Question regarding using the expression $dE_{int}$=nC_vdT$

[Harikesh_33]

The first law of Thermodynamics states that the change in Internal energy is equal to the sum of Heat gained or lost by the system and work done by the system or on the system .

$dE=Q-W$...(1).

In an Adiabatic process ,Q=0 .

Therefore $dE=-W$ .

Now (https://phys.libretexts.org/Bookshelves/University_Physics/Book:_University_Physics_(OpenStax)/Book:_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/03:_The_First_Law_of_Thermodynamics/3.07:_Adiabatic_Processes_for_an_Ideal_Gas)

here specific heat capacity at Constant volume is used instead of internal energy (ie) $dE=nC_vdT$ .

How can this specific heat be used here isn't the Volume changing (through Work done ?) .

How can $nC_vdT$=-dW$ be used ?

 

[Steve4Physics]

How can this specific heat be used here isn't the Volume changing (through Work done ?) .

How can $nC_vdT$=-dW$ be used ?

This is a common source of confusion. Here are a couple of links which might help:
https://engineering.stackexchange.c...-volume-used-in-non-constant-volume-processes
http://john.maloney.org/cruelest_equation.htm

 

[Steve4Physics]

@Harikesh_33, it's worth noting that to make Latex render correctly here, you enclose the code between a pair of double hash-signs ($\text {$Your Latex code here$}$).

For example, doing this for nC_vT=-dW gives $nC_vT=-dW$.

Or you can similarly use a pair of double dollar-signs to render the code on its own line. For example $$nC_vT=-dW$$Use the preview-toggle (top right on edit-toolbar) to check before posting.

 

[Chestermiller]

In thermodynamics, the correct equation to use for $C_v$ is in terms of the internal energy E rather than heat Q: $$C_v=\frac{1}{n}\left(\frac{\partial E}{\partial T}\right)_V$$In the case of an ideal gas, E is a function only of T, and not V. So, for an ideal gas, we can write that $$dE=nC_vdT\tag{ideal gas}$$So, for an adiabatic reversible process of an ideal gas, we have $$dE=nC_vdT=-dW$$It's as simple as that.