critical-phenomenaphase diagramphase-transitionstatistical mechanicstemperature
We know that $T_c$ is the temperature above which no amount of pressure could force a gas to liquefy.
But why is this? Somehow I don't buy the point that the gas molecules exert too much pressure back to get close and turn into a liquid. If we had tens of thousands of atmosphere pressure (such as on the inside of hot planets), we should be able to liquefy any gas at any temperature.
Let us assume we have a function, $f_{s}(\mathbf{x},\mathbf{v},t)$, which defines the number of particles of species $s$ in the following way: $$ dN = f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \ d^{3}x \ d^{3}v $$ which tells us that $f_{s}(\mathbf{x},\mathbf{v},t)$ is the particle distribution function of species $s$ that defines a probability density in phase space. We can define moments of the distribution function as expectation values of any dynamical function, $g(\mathbf{x},\mathbf{v})$, as: $$ \langle g\left( \mathbf{x}, \mathbf{v} \right) \rangle = \frac{ 1 }{ N } \int d^{3}x \ d^{3}v \ g\left( \mathbf{x}, \mathbf{v} \right) \ f\left( \mathbf{x}, \mathbf{v}, t \right) $$ where $\langle Q \rangle$ is the ensemble average of quantity $Q$.
If we define a set of fluid moments with similar format to that of central moments, then we have: $$ \text{number density [$\# \ (unit \ volume)^{-1}$]: } n_{s} = \int d^{3}v \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\ \text{average or bulk velocity [$length \ (unit \ time)^{-1}$]: } \mathbf{U}_{s} = \frac{ 1 }{ n_{s} } \int d^{3}v \ \mathbf{v}\ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\ \text{kinetic energy density [$energy \ (unit \ volume)^{-1}$]: } W_{s} = \frac{ m_{s} }{ 2 } \int d^{3}v \ v^{2} \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\ \text{pressure tensor [$energy \ (unit \ volume)^{-1}$]: } \mathbb{P}_{s} = m_{s} \int d^{3}v \ \left( \mathbf{v} - \mathbf{U}_{s} \right) \left( \mathbf{v} - \mathbf{U}_{s} \right) \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\ \text{heat flux tensor [$energy \ flux \ (unit \ volume)^{-1}$]: } \left(\mathbb{Q}_{s}\right)_{i,j,k} = m_{s} \int d^{3}v \ \left( \mathbf{v} - \mathbf{U}_{s} \right)_{i} \left( \mathbf{v} - \mathbf{U}_{s} \right)_{j} \left( \mathbf{v} - \mathbf{U}_{s} \right)_{k} \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\ \text{etc.} $$ where $m_{s}$ is the particle mass of species $s$, the product of $\mathbf{A} \mathbf{B}$ is a dyadic product, not to be confused with the dot product, and a flux is simply a quantity multiplied by a velocity (from just dimensional analysis and practical use in continuity equations).
In an ideal gas we can relate the pressure to the temperature through: $$ \langle T_{s} \rangle = \frac{ 1 }{ 3 } Tr\left[ \frac{ \mathbb{P}_{s} }{ n_{s} k_{B} } \right] $$ where $Tr\left[ \right]$ is the trace operator and $k_{B}$ is the Boltzmann constant. In a more general sense, the temperature can be (loosely) thought of as a sort of pseudotensor related to the pressure when normalized properly (i.e., by the density).
How can a Hot gas be Low Pressured?
If you look at the relationship between pressure and temperature I described above, then you can see that for low scalar values of $P_{s}$, even smaller values of $n_{s}$ can lead to large $T_{s}$. Thus, you can have a very hot, very tenuous gas that exerts effectively no pressure on a container. Remember, it's not just the speed of one collision, but the collective collisions of the particles that matters. If you gave a single particle the enough energy to impose the same effective momentum transfer on a wall as $10^{23}$ particles at much lower energies, it would not bounce off the wall but rather tear through it!
How can a High Pressured gas be Cold?
Similar to the previous answer, if we have large scalar values of $P_{s}$ and even larger values of $n_{s}$, then one can have small $T_{s}$. Again, from the previous answer I stated it is the collective effect of all the particles on the wall, not just the individual particles. So even though each particle may have a small kinetic energy, if you have $10^{23}$ hitting a wall all at once, the net effect can be large.
But if I choose to apply the same reasoning the other way, given that the saturation pressure for T = 200 ºC is 15,54 bar, I do not understand how this would tell me that the water is vapor.
At T = 200 ºC, your pressure of 5 bar is less than the saturation pressure of 15,54 bar. This means that the water is more compressed; the molecules are closer together (i.e., closer to being a liquid).
And I say still liquid because a lower pressure somehow translates as less energy, and less energy means to me that it still remains liquid.
Adiabatic expansion (lowering pressure without adding heat) will decrease energy, like you've said. This transition, however, is a process; you can't apply this intuition to 2 unrelated states without considering a process between those states.
If you adiabatically expand water (200 ºC, 15,54 bar) to 5 bar, its temperature will drop below 200 ºC. The final state will therefore not be steam.
You can see, therefore, that to go from water (200 ºC, 15,54 bar) to steam at the same temperature (200 ºC, 5 bar), we need to expand while adding heat. You need to add energy to make up for the energy loss of expansion, so that we now have an isothermal expansion. By lowering the pressure, however, we reduce the boiling point of water and a phase change occurs.
You can visualize this better on the $P-v-T$ surface for water below:
Check out the constant temperature line starting from the saturated state to the some lower pressure vapor state (this is our isothermal expansion). You can see that at constant temperatures, we can have steam at lower pressures. This is because it is easier to evaporate water at lower pressure, which leads to your next confusion:
It must increase it's energy (just like with temperature) before going for the phase-change.
Lower pressure actually makes it easier for liquids to evaporate. Can you visualize how molecules would more readily break out of the liquid phase into a vapor under low pressures? There is simply less pressure holding the molecules in place. This is why it's easier to boil water at high altitudes (lower pressure).
Best Answer
Your description of critical temperature isn't quite right.
If you increase the temperature of a liquid beyond the critical point, the atoms are moving so quickly that persistent structure fails to form and so you have something that behaves a lot like a very dense gas.
Similarly, if you increase the pressure of a gas beyond the critical point, it becomes very dense so that it's like a liquid but without persistent structure.
In other words, it's not so much that the liquid phase does not form, but rather the liquid and gas phases become indistinguishable (rather intuitively) and you end up with what's called a supercritical fluid.
Here's the phase diagram of CO$_2$ for clarity: