O 4/29/2016 10:55 PM A 86.3/100 4/22/2016 11:05 AM Gradebook Print Calculator Pe

Question

O 4/29/2016 10:55 PM A 86.3/100 4/22/2016 11:05 AM Gradebook Print Calculator Periodic Table Incorrect Question 17 of 28 Incorrect Map A sapling earning The mean free path is the average distance travelled by a molecule, atom, particle, etc. between a collision with another particle or molecule. Calculate the mean free path, A, of air at room temperature, T 72.6 F. Air is mostly nitrogen. Assume the colisions are between N2 molecules. The diameter of N2 is d 2.74 x 10-10 m and the gas is at atmospheric pressure, P J 101325 Pa. Number 14 1.87 x 10 m There is additional feedback available! View this feedback by clicking on the Incorrect. bottom divider bar. Click on the divider bar again to hide the additional feedback. Close Previous & Give Up & View Solution Try Again Next Exit Explanation

Explanation / Answer

Mean free path is lamda = KB*T/1.414*pi*d^2*p

kB is the Boltzmann constant = 1.38*10^-23 m^2 kg s^-2 K^-1

T is the temperature = 72.6 F = 295.7056 K

d is the diameter = 2.74*10^-10 m

P is the pressure of the gas = 101325 Pa

then lamda= (1.38*10^-23*295.7056)/(1.414*3.142*(2.74*10^-10)^2*101325)

lamda = 120.7*10^-9 m = 120.7 nm

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.