Relation between pressure $(P)$ and energy density $(E)$ of an ideal gas is-

P = \frac{2}{3} E

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P = \frac{3}{2} E

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P = \frac{3}{5} E

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P = E

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Solution

The correct option is A $P = \frac{2}{3} E$
$P V = n R T$

$E = \frac{3}{2} R T$

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

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

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Similar questions

Internal energy and pressure for unit volume of gas are related by

A.)P=2/3E. B.)P=3/2E. C.)P=1/3E. D.)P=2E

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

E

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