Gas pressure is created when gas molecules collide with the wall of the container creating a force.
Gas temperature is a measure of how fast the molecules are moving / vibrating.
However, they both seem to be concerned by "kinetic energy" of the molecules, or in other words, the "collision" they impose on the target.
How do we visualize the difference between pressure and temperature of gas?
Is there any obvious difference between the two?
The same question in another form:
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A gas is hot when the molecules collided with your measuring device.
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A gas have high pressure when the molecules collided with your measuring device.
So, what is the difference between the two "collisions" in the physical sense and how do we visualize the difference?
For Simplicity,
How can a Hot gas be Low Pressured? ( They are supposed to have High Kinetic Energy since it is Hot. Therefore should be High Pressured at all times! But no. )
How can a High Pressured gas be Cold? ( They are supposed to collide extremely frequently with the walls of the container. Therefore should be Hot at all times! But no. )
Best Answer
Background
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 $Q$.
Application
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).
Answers
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!
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.