Question

An apartment has a living room whose dimensions are 2.8 m x 4.1 m x 4.6 m. Assume that the air in the room is composed of 79% nitrogen (N2) and 21% oxygen (O2). At a temperature of 24 °C and a pressure of 1.01 x 105 Pa, what is the mass (in grams) of the air?

Answer 1

Using Ideal gas law:

PV = nRT

n = number of moles of air in the apartment

n = PV/RT

P = Pressure = 1.01*10^5 Pa

V = L*B*H = 2.8*4.1*4.6 m^3

R = 8.314

T = 24 C = 297 K

So,

n = 1.01*10^5*2.8*4.1*4.6/(8.314*297) = 2160 moles

Now average molar mass of air will be

Mw_air = 0.79*Molar mass of N2 + 0.21*molar mass of O2

Mw_air = 0.79*28 + 0.21*32 = 28.84 gm/mol

So mass of the air will be

mass = moles*Molar mass

mass = (2160 mol)*(28.84 gm/mol)

mass = 62294.4 gm = 6.23*10^4 gm

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