MHT CET · Physics · Kinetic Theory of Gases
KE per unit volume is \(E\). The pressure exerted by the gas is given by
- A \(\frac{E}{3}\)
- B \(\frac{2 E}{3}\)
- C \(\frac{3 E}{2}\)
- D \(\frac{E}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 E}{3}\)
Step-by-step Solution
Detailed explanation
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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