# How is it possible to equate the internal energy at constant volume with the internal energy of an adiabatic process?

I hope my question makes sense. My problem is that, I have read through numerous textbooks that $$nCdT = -PdV$$ (where $NC$ is constant volume) when deriving the relationship between $T$ and $V$ for an adiabatic process, though I can't seem to understand why this relation has been made. Surely, at constant volume, work done by the system should be zero? I am lead to believe that, at constant volume, no work is done by the system, but for an adiabatic process, work is only done by the system, not on it. So why is it possible to equate the two internal energies together when they describe two different systems?

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Great question; I myself got confused for a moment there. I'm gonna try to be somewhat thorough, so bear with me. First consider the differential form of the first law of thermodynamics which holds for any quasi-static process. $$dE = \delta Q - \delta W$$ For an adiabatic process, $\delta Q = 0$ by definition, so one obtains $$dE = -\delta W$$ On the other hand, suppose that we are considering a system (such as an ideal gas with fixed number of particles) whose thermodynamic state can be specified by $(T,V)$, the pair consisting of its volume and temperature. In this case, one obtains $$dE = \left(\frac{\partial E}{\partial T}\right)_V dT + \left(\frac{\partial E}{\partial V}\right)_TdV$$ It follows that for $\delta W = PdV$ like for an ideal gas, we get $$\left(\frac{\partial E}{\partial T}\right)_V dT + \left(\frac{\partial E}{\partial V}\right)_TdV = -PdV$$ Now we simply recall the definition of $C_V$ $$C_V = \left(\frac{\partial E}{\partial T}\right)_V$$ to obtain $$C_V dT + \left(\frac{\partial E}{\partial V}\right)_TdV = -PdV$$ We're almost there! Your concern basically boils down to how we can eliminate the $dV$ term on the left hand side seeing as how $dV \neq 0$ for an adiabatic process. Well, that's not the only way for that term to vanish. It also vanishes if the partial derivative of the energy with respect to the volume is zero, namely if the internal energy is volume-independent. For some systems, this is in fact the case (like for an ideal gas), in which case we get the desired relation: $$C_V dT = -PdV$$

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Excellent! Thank you so much! I don't know why they leave these out of textbooks. I've read through two and one set of lecture notes and they all quickly jump to the conclusion that we can equate the two without questions asked. Thank you so much for taking the time to explain. This has clarified a lot. :) – Sanyia Apr 25 '13 at 21:54
@Sanyia Sure thing. I also find it frustrating when physics books leave out logical details; I guess we just have to deal with it because we're not studying pure math :( – joshphysics Apr 25 '13 at 23:46
@joshphysics is quite right. I add two notes. One is that the resulting formula hold ONLY for ideal gases. The other is that in $w = -pdV$, $p$ is the internal pressure ONLY if the process is reversible, because only then are the internal and external pressures equal. – Paul J. Gans Apr 26 '13 at 1:48
@PaulJ.Gans Thank for the additions. I'm not sure I agree about the comment that this only holds for ideal gases. The internal energy of, for example, a hard core gas which obeys the equation of state $P(V-Nb) = NkT$ has the property that $\left(\frac{\partial E}{\partial V}\right)_T=0$, so I think that in this case the reasoning goes through just fine. – joshphysics Apr 26 '13 at 1:53

There are two approaches to the definition of $C_V$, one of which displays its physical interpretation---$C_V=T \frac{\partial S}{\partial T}|_V$, i.e., as the partial derivative of entropy with respect to temperature under constant volume, scaled up by the temperature. A second one is less intuitive, but much easier to calculate and avoids some of the usual pitfalls, namely as $f \frac {g_p}{f_p}$ (the subscripts denote partial derivatives with respect to $p$). This has the advantage that it works for any equations of state (try it on the van der Waals gas, better still try it for $C_p=f\frac {g_V}{g_V}$ for a wee surprise) for a general pair of equations of state $T=f(p,V)$ and $S=g(p,V)$. Thus for the ideal gas, we have $$T=f(p,V)= \frac 1 {RN} p V, \quad S= g(p,V)= \frac {RN}{\gamma-1}\ln (pV^\gamma)$$ and this easily gives the above formula. Note that it is necessary to know both equations of state to compute $C_V$, not just $T$ as a function of $p$ and $V$ as seems to be implied by some answers. In fact for any function $f$ and any expression for $C_V$ as a function of $p$ and $V$, there are a many entropy functions which gives this $C_V$, namely, those of the form $$g(p,V) ] \int \frac f{f_1} c_V(p,V) dp + h(V)$$ for an arbitrary function $h$ of $V$. In particular, there are many possible forms of the equations of state which lead to a constant $C_V$, not just those of the ideal gas. This is, of course, a mathematical result---there is no implication that all of such solutions are thermodynamically realistic---though I see no reason why this should not often be the case and one example can be found in the answers above.

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Thank you joshphysics for this wonderful answer. I myself was confused like Sanyia. I just wanted to add in a point on what caused this confusion. In the left-hand side of the equation, when we see the term $cv$, we automatically assume that the all terms should have this same condition. But this is wrong because joshphysics pointed out that $$(dE/dT)v dT + (dE/dV)t dV = -PdV$$ or that $$cv + (dE/dV)t dV = -PdV$$ This equation shows that even though the first term on the left-hand side ($cv$) has a constant volume condition, this condition does not apply to the second term of the left-hand side $(dE/dV)t dV$ or the term on the right-hand side ($-PdV$). Although this is not a rigorous mathematical proof, I hope this helps with your understanding.

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