Question

How many moles of xylene would be contained in an empty warehouse (200' x 50' x 20') with 10 ppmv xylene in the air @ 15 degrees celsius and pressure of 1 atm

( 0 C = 273 K; 1 ft^3 = 28.32 L)

MW of xylene = 106g/mol
R = 8.205E-5 atm-m^3 / mol-K

Answer 1

Solution:

Volume of warehouse V = 200in x 50in x 20in = 200000 in³ = 200000 in³/ (12³ in³ft^-3) = 115.741 ft³

Volume of warehouse in m³ = 115.741 ft³ x 28.32 Lft^-3 x 10^-3 m³L^-1= 3.2778 m³

Xylene in air = 10 ppmv

i.e., in 10⁶ m³ of air, xylene = 10 m³

Therefore, xyene present in warehouse = (3.2778/ 10⁶) x 10 ft³ = 3.2778 x 10^-5

Using ideal gas equation, total number of moles (N) of air and xylene present in the warehouse N=PV/RT

N = (1 atm x 3.2778 m³) /( 8.205 x10^-5 atm m³ mol^-1 K^-1) = 138.71 mol

Mole fraction of xylene = volume fraction of xylene = (3.2778 x 10^-5 m³) / (3.2778 m³) = 10^-5

Therefore, number of moles of xylene present in warehouse  

= mole fraction of xylene in warehouse x total number of moles of air and xylene in warehouse

= 10^-5 x 138.71 mol = 0.0013871 mol