Question

Use R = 8.2 × 10–5 m3 atm/mol K and NA = 6.02 × 1023 mol–1. The approximate number of air molecules in a 1 m3 volume at room temperature (300 K) and atomospheric pressure is:

Answer 1

The ideal gas law states:
PV=nRT
where
n=number of moles
R is constant given above (atm/mol K)
so
n=PV/RT
=1 atm. * 1 m³ /(8.2*10^-5 m³ atm / mol K * 300 K)
=40.65 moles

Check:
air weighs 1.293 g/l, or 1293 g/m³. Take an average molar mass equal to oxygen (32), we have
1293/32=40.406 moles ≈ 40.65
Since the temperature is 300 K and not at STP, the number of moles is expected to be lower.

The approximate number of molecules is obtained by multiplying the number of moles by the avogadro's constant (N_a)

Approximate number of molecules
=40.65 moles * 6.02*10^-23 mol^-1
=2.45*10²⁵

Answer 2

how can the notion of symmetry be used to estimate the number of marbles in a 1-liter jar