Can the internal energy of an ideal gas system increase as temperature decreases?

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!

• $U$ is the total kinetic energy, so is proportional to how much gas you've got and its temperature. – Rob Jeffries Feb 6 '16 at 9:27
• Are you adding extra gas at constant volume? – Anubhav Goel Feb 6 '16 at 9:32
• Yes at constant volume.. sorry for not mentioning – Tejas Menon Feb 6 '16 at 17:49

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.

• But isn't Q the work done by heating and therefore energy? – Tejas Menon Feb 6 '16 at 17:36
• I meant heat is energy as change in Temp. and not total energy of gas. – Anubhav Goel Feb 6 '16 at 18:01