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Question: Modern vacuum pumps permit the pressures down to $4.1 \times 10^{-14}$ atm to be reached at room tem...

Modern vacuum pumps permit the pressures down to 4.1×10−144.1 \times 10^{-14} atm to be reached at room temperature (27∘C)(27^\circ C). Assuming that the gas exhausted is nitrogen, find the mean distance between the gas molecules remained at this pressure. Diameter of nitrogen molecule is 1.5 Å

A

1 x 10^7 m

B

0.01 cm

C

1 x 10^-7 m

D

1 x 10^7 cm

Answer

1 x 10^7 m

Explanation

Solution

The mean distance between gas molecules at low pressure is the mean free path (λ\lambda), given by λ=kT2πd2P\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}. Given: P = 4.1×10−144.1 \times 10^{-14} atm =4.1×10−14×1.013×105= 4.1 \times 10^{-14} \times 1.013 \times 10^5 Pa ≈4.153×10−9\approx 4.153 \times 10^{-9} Pa T = 27∘C=27+273.15=300.1527^\circ C = 27 + 273.15 = 300.15 K ≈300\approx 300 K d = 1.51.5 Å =1.5×10−10= 1.5 \times 10^{-10} m k = 1.38×10−231.38 \times 10^{-23} J/K

Substituting the values: λ=(1.38×10−23 J/K)×(300 K)2×π×(1.5×10−10 m)2×(4.153×10−9 Pa)\lambda = \frac{(1.38 \times 10^{-23} \text{ J/K}) \times (300 \text{ K})}{\sqrt{2} \times \pi \times (1.5 \times 10^{-10} \text{ m})^2 \times (4.153 \times 10^{-9} \text{ Pa})} λ≈4.14×10−211.414×3.14159×(2.25×10−20)×(4.153×10−9)\lambda \approx \frac{4.14 \times 10^{-21}}{1.414 \times 3.14159 \times (2.25 \times 10^{-20}) \times (4.153 \times 10^{-9})} λ≈4.14×10−214.151×10−28≈0.997×107\lambda \approx \frac{4.14 \times 10^{-21}}{4.151 \times 10^{-28}} \approx 0.997 \times 10^7 m λ≈1×107\lambda \approx 1 \times 10^7 m