A spherical balloon of volume `4.00 xx 10^(3) cm^(3)` contains helium at a pressure of `1.20 xx 10^(5) Pa`. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is `3.60 xx 10^(-22) J`?

Options

A. `3.32 mol`
B. `2.16 mol`
C. `4.12 mol`
D. `2.8 mol`

Answer

A

Explanation

The gas temperature must be that implied by we know kinetic energy of a molecules is given by
`1/2m_(0) bar(v^2) = 3/2 k_(B)T implies 2/3 ((1/2 m_(0)bar(V^2))/(k_B))`
`implies T = 2/3 ((3.60xx10^(-22))/(1.38xx10^(-23)J//K)) = 17.4K`
Now `PV = nRT implies n = (PV)/(RT)`
`n = ((1.20xx10^(5)N//m^2)(4.00xx10^(-3)m^3))/((8.314 j//mol.K)(17.4K)) = 3.32mol`.