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Why is $dE=c_vdT$ valid for an ideal gas undergoing a reversible, adiabatic process? If the process is adiabatic, then there is no heat transfer, $dQ=0$, and from the first law $dE=dW$. This implies that there is work being done when changing the internal energy of the gas, ergo, that the volume changes (we don't consider work done against friction because the gas is ideal), $dW=-PdV$. However, the specific heat capacity in $dE=c_vdT$ indicates that the the change in internal energy is an isochoric process.

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In thermodynamics, we define the heat capacity Cv (a physical property of the material) in terms of the internal energy (a related physical property of the material), not in terms of the amount of heat transferred Q (not a physical property of the material). This corrects an error of what they taught us in freshman physics, where the heat capacity was described as being related to the amount of heat. The heat Q can be related to the heat capacity (and the change in internal energy) only if no work is done. This does not include most of the cases we encounter in thermodynamics.

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  • $\begingroup$ As I cite in the note I linked to above, the framework you describe tends to lead new thermo students (as exemplified by previous posts linked within) to conclude that if $C_V=(\partial U/\partial T)_V$, then analogously $C_P=(\partial U/\partial T)_P$, which is incorrect. If they defined $C_V=(\delta Q/\partial T)_V=(\partial U/\partial T)_V$, then they'd have a better chance of correctly obtaining $C_P=(\delta Q/\partial T)_P=(\partial H/\partial T)_P$. $\endgroup$ – Chemomechanics May 24 '18 at 20:27
  • $\begingroup$ I was only trying to keep the explanation simple, and at the level of the OP. I’m well aware the Cp is the partial derivative of H with respect to T. $\endgroup$ – Chet Miller May 24 '18 at 20:38
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Chester Miller is correct. Heat capacity $C_v$ is a property related to the internal energy property. For an ideal gas (closed system), the change in internal energy (I will use u instead of E) is given by

$\Delta u=c_v\Delta T$ (for constant $c_v$)

But just because it is given in terms of a constant volume process does not mean it only applies to a constant volume process. Similar confusion applies to entropy. A change in entropy is defined in term of a reversible transfer of heat divided by the temperature at which it is transferred. But just because it is defined for such a process does not mean entropy change requires heat transfer. A classic example is the free expansion of an ideal gas into a vacuum within a perfectly insulated rigid chamber. The important point is that internal energy (entropy, enthalpy, etc.) is a state property. Changes in state properties between two equilibrium states does not depend on the path that gets you there.

If we can show that $P\Delta v=c_v\Delta T$ for two different paths between to equilibrium states, I believe that should answer your question. Consider initial state 1 and final state 2.

For a constant volume process from 1 to 2 $\Delta u=c_v\Delta T$.

Let’s take a different path. Let’s do an isothermal compression of the gas from state 1 to some intermediate state, say 1A. We do the compression until the final pressure is $P_2$ (the same pressure as at the end of the constant volume process). For the isothermal process, $\Delta u=0$.

In order to get from state 1A to state 2 we can do a constant pressure expansion work to return the volume to that at state 2. The work done by the gas is

$w=P\Delta V$

Since there was no change in internal energy in going from state 1 to state 1A, the change in internal energy from state 1A to 2 must equal the change from state 1 to 2.

Ergo

$w=P\Delta V=c_v\Delta T=\Delta u$

Hope this helps

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