Which of the following relations is correct for an ideal gas regarding its pressure (P) and translational kinetic energy per unit volume (E)?

P = \frac{2}{3} E

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P = \frac{3}{2} E

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P = \frac{1}{2} E

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P = 2 E

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Solution

The correct option is A $P = \frac{2}{3} E$
According to kinetic theory of gases,
$E = \frac{1}{2} m u_{r m s}^{2} = \frac{1}{2} m N u_{r m s}^{2}$
Energy per unit volume $E = \frac{E}{V} = \frac{1}{2} \frac{m N u_{r m s}^{2}}{V}$
$= \frac{3}{2} \times \frac{1}{3} \frac{m N U_{r m s}^{2}}{V} = \frac{3}{2} P (a s P V = \frac{1}{3} m N u_{r m s}^{2})$
So $P = 2 / 3 E$

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Similar questions

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

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

E

P = \frac{2}{3}

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