The ideal gas law

Exercise 6

   

A very low pressure of $1.0\cdot 10^{-3}\;torr$ is attained in a gas with a vacuum pump. Calculate the number of molecules in $1.00\;mL$ of the gas at $20^oC$ ?     

$760 \;torr\;$ $ = $ $ 1 \;atm$ number of molecules = $n\cdot6\cdot 10^{23}$ ($n$: number of moles)

$P$ $=$ $\frac{10^{-3}}{760}atm$ $1 mL$ $ =$ $ 10^{-3}$ L $n$ $=$ $\frac{PV}{RT}$ $=$ $\frac{10^{-3}10^{-3}}{760\cdot 0.082\cdot 293}$ Number of molecules = $n\cdot 6\cdot 10^{23} $ $ = $ $\frac{10^{-3}10^{-3}6\cdot 10^{23}}{760\cdot 0.082\cdot 293}$ $ =$ $ 3.29\cdot 10^{13} $ molecules