The discussion revolves around the relationship between internal energy change (dU) and heat capacity at constant volume (C_V) in thermodynamics. Participants explore whether the equation dU = C_V dT holds true under various conditions, particularly when volume is held constant.
There is an ongoing exploration of the conditions under which the equation holds true, with some participants suggesting that it is valid for ideal gases while questioning its applicability for non-ideal gases. The discussion reflects differing interpretations of the relationship based on the nature of the gas involved.
Participants note the importance of holding volume constant and the implications of this condition on the derivation of the equation. There is also mention of specific gas behaviors, such as those described by the Van der Waals equation, which may affect the validity of the relationship.
Grand said:[tex]dU=C_VdT+\left(\frac{dU}{dV}\right)_TdV[/tex]
However, if we hold the volume constant, dV=0 and therefore the second term disappears, is this right?
http://en.wikipedia.org/wiki/Heat_capacityWikipedia said:If the process is performed at constant volume, then the second term of this relation vanishes and one readily obtains,
[tex]\left(\frac{\partial U}{\partial T}\right)_V=\left(\frac{\partial Q}{\partial T}\right)_V=C_V[/tex]
This defines the heat capacity at constant volume, CV.
Are you speaking about an ideal gas? If so, [itex]dU=C_VdT[/itex] 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.<br /> <br /> AM[/itex]Grand said:Homework Statement
The question asks whether it is always true that
[tex]dU=C_VdT[/tex]
and their answer is no, because from the 1st law of TD we derive that:
[tex]dU=C_VdT+\left(\frac{dU}{dV}\right)_TdV[/tex]
However, if we hold the volume constant, dV=0 and therefore the second term disappears, is this right?
By definition if volume is constant then dU = dQ = CvdT. That applies to any gas. The issue is whether it is always true that dU = CvdT, ie. in situations in which V is not constant. For an ideal gas it is always true.Grand said:But if the gas isn't ideal, then again dV=0 for constant volume and the second term vanishes.