CBSE Class 11-science Answered

Let n be the number of air molecules in the bulb.

Given : V₁ = 250 cm³, P₁ = 10^-3 mm of Hg, T₁ = 300 K

Using PV = nRT, where R is the universal gas constant, we get

10^-3 x 250 = nR x 300                        ...................(i)

At N.T.P, one molecule of air occupies a volume of 22400 cm³. So, at N.T.P,

V₂ = 22400 cm³, P₂ = 760 mm of Hg, T₂ = 273 K, n₂ = N = 6 x 10²³ molecules

Therefore, 760 x 22400 = 6 x 10²³ x R x 273 ......................(ii)

Dividing (i) by (ii), we get