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In the formula for the work done by adiabatic expansion of gas, why is the constant-volume heat capacity and not the constant-pressure heat capacity used in calculation? If the system is at constant volume, it is a isovolumetric process and hence no work can be done in that case.

The formula I'm referring to is w = Cv dT.

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Pretty simple, ∆U=nCv∆T no matter what the process is. This is called a state function it is independent of the path taken and only depends on the final and initial conditions here initial and final temperature. However Work done is path dependent so it depend on how you change pressure and volume. So ex: Let's say you take a gas from P1,V1,T1 to P2,V2,T2 you can say ∆U=nCv(T2-T1) however we need to know the process(isothermal,adoabetic,polytropic) for finding work done. Hope this helps :)

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  • $\begingroup$ One caveat... This only applies to an ideal gas. In a real gas, the internal energy is also a function of volume. $\endgroup$ – Chet Miller Oct 15 '17 at 12:08
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Part 1. Let's consider first law of thermodynamics.

$\delta Q = dU + \delta W$ (classical sign convention)

or

$\delta Q = dU - \delta W$ (IUPAC sign convention)

In the case of adiabatic process, $\delta Q = 0$.

$0 = dU - \delta W$

$dU = \delta W$

$dU = C_VdT$ (by definition)

$C_VdT = \delta W$

After integration, assuming CV is constant.

$C_V\Delta T = W$

Part 2. Why not Cp?

By definition, $\delta Q_p = dH = C_pdT$. Obviously, we don't have an isobaric process here.

P. S. Yes, those sign convention differences also make me feel bad.

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