We use the old Ideal Gas equation.....
PV=nRTPV=nRT, and now all the given measurements are kosher except for the PRESSURE measurement. Pressure, "force per unit area"force per unit area is something that is hard and non-intuitive to measure. Physical scientists, however, have long known that "1 atmosphere"1 atmosphere of pressure will support a column of mercury that is 760*mm760⋅mm high (and "1 Torr"-="1 mm Hg"1 Torr≡1 mm Hg). And thus we can use a measurement of length to give the pressure in atmospheres. Are you with me.....?
So here P=(407*mm*Hg)/(760*mm*Hg*atm^-1)=0.536*atmP=407⋅mm⋅Hg760⋅mm⋅Hg⋅atm−1=0.536⋅atm.
The problem is almost done now.........
PV=nRTPV=nRT; and so n=(PV)/(RT)n=PVRT
n=(0.536*atmxx3.93*L)/(294.15*Kxx0.0821*L*atm*K^-1*mol^-1)n=0.536⋅atm×3.93⋅L294.15⋅K×0.0821⋅L⋅atm⋅K−1⋅mol−1
=??*atm=??⋅atm
Note the dimensional consistency of the answer.....
n=(0.536*cancel(atm)xx3.93*cancelL)/(294.15*cancelKxx0.0821*cancelL*cancel(atm*K^-1)*mol^-1)
=1/(mol^-1)=1/(1/(mol))=mol as required...............this is what we want, and the consistent unit is an internal check on our calculation......
