Question

The surface area of the floor of a room is $30m^2$ and the walls are 3m high. If the room temperature is ${27}^{\circ }$C and pressure is 1 atm, then find the mass of the air inside the room
[Given: Molecular weight of air = 29, 1 atm = 1.013 $\times {10}^5$ Pa]

• A. 105.85 kg
• B. 29 kg
• C. 74.37 kg
• D. 135.35 kg

Answer 1

The correct option is A 105.85 kg

PV = nRT
$(1.013\times {10}^5)(30\times 3)=\stackrel{no.ofmoles}{n}\times 8.31\times \stackrel{Kelvin}{(27+278)}$
$n=3.65\times {10}^{3}moles$
Mass of air inside $=nM=3.65\times {10}^3\times 29g$