Does hot air really rise?

The mechanism responsible for the rising of hot air is flotation: Hot air is less dense than cold air and hence air pressure will exert an upwards force, in the same way air rises in water. Now if cold air was magically unaffected by gravity, then it would not be able to exert pressure on the hot air and thus it would not rise.

The statement that "heat rises", by the way, is not universally true. Look at water. Here, it is the cold water that is less dense than warm water (at least in the temperature regime of importance to freezing). In winter, when water gets colder, the cold water raises to the top and eventually will freeze, while the water below remains liquid for the moment.

Buoy, does it ever. To contrast the previous answers I will give a mathematical description and a concrete example to bolster the intuitive understanding.

Ideal Gas Law

From thermodynamics we know that pressure, $P$, temperature $T$, and density $\rho$ (or specific volume $v=1/\rho$) are related through an equation of state. For suitable gases (including air at atmospheric conditions) this equation is the ideal gas law:

\begin{equation} \tag{1} \label{igas} P = \rho R T \end{equation}

where $R$ is the specific gas constant, which can be determined by the chemical makeup of the gas under consideration (e.g. $R_{air}=287.058 \:\mathrm{J kg^{−1} K^{−1}} $).

Bouyancy

As already mentioned by Helder Velez, Archemides' Principle informs us that an object immersed in a fluid will experience an upward force equal to the weight of the displaced fluid, where 'up' is the direction of decreasing density gradient.¹ Mathematically, this may be stated as:

\begin{equation} \tag{2} \label{buoy} \mathbf{F_b} = -\rho V \mathbf{g} \end{equation}

where $\mathbf{g} = -g\mathbf{\hat{k}}$ is the body force vector (usually gravity).

Air Bubble in water

Consider a small air bubble, initially at rest near the bottom of a pool, at thermal equilibrium (same temperature) as the pool water. The buoyancy force acting on the bubble is given by equation \ref{buoy}, and the weight of the bubble is given by $\mathbf{F_g} = m\mathbf{g}$. The subscript $w$ refers to water and the subscript $a$ refers to the air in the bubble. Applying Newton's second law yields:

\begin{align} m_a \mathbf{a} &= \sum \mathbf{F} \\ m_a \mathbf{a} &= \mathbf{F_g} + \mathbf{F_b} \\ m_a \left( a_x \mathbf{\hat{i}} + a_y \mathbf{\hat{j}} + a_z \mathbf{\hat{k}} \right) &= -m_a g\mathbf{\hat{k}} + \rho_w V_b g\mathbf{\hat{k}} \\ m_a a_z &= \rho_w V_a g - m_a g \\ a_z &= g\left(\frac{\rho_w V_a}{m_a} - 1\right) \\ a_z &= g\left(\frac{\rho_w V_a}{\rho_a V_a} - 1\right) \\ a_z &= g\left(\frac{\rho_w}{\rho_a} - 1\right) \\ \end{align}

where I have used $m_a = \rho_a V_a$. Here, it can be seen that the bubble will accelerate upward whenever $\rho_w > \rho_a$. Leveraging the fact the pressure varies linearly with depth in a static fluid, you can prove to yourself that $\rho_w \gg \rho_a$ for bubbles in most pools.

Answer

Parcel of Air

Now consider a similar scenario, where instead of a pool we have a room full of air at uniform temperature $T_\infty$ and the bubble is now a parcel of air which has been heated to a slightly elevated temperature $T_\infty + \Delta T$. I will use the subscripts $c$ for the air in the cool room, and $h$ for the air within the hot parcel.

If we perform a similar analysis to the bubble in the pool, we will go through the same motions as the derivation above, and end with a similar expression for the initial acceleration of the hot parcel:

\begin{equation} a_z = g\left(\frac{\rho_c}{\rho_h} - 1\right) \end{equation}

In this case however, we can use equation \ref{igas} to further simplify the result:

\begin{align} a_z &= g\left(\frac{P/(R_{air} T_\infty)}{P/\left(R_{air} \left[T_\infty + \Delta T\right]\right)} - 1\right) \\ a_z &= g\left(\frac{T_\infty + \Delta T}{T_\infty} - 1\right) \\ a_z &= g\left(\frac{\Delta T}{T_\infty}\right) \\ \end{align}

I can think of no better mathematical affirmation of the adage Hot air rises than the above equation. Wherever $\Delta T > 0$, $a_z$ will be also. Conversely a cooler parcel will fall: $\Delta T < 0 \rightarrow a_z < 0$.

Cleanup

You might wonder:

Why is it that the bubble in the pool is so straightforward, yet the air parcel rising is not immediately obvious?

Three reasons come to mind:

1. The bubble is well defined. It has a clear spherical boundary which is more or less maintained during its' ascent. On the other hand, our parcel is not visible, and even if it is spherical initially, it can stretch and morph at the mercy of any local air currents.
2. The ratio $\rho_w/\rho_a$ is usually much bigger than $\rho_c/\rho_w$, making the motion of the bubble much more pronounced than that of the hot air parcel.
3. The air parcel is subject to heat transfer. Imagine we wrapped our little air parcel in a tiny balloon. Even if its shape is maintained, the air parcel will transfer heat to the surrounding air as it rises, the temperature will drop so that $a_z \rightarrow 0$, and viscous forces will slow it to a halt.

Also note: the magnitude of acceleration is independent of the pressure. Whether we are in a pressure chamber at $10 \:\mathrm{atm}$ or on mount everest at $0.333 \:\mathrm{atm}$ it will always divide through.

Finally, I will point out that, even though the ideal gas law gives us a very elegant expression for acceleration, all other fluids (which I can think of) have equations of state with negative correlations between $T$ and $\rho$, meaning that a fluid parcel with an elevated temperature relative to a quiescent fluid of the same thermodynamic makeup will always have a buoyancy force of greater magnitude than its weight.

¹For hydrostatic fluids and many flows the pressure gradient $\nabla P$ is nearly always aligned with the density gradient $\nabla \rho$. Specifically the direction of the body force vector $\mathbf{g}$ is opposite the density gradient.

Heat does only 1 thing in a closed system, and that is evenly distribute itself about the system as it reaches thermodynamic equilibrium. I dont think this is what you asking about though. I assume you are talking about hot air (hot being a relative term just meaning it is hotter then the surrounding air). This hot air will be less dense then the surrounding air, and will therefore want to be above the more dense, colder air. If you want to actualy see this, get a beaker of water and add some oil, this is same thing that happens with air (as both cases involve 2 liquids of different densities)

To answer the question exactlty, hot air does rise, and it is also displaced by cold air (though often from the side, not directly above it). And yes, gravity is the reason less dense liquids like to sit on top of more dense liquids