In an Adiabatic process ,$Q=0$ (heat added or removed from the system is 0).

First law of thermodynamics states that , $dU=Q-W$...(1)

$Q=0$ .

Therefore $dU=-W$.

Now my question is why $dU=nC_vdT$ used here ,isn't the Volume changing ?

Why should the work done be equal to $nC_vdT=-dW$.


2 Answers 2


In thermodynamics, the correct equation to use for $C_v$ is in terms of the internal energy U rather than heat Q: $$C_v=\frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_V$$In the case of an ideal gas, U is a function only of T, and not V. So, for an ideal gas, it doesn't matter whether the volume is varying or not, and we can always write for an ideal gas in that $$dU=nC_vdT\tag{ideal gas}$$So, for an adiabatic process of an ideal gas, we have $$\Delta U=nC_v\Delta T=-W$$It's as simple as that.


Yes you are right for the adiabatic process, the volume will definitely change, which may aptly make you question the validity of using $dU=nC_vdT$

Lets say in the given adiabatic process which you are mentioning, the parameters change from (P1,V1,T1) to (P2,V2,T2) and due to which some internal energy change occurs. If you plot this two parameters on a P-V graph, these are exactly two points on the graph. You can form infinite number of curves connecting these two points, where each of the curves represent different process to change the parameters. Recall that internal energy is a state function which is it depends only an initial and final states. So if we find $dU$ for any one of the curves connecting the two points, we will find $dU$ for the adiabatic process also.

One of such curves is when first there is an isobaric process from (P1,V1,T1) to (P1,V2,Tintermediate) and then an isochoric process from (P1,V2,Tintermediate) to (P2,V2,T2). Total $dQ$ here is $nC_p$(Tintermediate-T1) + $nC_v$(T2-Tintermediate). Total $dW$ = -P1(V2-V1). Using first law of thermodynamics and some algebraic manipulations, you will get $dU=nC_v$(T2-T1).(I suggest you to draw a graph from the data and visualise it out, I couldn't add graphs due to shortage of time from my side)

The later part of the question where you ask why $dW=-nC_vdT$ in the adiabatic process, the answer is again first law of thermodynamics. In adiabatic process $dQ=0$, so if internal energy changes by some amount and no heat is exchanged, the gas has to do the work equal to the amount of internal energy change.

Edit - I am adding the algebra part which I initially omitted in my answer

$dQ$ = $nC_p$(Tintermediate-T1) + $nC_v$(T2-Tintermediate)

$dQ$ = $n(C_v+R)$(Tintermediate-T1) + $nC_v$(T2-Tintermediate)

= $nR$(Tintermediate-T1)+ $nC_v$(T2)-$nC_v$(T1)

$dW$ = P1(V2-V1)

= -nR(Tintermediate-T1)

$dU = dQ + dW$ = $nC_v$(T2)-$nC_v$(T1)

= $nC_v$(T2-T1)

  • $\begingroup$ I have attached the algebra in the following pic ,could you say where have I gone wrong .imgur.com/a/n2LTbCg $\endgroup$
    – Harry Case
    Sep 10, 2022 at 5:53
  • $\begingroup$ I have added the algebra in my latest edit. Feel free to ask if you don't understand any point. $\endgroup$
    – Sam
    Sep 10, 2022 at 7:32
  • $\begingroup$ Thank you ,I have got it . $\endgroup$
    – Harry Case
    Sep 10, 2022 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.