Question

Van der Waals constant 'b' of a gas is 1250 litre/mole. How near can be the centres of the two molecules approach each other?

• A. $1.0\times {10}^{-7}$ cm
• B. $5\times {10}^{-8}$ cm
• C. $2\times {10}^{-7}$ cm
• D. $3\times {10}^{-8}$ cm

Answer 1

The correct option is A $1.0\times {10}^{-7}$ cm

$b=N_A\times 4\times \frac{4}{3}\pi r^3$
$\therefore r^3=\frac{3\times 1250}{6\times {10}^{23}\times 4\times 4\times 3.14}\cong 125\times {10}^{-24}$
$\therefore r=5\times {10}^{-8}$cm

So, collision diameter (nearest possible distance)
$2r=2\times 5\times {10}^{-8}$ cm =${10}^{-7}$ cm