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My question differs from these questions:

  1. When is $\Delta U=nC_V \Delta T$ true?

Because here he asks to distinguish between $C_P$ and $C_V$ and not why it's always $C_V$

  1. Work done in adiabatic process

Here the answer is "just because it is always the case"

Is there no mathematical proof for this? What is the real physical intuition for this being the case?

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  • $\begingroup$ Are you looking for mathematical proof that $C_v$ applies for any process involving an ideal gas? $\endgroup$ – Bob D May 9 '20 at 14:59
  • $\begingroup$ That or some intuitive explanation of why it is so? $\endgroup$ – Buraian May 9 '20 at 15:17
  • $\begingroup$ More discussion in The cruelest equation in introductory thermodynamics. Brief answer: we can always write dU in terms of dT and dS, dT and dP, or dT and dV for a closed system. (The reason there are two variables is that there are two ways to add energy to the system: heat it and do work on it.) The ideal gas is unique in that its coefficients for dS, dP, and dV are all zero. Thus, U~T is one of its equations of state. The reason it's the constant-volume heat capacity is that volume is the natural nonthermal variable for the energy U. $\endgroup$ – Chemomechanics May 30 '20 at 0:54
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There are two criteria. First, $c_V=nC_V$ must be constant; second, we must have that $P = f(V) T$ for some function $f$. Both criteria hold in the specific case of an ideal gas, but neither holds for a general thermodynamical system. I'll give the mathematical explanation first, and then the physical explanation second.


Starting from the perspective that $U=U(S,V)$ and $T= \left(\frac{\partial U}{\partial S}\right)_V (S,V)$, note that a small change in $S$ and $V$ will cause small changes $$dU=\left(\frac{\partial U}{\partial S}\right)_V dS + \left(\frac{\partial U}{\partial V}\right)_S dV \equiv TdS - PdV$$

and $$dT = \left(\frac{\partial T}{\partial S}\right)_V dS + \left(\frac{\partial T}{\partial V}\right)_SdV$$

Solving the second equation for $dS$ and substituting it in the first equation yields

$$dU = T \frac{1}{\left(\frac{\partial T}{\partial S}\right)_V}dT- \left[T\frac{\left(\frac{\partial T}{\partial V}\right)_S}{\left(\frac{\partial T}{\partial S}\right)_V} + P\right]dV$$ $$= c_V dT+ \left[T\frac{\left(\frac{\partial P}{\partial S}\right)_V}{\left(\frac{\partial T}{\partial S}\right)_V} - P\right]dV$$

Where we've used that $\left(\frac{\partial T}{\partial V}\right)_S = \frac{\partial^2 U}{\partial V\partial S} = -\left(\frac{\partial P}{\partial S}\right)_V$, and that the definition of the specific heat at constant volume is $c_V \equiv T \left(\frac{\partial S}{\partial T}\right)_V$. Finally, note that

$$\frac{\left(\frac{\partial P}{\partial S}\right)_V}{\left(\frac{\partial T}{\partial S}\right)_V} \equiv \left(\frac{\partial P}{\partial T}\right)_V$$

so finally $$ dU = c_V dT + \left[T\left(\frac{\partial P}{\partial T}\right)_V - P \right]dV$$

Assuming that we are not dealing with variable numbers of particles, what has been written here is completely general, so your question boils down to asking when the second term is zero. The answer is that

$$\left(\frac{\partial P}{\partial T}\right)_V = \frac{P}{T} \implies P = f(V) T$$

for some function $V$. If this is the case, then the second term on the right of the preceding equation vanishes, and we have

$$dU = c_V dT \implies \Delta U = \int c_V dT = \int nC_V dT$$ since $C_V$ is the specific heat per mole. If $C_V$ is constant, then this just becomes

$$\Delta U = nC_V \Delta T$$


From a physical standpoint, the answer is that the energy of an ideal gas is purely kinetic - the gas particles do not have any long-range interactions with each other at all. As a result, since the temperature can be shown to be a measure of the average kinetic energy of the ideal gas particles, the internal energy of the system is unaffected by changes in volume, as long as the temperature is fixed.

This would not be the case if the particles attracted each other, for example. Putting such a system in a larger box with the same amount of kinetic energy would result in a larger average spacing, and therefore a less negative potential energy (remember that attractive potential energies are negative). Therefore, larger box $\implies$ more energy, even if the kinetic energy didn't change.

As I showed, $$ dU = c_V dT + \left[T\left(\frac{\partial P}{\partial T}\right)_V - P \right]dV$$

The first term on the right describes the change in energy due to change in temperature while holding the volume fixed; the second describes the change in energy due to a change in volume while holding temperature fixed. Because of the lack of interaction between gas particles, the second term goes away, leaving only the first, and so

$$dU = c_V dT = nC_V dT$$

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  • $\begingroup$ I would just like to say this answer was amazing except for the differential work part. Did you use the conjugate variable transform on U ? $\endgroup$ – Buraian May 9 '20 at 15:38
  • $\begingroup$ @DDD4C4U I’m not sure to which part you are referring. I didn’t do any Legendre transforms, though. $\endgroup$ – J. Murray May 9 '20 at 15:41
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The average kinetic of the molecules in a gas determines the temperature. This means that the total energy depends only on the temperature and the number of molecules (or moles). In a constant volume process, no work is done, and the heat added is a measure of the change in internal energ. If no head is added or removed, then the change in internal energy can depend only on the work done.

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