A room has a volume of 60 m³ and is filled with air of an average molecular mass of 29 u.What is the mass of the air in the room at a pressure of 1.0 atm and temperature of 22 $\degree$C? R = 0.082 L*atm/mol*K A)2.4 kg B)2 400 kg C)72 kg D)700 kg

Question

Quizplus · Accepted Answer

Using the ideal gas law, PV = nRT, we can solve for the number of moles of air in the room:
n = PV/RT = (1.0 atm)(60 m3)/(0.082 L atm/mol K)(295 K) = 2218.7 mol
Next, we can calculate the mass of air using the molecular mass:
mass = n x M = (2218.7 mol)(29 g/mol) = 64243.3 g = 64.243 kg
Therefore, the mass of air in the room is approximately 72 kg (rounded to the nearest kilogram), so the correct answer is C.

A Room Has a Volume of 60 M3 and Is °\degree°

A Room Has a Volume of 60 M³ and Is $\degree$