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When I took this class years ago, I simply accepted it as fact. However, now that I'm teaching it, the following bugs me....a lot.

Why do they use Cv to describe the change of internal energy during the carnot cycle?

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/Thermodynamic_Cycles/Carnot_Cycle

There is obviously a volume change. But it is defined that $\Delta U = n*Cv*\Delta T$. How do you justify the use of Cv when there's a volume change?

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It is always true that, for an ideal gas, $\Delta U = C_V \Delta T$, regardless of the process. Remeber, we define $C_V=(\delta Q/dT)_V$. Since this is happening at constant volume (aka $\delta W=0$), we have $C_V=(\delta Q/dT)_V=(dU/dT)_V$. Then, since $U$ doesn't depend on volume for an ideal gas, we have that $C_V=dU/dT$ even if volume is changing. So $dU=C_V dT$.

Another way to think about it, just using algebra: for an ideal gas, $U=\alpha NkT$. Thus $\Delta U=\alpha NK\Delta T$. But of course $\alpha NK$ is just $C_V$ for an ideal gas, proving the statement.

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  • $\begingroup$ I understand now how there is no constant volume dependence for the ΔU case. Thanks. Wouldn't it be easier as a teacher to not introduce the definition of Cv=(δQ/dT)V and Cp=(δQ/dT)P? Just go directly to ΔU=CvΔT and ΔH=CpΔT. I know it shows competence in Thermodynamics algebra to convert from one to the other, but I'd rather just define Cv and Cp on the equations engineers will use the most. Is there a fault for skipping the Cv=(δQ/dT)V and Cp=(δQ/dT)P definition all together? $\endgroup$
    – Ted Yu
    Aug 27, 2015 at 6:34
  • $\begingroup$ @TedYu I think the usual definition of $C_V$ helps motivate the definition a little, and certainly helps explain the name of the constant! $\endgroup$ Aug 27, 2015 at 13:32
  • $\begingroup$ BTW, in Atkins physical chemistry, it says explicitly that ΔH=n∗Cp∗ΔT only applies at constant pressure, which I believe from our discussion is incorrect. $\endgroup$
    – Ted Yu
    Aug 31, 2015 at 20:44
  • $\begingroup$ You might be interested in looking at this question too: physics.stackexchange.com/questions/203605/delta-h-c-p-delta-t The first answerer seems to be contradicting what you just answered. $\endgroup$
    – Ted Yu
    Aug 31, 2015 at 22:28
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In an adiabatic expansion there is not heat exchange, technically the gas cools down because it does work. We are using $C_v$ here only as means to account for the change in internal energy, which happens to be equal than in a constant volume process. So we used $C_v$ here not because there is a constant volume (there is not) but because is equivalent to a process carried out at constant volume where there is the same temperature change.

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