andrew Mason