I have a doubt in understanding the concept of internal energy. The internal energy of a system is equal to U=f/2nrT, where T is the temperature. Now, also we know that dU=Q-W Then if heat is zero, magnitude of work done should be equal to that of internal energy. But we see that, W=nRdT Using it both, we get: f/2nRdT=nRdt Now,that's what confuses me. Can anyone clarify where my concept lack?

  • $\begingroup$ Can you explain why $W=nRdT$? $\endgroup$
    – velut luna
    Feb 7, 2018 at 18:31
  • $\begingroup$ Well, if volume and temperature both change then, W=PdV=nRdT. Aint it? $\endgroup$
    – 123IR
    Feb 8, 2018 at 6:18
  • 1
    $\begingroup$ I feel it is worth commenting that, while it is true in the case of a classical idea gas, not all systems obey the equipartition theorem and so the equation $U = \frac{f}{2}nRT$ is not universally applicable. $\endgroup$ Oct 5, 2018 at 11:23
  • $\begingroup$ Please define your symbols. $\endgroup$
    – my2cts
    Apr 13, 2019 at 10:03

2 Answers 2


If heat exchange is zero then the process will be adiabatic and in that case the work done by the gas will NOT be equal to


rather it will be equal to the expression given in the image. if you use the expression given in the image then you get the answer. I am also posting the proof calculation of work done in adiabatic process.

So that is where the problem is lying in your solution.

enter image description hereenter image description here


First of all, if we have an ideal gas then $pV=nRT$, so $p=(nRT)/V$. Then, $W = (nRT)/V dV$. Your Work equation is wrong.

To clarify on how to develop from $dU=dW$ in an alternative way, we can obtain some interesting properties.

We have $dU = dW \rightarrow U = W$ , for an adiabatic process. (It is also known that $C_p=(f/2)nR+nR$ and $C_v=(f/2)nR$ ; Heat capacities for constant pressure and volume, respectively)


$dU = (f/2)nRdT = (C_v)dT , as C_v=(f/2)nR$

Now we can develop the equation:

$(C_v)dT = -((nRT)/V)dV <-> $

$(C_v)dT/T = - ((C_p-C_v)/V)dV$

Integrating both terms, (assuming $T_0$,$V_0$ as initial $T$ and $V$, respectively)

$C_v*ln(T/T_0) = -(C_p-C_v)*ln(V/V_0) <->$

$ln((T/T_0)^{C_v}) = ln((V/V_0)(C_v-C_p)) <->$

$T/T_0 = (V/V_0)^{((C_v-C_p)/C_v)}$

And as $\gamma=(C_v/C_p)$: $T/T_0 = (V/V_0)^{(1-\gamma)}$ ; where $\gamma$ is a greek letter, which represents the adiabatic constant.

We can say that, $TV^{(\gamma-1)} = T_0 V_0 = constant$ We can also obtain $PV^{\gamma}$ = constant applying the Ideal gas law in $TV^{(\gamma-1)} = constant$.

  • $\begingroup$ If the volume and temperature both change then wouldn't work done =nrdT? $\endgroup$
    – 123IR
    Feb 28, 2018 at 12:14
  • $\begingroup$ The process you are mentioned is an adiabatic process and in that process pressure and volume both change and in that case the work done will not be equal to nRdT. $\endgroup$ Dec 18, 2019 at 3:52

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.