1
$\begingroup$

I read that heat supplied at constant volume to a system containing ideal gas changes the internal energy of system, i.e $q=nC_v\Delta T=\delta U$

But what happens if I heat a system and change its temperature from $T_1$ to $T_2$ with system not being at constant volume, can I still say that $\delta U=nC_v\Delta T$? Is there any proof regarding this?

$\endgroup$
4
$\begingroup$

Yes, you can apply the formula $\Delta U$=nCv$\Delta T$ for an ideal gas in a container whether or not the process is isochoric.

This is because the change in internal energy for an ideal gas undergoing some mechanical process is mainly affected by the change in kinetic energy of the gas. It does not depend on the change in volume of gas.So as long as the temperature difference is same the change in internal energy will be same. Heat capacity is the amount of heat required to raise the temperature of a substance by 1 unit so $\Delta U$ is directly proportional to heat capacity and $\Delta T$ along with the number of moles of gas.

When the process is carried with constant volume only then $\Delta W$ is $0$ so in that case $\Delta Q$=$\Delta U$ whereas $\Delta U$=nCv$\Delta T$ holds good always for an ideal gas under some mechanical process.

$\endgroup$
  • 1
    $\begingroup$ To elaborate on this, the internal energy of an ideal gas depends only on temperature, and not pressure (or specific volume). So, once you change the temperature from state A to state B at constant volume, you can then go from state B to state C at constant temperature but a different volume, with no change in internal energy for this step. So, overall you have gone from state A to state C with the same change in internal energy as from A to B. The reason we call Cv the heat capacity at const. volume is that to measure it, we measure the heat transferred Q in a const. volume experiment. $\endgroup$ – Chet Miller Mar 20 '16 at 20:07

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.