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Relation between pressure (P) and energy density (E) of an ideal gas is-
A) $P = \dfrac{2}{3}E$
B) $P = \dfrac{3}{2}E$
C) $P = \dfrac{3}{5}E$
D) $P = E$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Use the relation of ideal gas $PV = RT$ to generate the relation between pressure and the energy density of an ideal gas.

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 behavior of real gases

We know for one mole of ideal gas,
 For an ideal gas
$PV = RT$ ………(i)
And the internal energy of an ideal gas:
${E_{in}} = \dfrac{3}{2}PV$ ………(ii)
From equation i and equation ii
We get,
${E_{in}} = \dfrac{3}{2}RT$
$\dfrac{{{E_{in}}}}{V} = \dfrac{3}{2}P$

Energy density $E = \dfrac{3}{2}P$
$P = \dfrac{2}{3}E$


Alternative method-
Kinetic energy$ = \dfrac{1}{2}MV_{rms}^2$
$ \Rightarrow P = \dfrac{2}{3}E$

Where
M= molar mass
${V_{rms}} = \sqrt {\dfrac{{3KT}}{m}} = \sqrt {\dfrac{{3RT}}{M}} $
$ \Rightarrow KE = \dfrac{3}{2}RT$
$ \Rightarrow KE = \dfrac{3}{2}PV$ $\left[ {PV = RT} \right]$
$ \Rightarrow \dfrac{{KE}}{V} = \dfrac{3}{2}P$
$E = \dfrac{{3P}}{2}$ [E = Energy density]
$ \Rightarrow P = \dfrac{2}{3}E$

Additional Information:
 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.

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.