I have looked at the proof of this relation $PV^\gamma = C$; (where $P$ is pressure and $V$ is volume) in quite some places but I am not able to understand the logic behind the third step.


$W= -P\Delta V$

If $\Delta T$ is the fall in temperature then $C_V\Delta T = -P\Delta V$, where $C_V$ means specific heat at constant volume. This is where I am having problem. I know here adiabatic expansion is there so $q=0$ that is $\Delta U$ or $\Delta E = W$ ,but how can we write $W = C_V\Delta T$? (Doubt 1)

Also even though I don't understand how this is written but I still know $C_V$ is heat capacity at constant volume but volume is changing here(expansion), so how can we use $C_V$ which is meant to be used at a specific volume only?? (Doubt 2)


3 Answers 3


The internal energy of a mono-atomic gas is given by:


Where $n$ is the number of moles and $R$ is the gas constant and $T$ is the temperature.

The statement of conservation of energy is given by:


The work done by the gas is given by $W=-\int PdV$.

For a gas undergoing temperature variation at constant volume, the work $W$ done by it, is of course zero, therefore

$\Delta E_{\text{int}}=Q=\dfrac{3}{2}nR\Delta T$.

Now we define the molar heat capacity at constant volume $C_V$ as the amount of heat $Q$ required to raise the temperature of one mole of a gas by $1$ degree at constant volume. Therefore it follows from the last equation that:

$C_V=\dfrac{3}{2}R$ for one mole of any monoatomic gas.

For an adiabatic process applied to one mole of a gas, by definition $Q=0$ therefore it follows from $E_{\text{int}}=Q+W$ that

$\Delta E_{\text{int}}=W=\dfrac{3}{2}R\Delta T=-P\Delta V$.

Since $\dfrac{3}{2}R=C_V$.


$C_V\Delta T = -P\Delta V$.

For any given mono-atomic gas, since it's always the case that $C_V=\dfrac{3}{2}R$, therefore you can use it whenever you like, whether the process under question is isovolumetric or not.

  • $\begingroup$ Can you please tell me How on or what basis did you write E =3/2nRT in the first line I only know ∆U or E =q+w and ,if you would tell me why you wrote E=3/2NRT I think your answer will start making much more sense for me!! Or at least give me a link: on where I can study why for a monoatomic gas E=3/2NRT??any help ?? $\endgroup$
    – Freelancer
    Oct 16, 2015 at 2:14
  • $\begingroup$ Also you have talked only about monoatomic gas,and even when you extend it to diatomic or polyatomic molecules still this is a sort of example while I think c(v)×∆T=P×∆v is a general result and it is applicable to any molecule??right?? If no why?? And if yes ,,can you give the general proof**one which doesn't use advanced calculus ?? $\endgroup$
    – Freelancer
    Oct 16, 2015 at 5:38
  • $\begingroup$ It's a general result that the internal energy of any gas is proportional to its temperature, that is $E_{\text{int}}=\dfrac{f}{2}nRT$, with $f$ being the degrees of freedom the gas have. A mono-atomic gas has 3 degrees of freedom(it execute translatory motion along three axes), hence $f=3$. One can prove this result assuming the ideal gas law $PV=nRT$ and kinetic theory of heat. Here's a link to the result:galileo.phys.virginia.edu/classes/252/kinetic_theory.html. $\endgroup$
    – Omar Nagib
    Oct 16, 2015 at 8:12
  • 2
    $\begingroup$ @Freelancer My advice to you is that you should start reading thermodynamics from some textbook. You lack the background(both physical and mathematical) to follow the proof. $\endgroup$
    – Omar Nagib
    Oct 16, 2015 at 8:16
  • $\begingroup$ @OmarNagib My thoughts exactly $\endgroup$ Oct 16, 2015 at 10:29

We know that for an isochoric process $C_v= (\frac{\partial U}{\partial T})_v$; i.e. $dU=C_vdT$ at constant volume.

Now first law of thermodynamic is $\bar{d}Q=dU+dW;$ For adiabatic process $\bar{d}Q=0$.
$\therefore$ $dU=-dW$ or, $dU=-pdV$

Now $dU=C_vdT;$

If we allow to change the volume of the system then the system doing a work against the surrounding pressure and for this work we have to supply heat form out side. In case of adiabatic process the energy for work provides the internal energy of the gas and that's why the temperature of the gas reduces. Now this reduce of internal energy must be, $dU=C_vdT;$
So, $dU=C_vdT=-pdV$

If we rise the temperature of the gas by $dT$ keeping volume constant then we have to supply $C_vdT$ amount of heat which would increase the internal energy of the gas by $dU$. Now if I expand this gas adiabatically and the temparature of the gas reduced by $dT$ then we can say that the externally supplied heat in isochoric process is converted in work to expand the gas. So the change of internal energy $dU$ must be $C_vdT$

  • $\begingroup$ Sorry @rajesh but I think you couldn't get what I was asking //please read the question //I didnt want a proof the relation I just wanted someone to clarify the part I was asking but I don't know what all you have written and am not able to understand it $\endgroup$
    – Freelancer
    Oct 15, 2015 at 18:23
  • $\begingroup$ Also on the very first line you talked of isochoric process and have used the relations till the end but the process here is not isochoric !!otherwise ∆v would have been equal to zero and so would have work , but it is not!! $\endgroup$
    – Freelancer
    Oct 15, 2015 at 18:26
  • $\begingroup$ I've explain why $C_vdT=dU$ in last para. If you understand that I think you will able to know why $C_v$ come in adiabatic derivation. $\endgroup$ Oct 15, 2015 at 18:35
  • $\begingroup$ Sorry but with so many derivatives it becomes really difficult to see things will try reading the answer again and again till I don't understand it,Thnx for the effort but it will become easy for me if you put in some more words instead of using so may symbols like that c(v) =something ..something in first line I havent seen that odd looking reversed 6 symbol anywhere what does it mean?? $\endgroup$
    – Freelancer
    Oct 15, 2015 at 18:44
  • $\begingroup$ That symbol defines partial derivative at constant volume. I can't able to write that properly. And I'm sorry also for making my ans complicated. I'll deleted my ans if you want. $\endgroup$ Oct 15, 2015 at 18:49

The proof for the adiabatic process can be understood much more simply if you use calculus, specifically differentials of multivariable functions, as demonstrated by the answer from Rajesh Sardar.

For example, $W=-P\Delta V$ is incorrect in this case since pressure isn't constant during the process (a fact you can check by simply looking at an adiabat in a P-V diagram). It's just a special case (for the isobaric process) of the general formula $\delta W=-pdV$ and the first law of thermodynamics. The standard theory of thermodynamics also explains why for an ideal gas we always have $U=C_VT$, even when the process isn't isochoric.

What you are trying to do (derive the formula for the adiabatic process without calculus) is very hard, if not impossible. I suggest you delve more deeply into the mathematics of it all and try figuring out the problem in a few years.

  • $\begingroup$ So you mean to say I should write w=∆pv and that w=p∆v is wrong because process is reversible $\endgroup$
    – Freelancer
    Oct 15, 2015 at 18:35
  • $\begingroup$ No, $W=P\Delta V$ is wrong because the process isn't isobaric. It has nothing to do with reversibility. $\endgroup$ Oct 15, 2015 at 18:36
  • $\begingroup$ And that there is a special reason why or how U=c(v)×T can you tell me something about this reason or at least provide a link where I can study it?? $\endgroup$
    – Freelancer
    Oct 15, 2015 at 18:36
  • $\begingroup$ For your second comment why cant I say w=p×∆v is wrong on basis of reversibility because as far as I understand it in reversible process external pressure is not constant it changes continuously isn't this and your isochoric claim just the same?? $\endgroup$
    – Freelancer
    Oct 15, 2015 at 18:38
  • 1
    $\begingroup$ @Sobanoodles I'm really feeling deep sympathy for Freelancer at this point. He (or she?) is asking for a more firm definition and the physics behind the math - not to be buried more in the math. Freelancer - doesn't Omar's answer better approach what yo9u are looking for? $\endgroup$
    – docscience
    Oct 15, 2015 at 19:20

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