Ans. 290

Sol. P = RT

Let molar volume at 10000 atm and 300 K = V m

10 4 = 0.082 × 300 ….(I)

∴ 1 × (100 V m –b) = 0.082 × 300 ….(II)

(I) ÷ (II) gives.

= 1

or 10 4 V m – 10 4 b = 100 V m –b

or (10 4 –100) V m = (10 4 + 1) b

or V m = × b = 1.0102 b

∴ substituting the value of b is equation (II)

1 × (101.02 –1) b = 0.082 × 300

or b × 100.02 = 0.082 × 300

or b = 0.246

or b = .246 cm 3 /mol

b = 4 × N A × π r 3 ∴

4 × 6 × 10 23 × π × r 3 or r

3 = × 10 –23 = × 10 –24 or

r 3 = 24.48 × 10 –24 or r = 2.9 × 10

–8 cm = 2.9 × 10 –10 m.

or r = 290 pm.