The number of molecules in one litre of an id | vedclass

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(C) Given:Volume $V = 1 \, L = 10^{-3} \, m^3$Temperature $T = 300 \, K$Pressure $P = 2 \, atm = 2 	imes 1.013 	imes 10^5 \, Pa \approx 2.026 	imes

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$1.5$

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(C) Given:Volume $V = 1 \, L = 10^{-3} \, m^3$Temperature $T = 300 \, K$Pressure $P = 2 \, atm = 2 	imes 1.013 	imes 10^5 \, Pa \approx 2.026 	imes 10^5 \, Pa$Mean kinetic energy per molecule $\bar{E} = 2 	imes 10^{-9} \, J$From the ideal gas law,$PV = NkT$,where $N$ is the number of molecules and $k$ is the Boltzmann constant $(1.38 	imes 10^{-23} \, J/K)$.However,we are given the mean kinetic energy $\bar{E} = \frac{3}{2}kT$. Thus,$kT = \frac{2}{3}\bar{E}$.Substituting this into the ideal gas equation:$N = \frac{PV}{kT} = \frac{PV}{\frac{2}{3}\bar{E}} = \frac{3PV}{2\bar{E}}$$N = \frac{3 	imes (2.026 	imes 10^5) 	imes 10^{-3}}{2 	imes (2 	imes 10^{-9})}$$N = \frac{6.078 	imes 10^2}{4 	imes 10^{-9}} \approx 1.5195 	imes 10^{11}$Rounding to the nearest provided option,the value is $1.5 	imes 10^{11}$.

The number of molecules in one litre of an ideal gas at $300 \, K$ and $2$ atmospheric pressure with mean kinetic energy $2 \times 10^{-9} \, J$ per molecule is $.... \times 10^{11}$.

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