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

Maximum pressure P = 4.9 × 10⁻⁵ Pa

Step-by-step explanation:

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

`\lambda = (RT)/(√(2) \pi * d^2 * N_A * P)`

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

`P= (RT)/(√(2) \pi * d^2 * N_A * \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= (RT)/(√(2) \pi * d^2 * N_A * \lambda )`

`P= (8.314 \ J/mol.K * 298 \ K)/(√(2)* (3.14) * (364 * 10^(-12) \ m) ^2 * 6.023 * 10^(23)/mol * 1.0 \ m )`

P = 0.007 kg/m.s²

P = 0.007 Pa

`P = 0.007 Pa * (0.007 \ torr)/(1 \ Pa)`

P = 4.9 × 10⁻⁵ Pa