Heat capacity at constant volume

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  • #1
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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?


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  • #2
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[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?

From Wikipedia:
Wikipedia 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.
http://en.wikipedia.org/wiki/Heat_capacity

This is consistent with your assumption.
 
  • #3
Andrew Mason
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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?
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.

AM
 
  • #4
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But if the gas isn't ideal, then again dV=0 for constant volume and the second term vanishes.
 
  • #5
Andrew Mason
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But if the gas isn't ideal, then again dV=0 for constant volume and the second term vanishes.
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.

AM
 

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