What temperature actually measures

These are two different questions and a gas answers them separately.

1. How energetic is a typical particle? That is temperature: $K_{\text{avg}} = \tfrac{3}{2}k_B T$, with $k_B = 1.38\times10^{-23}$ J/K. Nothing in that expression counts particles.
2. How much energy is in the system? That is roughly $N K_{\text{avg}}$, so it scales with the particle count as well as the temperature.

A welding spark at $1200$ K and a bathtub at $320$ K land on opposite sides of those two rankings. The spark wins the first comparison by a factor of about four. The tub wins the second by something like $10^{20}$, because it holds that many more molecules, which is why the spark that lands on your arm stings for an instant and the bath does not.

So "hotter" answers only the per-particle question. Any time a problem asks which system will heat the other, or which one carries more energy, decide first which of the two quantities the question is about.

A particle of mass $m$ bounces off a wall and reverses its perpendicular velocity component, so the wall receives momentum. Do that billions of times a second over an area $A$ and the average perpendicular force per area is the pressure:

$$P = \frac{F_\perp}{A}, \qquad \text{in pascals}, \ 1\ \text{Pa} = 1\ \text{N/m}^2.$$

Nothing in that construction requires the surface to be a container. Slide an imaginary sheet of paper into the middle of the room and particles strike both faces at the same rate, delivering the same force per area. Pressure is defined at every point in the gas, and the walls are simply where a gauge is easy to attach.

Reading the mechanism forward tells you what raises pressure: more particles per volume gives more collisions per second, and higher temperature gives both faster arrivals and harder hits. Both routes lead to $PV = Nk_BT$, which is the next topic.

Set the per-particle kinetic energy equal to its kinetic-theory value:

$$\tfrac{1}{2}mv_{\text{rms}}^2 = \tfrac{3}{2}k_BT \quad \Longrightarrow \quad v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}}.$$

Energy is proportional to $T$. Speed is proportional to $\sqrt{T}$. Those are not the same scaling and the difference is tested constantly:

1. Double the absolute temperature and $v_{\text{rms}}$ rises by $\sqrt{2}$, about $41\%$.
2. To actually double the speed you need four times the absolute temperature.
3. Every one of these is in kelvins. Going from $20^\circ$C to $40^\circ$C is a factor of $1.07$, not $2$.

The mass in that formula does its own work. Two gases sharing a container share a temperature, so they share $K_{\text{avg}}$ per particle, and the heavier species is therefore slower by $\sqrt{m_2/m_1}$. Argon is about ten times the mass of helium, so argon atoms move at roughly a third the speed of the helium atoms right beside them.

$v_{\text{rms}}$ is a summary statistic, not an assignment. At any instant a gas at one temperature contains particles nearly at rest, particles far above $v_{\text{rms}}$, and everything between. The Maxwell-Boltzmann curve plots how many sit at each speed: it peaks below $v_{\text{rms}}$ and trails off slowly on the high side.

Heating the gas does not lift the whole curve. The area under it is the number of particles, and that number did not change. What happens instead is a redistribution:

1. The low-speed region empties out, so the curve drops on the left.
2. The peak slides to the right and gets shorter.
3. The high-speed tail fattens, which is what makes evaporation and reaction rates so temperature sensitive.

A curve that rose everywhere would be describing a container that gained particles. If the problem says the gas was sealed, that is not on the table.