Why can we warm up the air? When I heat up a room, this is basically both an isobaric and isochoric process. Since air can be thought of as an ideal gas, the equation of state $pV = NkT$ and thus $NkT$ is also constant. Because the energy of air can be expressed as $E = \frac32 NkT$, $E$ is also constant. How come that heating up the room works in real life then?
 A: The way to make the process isobaric would be to expose it to the atmosphere. But if you expose it to the atmosphere then particles must be allowed to leave the room and $N$ can change. As $T$ goes up $N$ goes down so that $P$ and $V$ can remain constant.
If, on the other hand, you seal the room so no particles can escape then there is nothing to stop the pressure from rising. As $T$ goes up $P$ goes up and $V$ and $N$ remain constant.
A: If $N$ is constant, then the room is a closed system (no air can enter or exit). The first law for a closed system is
$$\Delta E=Q-W$$
where $\Delta E$ is the change in internal energy of the gas (the $E$ in your equation), $Q$ is heat transfer and is positive if added to the air, and $W$ is positive if the gas does work by expanding.
If, as you say, $E$ is constant, then the heat added to the room must equal expansion work done by the gas on the room, or $Q=W$. However, if the room walls are rigid, i.e., they cannot expand, then $V$ is constant no work can be done, i.e., $W=0$. Therefore $E$ can't be constant, and from the first law $\Delta E=Q$. Then
$$\Delta E=Q=\frac{3}{2}K\Delta T$$
So the temperature increases.
Finally, since $V$ of the room is constant, then from the ideal gas law
$$pV=nKT$$
$$\Delta P=\frac{nk\Delta T}{V}$$
I.e., the pressure has to increase. So the process can't be isobaric.
Hope this helps.
A: Assume that the number of particles in the room is constant and try to see where the conservation of energy leads:
$$dU = \delta Q - pdV.$$
For constant volume $\delta Q = dU$;  the added heat will increase the room's temperature. Insisting on constant pressure, on the other hand,
$$\delta Q = d(U+pV)$$
Now the equation of state for the ideal gases will yield
$$\delta Q \sim \alpha T$$
Both in isochoric and isobaric processes, adding heat to the room will increase its temperature.
