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KE per unit volume is $E$. The pressure exerted by the gas is given by
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2478 Upvotes
Verified Answer
The correct answer is:
$\frac{2 E}{3}$
The pressure exerted by the gas,
$$
\begin{array}{l}
p=\frac{1}{3} \rho \bar{c}^{2} \\
=\frac{1}{3} \frac{m}{V} \bar{c}^{2} \\
=\frac{2}{3}\left(\frac{1}{2} m \bar{c}^{2}\right)
\end{array}
$$
$\left(\because \frac{1}{2} m \bar{c}^{2}=\frac{E}{V}=\right.$ energy per unit volume, $V=1$ )
$p=\frac{2}{3} E$
$$
\begin{array}{l}
p=\frac{1}{3} \rho \bar{c}^{2} \\
=\frac{1}{3} \frac{m}{V} \bar{c}^{2} \\
=\frac{2}{3}\left(\frac{1}{2} m \bar{c}^{2}\right)
\end{array}
$$
$\left(\because \frac{1}{2} m \bar{c}^{2}=\frac{E}{V}=\right.$ energy per unit volume, $V=1$ )
$p=\frac{2}{3} E$
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