Question

On a hot summer day, the density of air at atmospheric pressure at 32.0°C is 1.1346 kg/m³.

(a) What is the number of moles contained in 1.00 m³ of an ideal gas at this temperature and pressure?
mol

(b) Avogadro's number of air molecules has a mass of 2.88 ✕ 10⁻² kg. What is the mass of 1.00 m³ of air? (Assume air is an ideal gas.)
kg

(c) Does the value calculated in part (b) agree with the stated density of air at this temperature? (Consider that it does if the values are within 10% of each other.)

Answer 1

Solution :-

(a). Pressure P = 1 atm = 1.01 x10^ ⁵ Pa

Temprature T = 37 ^o C = 32+ 273 = 305K

Density = 1.1346 kg / m³

Volume V = 1 m ³

From the relation PV = nRT

Number of moles n = PV /RT

Where R = Gas constant = 8.314 J / mol K

Substitute values you get n = (1.01 x10 ^⁵)(1) /(8.314 x 305)

                                         = 39.83 mol

(b). Avogadro's number of air molecules has a mass of 2.88✕ 10⁻² kg

i.e., molar mass M = 2.88x10 ^-2 kg

Therefore the mass of 1.00 m³ of air is m = nM

                                                              = 1.147 kg

(c). Density of the air ' = m / V

                                    = 1.147kg / 1 m³

                                    = 1.147g/ m³

Percentage difference in density = [( ' - ) / ] x100

                                               = 1.093%

Yes