So we need "(i) the volume of the room.........."
And "(ii) the number of air molecules..........".
"(i) Volume" = 13.5xx12.0xx10.0*ft^3=1620*ft^3. (I never thought I would use "feet and inches"; I'll be bidding "guineas" at an auction next!)
"Volume" = 1620*ft^3xx28.2*L*ft^-3xx10^-3*m^3*L^-1=45.68*m^3.
And now we need the molar volume at 20 ""^@C, i.e. 293*K
V=(nRT)/(P)
=(1*molxx0.0821*L*atm*K^-1*mol^-1xx293*K)/(1*atm)=
=24.05*L
And so if we divide the "volume" by the "molar volume", and then multiply this number by the "Avocado number" (yes, I know but I am partial to avocadoes), we should reasonably get the number of air molecules in the room. It is a fact that 10^3L-=1*m^3.
So (45.68*m^3xx10^3*L*m^-3)/(24.05*L*mol^-1)~=1900*mol
And thus (ii) there are 1900*molxx6.022xx10^23*mol^-1=1.14xx10^27 "air molecules"
Please check my calculations,
"all care taken but no responsibility admitted".
