Question

QUESTION 4 What is the maximum pressure (in Torr) that will afford a N2 molecule a mean-free-path of at least 1.00 m at 25 oC

Answer 1

Answer:

Maximum pressure P = 4.9 × 10⁻⁵ Pa

Explanation:

From the information given, the mean free path can be expressed with the formula:

$\lambda = \dfrac{RT}{\sqrt{2} \pi \times d^2 \times N_A \times P}$

Making Pressure P the subject of the formula because we intend to find the maximum pressure, we have:

$P= \dfrac{RT}{\sqrt{2} \pi \times d^2 \times N_A \times \lambda }$

At standard conditions

R = gas constant = 8.314 J/mol.K

T = temperature at 25°C = (273 + 25) = 298 K

π = pi = 3.14

d = (364× 10⁻¹²m)²

$N_A$ = avogadro's number = 6.023 × 10²³

λ = mean free path = 1.0 m

$P= \dfrac{RT}{\sqrt{2} \pi \times d^2 \times N_A \times \lambda }$

$P= \dfrac{8.314 \ J/mol.K \times 298 \ K}{\sqrt{2}\times (3.14) \times (364 \times 10^{-12} \ m) ^2 \times 6.023 \times 10^{23}/mol \times 1.0 \ m }$

P = 0.007 kg/m.s²

P = 0.007 Pa

$P = 0.007 Pa \times \dfrac{0.007 \ torr}{1 \ Pa}$

P = 4.9 × 10⁻⁵ Pa