Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 × 10-15 atm at room temperature (300 K). Taking R = 8.3 JK-1 mole-1, 1 atm = 105 Pa and N textAvogadro = 6 × 1023 mole-1, the mean distance between molecules of gas in an evacuated vessel will be of the order of :

Question

Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 × 10-15 atm at room temperature (300 K). Taking R = 8.3 JK-1 mole-1, 1 atm = 105 Pa and N textAvogadro = 6 × 1023 mole-1, the mean distance between molecules of gas in an evacuated vessel will be of the order of : (A) 0.2 μ m (B) 0.2 mm (C) 0.2 cm (D) 0.2 nm

Tardigrade Admin · Accepted Answer

λ =( kT /√2 π d 2 P ) =(1.38 × 10-23 × 300/√2 π × 10-20 × 4 × 10-10) =(1.38 × 3/√2 × 4 π) × 10-9 = 0.2 nm

$0.2 μ m$

$0.2 mm$

$0.2 c m$

$0.2 nm$

$λ = \frac{k T}{2 π d ^{2} P}$
$= \frac{1.38 \times 1 0 ^{- 23} \times 300}{2 π \times 1 0 ^{- 20} \times 4 \times 1 0 ^{- 10}}$
$= \frac{1.38 \times 3}{2 \times 4 π} \times 1 0^{- 9}$
$= 0.2 nm$

Q. Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0×10−15atm at room temperature (300K). Taking R=8.3JK−1mole−1,1atm=105Pa and NAvogadro​=6×1023mole−1, the mean distance between molecules of gas in an evacuated vessel will be of the order of :

0.2μm

0.2mm

0.2cm

0.2nm

λ=2​πd2PkT​ =2​π×10−20×4×10−101.38×10−23×300​ =2​×4π1.38×3​×10−9 =0.2nm

$0.2 μ m$

$0.2 mm$

$0.2 c m$

$0.2 nm$

$λ = \frac{k T}{2 π d ^{2} P}$
$= \frac{1.38 \times 1 0 ^{- 23} \times 300}{2 π \times 1 0 ^{- 20} \times 4 \times 1 0 ^{- 10}}$
$= \frac{1.38 \times 3}{2 \times 4 π} \times 1 0^{- 9}$
$= 0.2 nm$