How does temperature affect gas pressure?

Hint: The relation between the temperature and the kinetic energy of gas has to be determined. Since the kinetic energy affects the velocity of the gas molecules the force is also affected by the energy changes. The force on the wall of the container in which the gas is kept is known as the pressure of the gas. Therefore, by determining the relation between energy and the velocity of the gas molecules the affection of the temperature on the gas pressure can be known.

Complete step by step answer:
 The gas gains heat, then i) the temperature $(T)$ of the gas increases and, ii) since, the heat is a form of energy, hence it converts into another energy inside the gas. This ‘another energy’ is the kinetic energy of the gas molecules, i.e the total energy of the gas is $E$.
The temperature is one of the properties of a gas that is directly proportional to the total energy of the gas molecules.
Therefore, $T \propto E$
Now, it is known that according to the kinetic theory, the potential energy of gas molecules is zero. Therefore, the total energy means kinetic energy.
The kinetic energy can be represented by, $E = \dfrac{1}{2}M{c^2}$
$M = $ the total mass of the gas
$c = $the r.m.s velocity of the gas molecules.
The RMS velocity has an Impact on the force of molecules. And, the pressure is none other than the force by the gas molecules on the wall of the container in which the gas is kept.
From this concept, we get the pressure-equation as, \[p = \dfrac{1}{3}\rho {c^2}\]
\[\rho \]is the density.
Hence, $E = \dfrac{3}{2}\dfrac{{Mp}}{\rho }$
$ \Rightarrow E \propto p$[ since $M,\rho $are constants for a gas]
Therefore we can say that the temperature is also directly proportional to the pressure of the gas.
Hence, if the temperature increases the pressure also increases in a gas, and when decreases the pressure also decreases.

Note:
When the mass of \[1\] mole of a gas i.e the molecular mass is $M$ , the density will be \[\rho = \dfrac{M}{V}\], \[V\] is the volume of the gas. Now from \[p = \dfrac{1}{3}\rho {c^2}\]we get,
${c^2} = \dfrac{{3p}}{\rho }$
$ \Rightarrow {c^2} = \dfrac{{3pV}}{{\rho V}}$
$ \Rightarrow {c^2} = \dfrac{{3RT}}{{\rho V}}$[since the ideal gas equation $pV = RT$]
$ \Rightarrow {c^2} = \dfrac{{3RT}}{M}$
$ \Rightarrow c = \sqrt {\dfrac{{3RT}}{M}} $
Hence, the RMS velocity of the gas molecules is directly proportional to the square root of the temperature.