Question

Relation between pressure ( $ P $ ) and energy density ( $ E $ ) of an ideal gas is:
(A) $ P = 2/3E $
(B) $ P = 3/2E $
(C) $ P = 3/5E $
(D) $ P = E $

Answer 1

Hint: We need to use the relation of ideal gas $ PV = nRT $ to generate the relation between pressure and the energy density of an ideal gas. Energy density is the amount of energy stored in a given system or region of space per unit volume.

Complete Step by Step Solution
Gases are complicated. They're full of billions and billions of energetic gas molecules that can collide and possibly interact with each other. Since it's hard to exactly describe a real gas, people created the concept of an Ideal gas as an approximation that helps us model and predict the behaviour of real gases.
We know for one mole of ideal gas
 $ PV = RT $ where $ P $ is the pressure and $ V $ is the volume of the ideal gas, $ R $ is the universal gas constant and $ T $ is the temperature.
The internal energy of an ideal gas $ {E_{\operatorname{int} }} = \dfrac{3}{2}PV $ , which can also be expressed as,
From the equations stated above, we get,
 $ \dfrac{{{E_{\operatorname{int} }}}}{V} = \dfrac{3}{2}P $ .
Thus, the energy density is given by,
 $ E = \dfrac{3}{2}P $
 $ \Rightarrow P = \dfrac{2}{3}E $
This can also be solved using an alternative method.
Kinetic energy of the molecules is given by, $ \dfrac{1}{2}M{V_{rms}}^2 $ where $ M $ is the molar mass and $ {V_{rms}} $ is the root mean square velocity.
 $ P = \dfrac{2}{3}E $ .
The root mean square velocity is the square root of the average of the square of the velocity. As such, it has units of velocity. The reason we use the rms velocity instead of the average is that for a typical gas sample the net velocity is zero since the particles are moving in all directions.
 $ {V_{rms}} = \sqrt {\dfrac{{3KT}}{m}} $ where $ k = \dfrac{R}{{{N_A}}} $ then, $ \dfrac{k}{m} = \dfrac{R}{M} $ .
 $ {V_{rms}} = \sqrt {\dfrac{{3KT}}{m}} = \sqrt {\dfrac{{3RT}}{M}} $ .
Kinetic energy $ KE = \dfrac{3}{2}RT = \dfrac{3}{2}PV $ since, $ PV = RT $ .
 $ \dfrac{{KE}}{V} = \dfrac{3}{2}P $ .
Thus energy density, $ E = \dfrac{3}{2}P $ .
Hence, the correct answer is Option B.

Note
There are two ways to solve this question first by using ideal gas and internal energy relation second by using ideal gas and kinetic energy relation.
Ideal gas molecules do not attract or repel each other. The only interaction between ideal gas molecules would be an elastic collision upon impact with each other or an elastic collision with the walls of the container. Ideal gas molecules themselves take up no volume. The gas takes up volume since the molecules expand into a large region of space, but the Ideal gas molecules are approximated as point particles that have no volume in and of themselves.