When we try to establish a relation between the pressure and temperature in adiabatic process we come across a equation..

$dU = dq - PdV$

$dq=0$ (Adiabatic process) and,

$dU=C_v.dT$ (Heat capacity at constant volume)

Therefore, $C_v.dT = -PdV$$\tag1$

In this equation we are using heat capacity defined at constant volume but their still is some work done by the system (i.e, $PdV$ is not $0$ or $dV$ not equals to $0$).

The first part of the equation $(1)$ is implying that the volume is constant but the second part is implying that the volume is not constant (if it was there would be no work done).

Then why there is this contradiction ?


Even though we call $C_v$ the heat capacity at constant volume, what we really mean by the subscript v is that this is the way we measure $C_v$. At constant volume, we can determine the heat capacity of the material by measuring the heat transferred $dQ=dU=C_vdT$.

But this same heat capacity also applies to all other situations for an ideal gas if we recognize that, for an ideal gas, $U$ depends only on temperature, such that $dU=C_vdT$. It is just that, in these other situations (involving work), dQ is not equal to $C_vdT$.

  • $\begingroup$ According to this , since you say "heat capacity is same for all situations" (quoted from your answer) then Cp should be equal to Cv since Cp becomes one of the situation where volume is changing. But this disagrees with Cp-Cv=R. I'm a little too confused at this point. Could you please explain in better terms? $\endgroup$
    – hegde98
    Apr 16 '17 at 10:19
  • 1
    $\begingroup$ $C_p$ and $C_v$ are not the same, and, in thermodynamics, neither of them is defined in terms of dQ. The proper definitions in thermodynamics are $C_p=(\partial H/\partial T)_p$ and $C_v=(\partial U/\partial T)_v$. For an ideal gas (whose internal energy and enthalpy depend only on temperature), $dU=C_vdT$ and $dH=C_pdT$ for any arbitrary process path. $\endgroup$ Apr 16 '17 at 11:12
  • $\begingroup$ Would you mind editing the answer here ? I like the explanation in your new post. (+2) $\endgroup$
    – Mitchell
    Sep 19 '17 at 20:51
  • $\begingroup$ At different times, I've tried to explain this in different ways. Some work better than others. For some reason, this is a very difficult concept to explain. There are other posts I've written that provide even more detail. Since this is from almost a year ago, I don't think many people go back to it. I'm unmotivated to go back now and make changes. $\endgroup$ Sep 19 '17 at 22:13

I will just expand on what Chester Miller had already said.

The molar specific heat at constant volume $C_v$ is basically the amount of heat required to raise the temperature of 1 mol of the gas by $1$ K while keeping the volume constant.

When we apply heat to an ideal gas, some of that heat is used to increase the internal energy $U$ of the gas and the rest is eventually used up to do the work to increase its volume. (Note that increasing U essentially means increasing the temperature since U is a function of T for any ideal gas.) Now if we were to keep the volume constant somehow, then the total applied heat would be used only to increase the internal energy or so to speak the temperature and nothing else.

So a different way to define $C_v$ is to say, it is the amount of heat required to raise the temperature of 1 mol of the gas by $1$ K, where the said heat is used to only to change the internal energy U or so to speak the temperature and nothing else.

Now let's consider a case where change in volume is allowed. Say we applied $dQ$ amount of heat to the a gas. Now we know some of that heat is going to account for the change in internal energy, let's say that amount is $dQ_u$. And the rest of the heat will eventually account for the change in volume, and say that is $dQ_w$. So instead of writing $dQ = dU + dW$ we write, $$dQ = dQ_u +dQ_w$$

So we can see that the heat $dQ_u$ is only used to change the internal energy and not the volume and so $dQ_u = dU$ and we can say according to the definition of $C_v$ the following $$C_v = \frac{dQ_u}{mdT}\ \Rightarrow\ dQ_u = mC_vdT\ \Rightarrow\ dU=mC_vdT$$

So, you see $dU = mC_vdT$ whether or not there is any change in volume. The same argument goes for the adiabatic process you mentioned.

  • $\begingroup$ Greatly done. Thanks for giving some time to this question. $\endgroup$
    – Mitchell
    Jan 8 '18 at 7:43

Thermal capacity at constant volume $\textrm C_V$ is defined as

$$\mathrm C_V ~=~\left(\frac{\partial U}{\partial T}\right)_V$$ where $U$ is the internal energy of the system.

For an ideal gas, $U=U(T);$ so, $$\mathrm C_V~\mathrm dT ~=~ \mathrm dU\,.\tag I$$

Substituting $\mathrm{(I)}$ in the First Law of Thermodynamics,

$$\mathrm C_V~\mathrm dT +đw~=~ đQ\tag{II} $$

From which, for an adiabatic process, $$\mathrm C_V~\mathrm dT ~=-~P~\mathrm dV,$$

which is actually $$\mathrm dU ~= -~P~\mathrm dV\,.$$

Nothing is contradictory here.

  • 1
    $\begingroup$ In your second line(the first equation) you've taken (dU/dT) at constant volume (you have specified using subscript 'v' which is why you get Cv). Now since volume is defined to be a constant, dV will be 0. Thus giving you work done=0. Isn't that contractory? $\endgroup$
    – hegde98
    Apr 16 '17 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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