Question

# 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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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]
$\Rightarrow P = \dfrac{2}{3}E$