An ideal gas in a container of volume 500 tex | vedclass

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(D) The pressure $P$ and volume $V$ of an ideal gas are related to the number of molecules $N$ and the average kinetic energy per molecule $\langle K

Vedclass · Accepted Answer

$2.5 	imes 10^{22}$

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(D) The pressure $P$ and volume $V$ of an ideal gas are related to the number of molecules $N$ and the average kinetic energy per molecule $\langle K angle$ by the relation:$PV = \frac{2}{3} N \langle K angle$Given:$V = 500 	ext{ c.c.} = 500 	imes 10^{-6} 	ext{ m}^3 = 5 	imes 10^{-4} 	ext{ m}^3$$P = 2 	imes 10^5 	ext{ N/m}^2$$\langle K angle = 6 	imes 10^{-21} 	ext{ J}$Rearranging the formula to solve for $N$:$N = \frac{3PV}{2\langle K angle}$Substituting the values:$N = \frac{3 	imes (2 	imes 10^5) 	imes (5 	imes 10^{-4})}{2 	imes (6 	imes 10^{-21})}$$N = \frac{3 	imes 10^2}{12 	imes 10^{-21}} = \frac{300}{12} 	imes 10^{21} = 25 	imes 10^{21} = 2.5 	imes 10^{22}$Thus,the number of gas molecules is $2.5 	imes 10^{22}$.

An ideal gas in a container of volume $500 \text{ c.c.}$ is at a pressure of $2 \times 10^5 \text{ N/m}^2$. The average kinetic energy of each molecule is $6 \times 10^{-21} \text{ J}$. The number of gas molecules in the container is:

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