Question

1. Calculate the mass of 6,72 m3
of nitrogen (t = 0 °C, p = 101,3 kPa)

Answer 1

volume of nitrogen, V = 672 m³

we know, 1 m³ = 1000L

so, volume of nitrogen in L , V = 672000 litres

Temperature , T = 0°C = 273K

pressure , P = 101.3 kPa = 1.013 × 10^5 Pa = 1 atm [ as we know , 1atm = 1.013 × 10^5 pa ]

using Gas equation, PV = nRT

where, n is number of mole,

we can write , n = m/M

where m is mass of nitrogen and M is molecular mass of nitrogen,i.e.,M = 28 g/mol

now, PV = mRT/28

1atm × 672000Litres = m × 0.082litre.atm/mol.K × 273K

or, 672000 = m × 0.082 × 273

or, m = 672000/(0.082 × 273)

m = 30018g or, 30.018 kg

hence, mass of nitrogen = 30.018 kg