The relation between the gas pressure P and average kinetic energy per unit volume E is

Engineering

Physics

Degree of Freedom and Internal Energy of Gas

Question

$P = \frac{1}{2} E$

P = E

$P = \frac{3}{2} E$

$P = \frac{2}{3} E$

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We know that, Kinetic Energy $= \frac{3 RT}{2}$ (Since, PV = nRT and n =1)

$KE = \frac{3 PV}{2}$
Therefore, $\frac{(KE)}{V} = \frac{3 P}{2} - eq (1)$

$According to question : E (Enogy per) = \frac{Kinetic Energy}{unit volume}$

$E = (\frac{3 P}{2})$

So, $P = \frac{2}{3} E$ Option (D)

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