If $\Delta U = Q + W $

Then adding $x$ moles of an ideal gas of lower temperature into a system of internal energy $U$ would increase $U$ as the kinetic energies of the molecules of the colder gas would be added? And would this be a change of $+Q$?
Going further, this should therefore increase the temperature as $U\propto$ temperature of an ideal gas? This is probably wrong as addition of a colder gas makes the overall temperature colder logically but I don't know why...
Another reason for the temperature to increase would be the work done by compression on the original gas molecules due to the input of more molecules? But this is obviously not the case.

Finally, how does all this tie in with the Ideal Gas Equation? Because an increase of $n$ in $PV=nRT$ should increase $T$ right?

Just wanted a clarification on where my understanding is going wrong...

And what happens if the gas added is warmer or the same temperature as the original gas?

P.S I'm just a 11th grader so please no Quantum mechanics!

  • $\begingroup$ $U$ is the total kinetic energy, so is proportional to how much gas you've got and its temperature. $\endgroup$
    – ProfRob
    Feb 6, 2016 at 9:27
  • $\begingroup$ Are you adding extra gas at constant volume? $\endgroup$ Feb 6, 2016 at 9:32
  • $\begingroup$ Yes at constant volume.. sorry for not mentioning $\endgroup$ Feb 6, 2016 at 17:49

1 Answer 1


Is you are adding gas at constant volume.

Then ,

$$W = P\Delta V + V\Delta P.$$

Since $\Delta V = 0\;_,$

$$W = V\Delta P$$

Then extra gas you put in container increase pressure, increasing Work and hence increasing internal Energy.

$$\Delta U = Q + W = Q + V\Delta P$$

Now, $Q$ here is heat inside system which is a measure of Temperature and not energy of system.

Temperature decreases and hence $Q$ decreases.

Now, $$\Delta U = nC_{V}\Delta T$$

Since $\Delta T$ decreases but $n$ moles inside systems increase which compensate more than decrease in Temperature and hence internal energy of system increases.

Finally, Ideal Gas law

$$PV = nRT$$

As $n$ increases, $T$ decreases and $P$ increases.

Increase in $n$ is more than decrease in $T$ and hence pressure increased.

Try yourself for same Temp.

  • $\begingroup$ But isn't Q the work done by heating and therefore energy? $\endgroup$ Feb 6, 2016 at 17:36
  • $\begingroup$ I meant heat is energy as change in Temp. and not total energy of gas. $\endgroup$ Feb 6, 2016 at 18:01

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.