This is where we get into some of the complexities of how we actually define the 'total' energy. As far as gravity is concerned, physics only depends on changes in the potential energy, not its absolute value; we can add a constant to the potential energy and get the same predictions! So the 'total potential energy' is not an entirely well-defined quantity.
As far as ideal gas theory is concerned, we treat gas particles as rigid balls and ignore any intermolecular interactions. So in this approximation, we ignore any electromagnetic forces between particles. 28
(If we wanted to add them, we would reach something like the <PERSON> model which I'm sure will come up later in the book you're working through, or one of many others!)
But what about gravitational potential energy? This is defined up to a constant, but there might still be differences in gravitational potential energy between the two situations, because the distribution of air density changes.
Just how much?
Since we're assuming uniform temperature, and assuming a uniform gravitational field, we can calculate the distribution of density in hydrostatic equilibrium. In hydrostatic equilibrium, the pressure and density are functions of height, but we still take them to be related by the ideal gas law $mP(z)=\rho(z) k_BT$. We have $$\frac{dP}{dz}=-\rho(z)g=-\frac{mg}{k_BT}P(z)$$ with the solution $$\rho(z)=\rho_0\exp\left(-\frac{mgz}{k_BT}\right)$$We set the density of air at ground level, $\rho_0$ by requiring that the total mass of particles in the gas is constant: $$\int_\text{room} \rho(z) dV = M$$ If we assume the room has floor area $A$ and height $h$, this becomes $$M=A\int_0^h \rho(z) dz = \frac{A\rho_0 k_B T}{mg}\left(1-\exp\left(-\frac{mgh}{k_BT}\right)\right)$$We can rearrange this to find $\rho_0$ as a function of temperature: $$\rho_0=\frac{Mmg}{Ak_B T \left(1-\exp\left(-\frac{mgh}{k_BT}\right)\right)}$$
Having found the density as a function of height, we can calculate the potential energy of the gas.
The gravitational potential energy density of a piece of gas at height $z$ above the zero-potential-energy reference level is just $u(z)=\rho(z) g z$. So the total potential energy of the whole room is $$U=\int_\text{room} u(z)\,dV=\int_\text{room} \rho(z) g z\thinspace dV=\int_\text{room}gz\rho_0\exp\left(-\frac{mgz}{k_BT}\right)dV$$ and with our area $A$ and height $h$, this becomes $$U=Ag\rho_0\int_0^h z\exp\left(-\frac{mgz}{k_BT}\right)\,dz$$
To calculate this, we can use a standard integral result: $$\int_0^h z \exp \left(-\frac{z}{C}\right)=C^2\left(1-\left(1+\frac{h}{C}\right)\exp\left(-\frac{h}{C}\right)\right)$$which we can use with $C=\frac{k_B T}{mg}$to find (deep breath)... 707