Q1-How many collisions per second does an N2 (= 0.43 nm) molecule make at an alt

Question

Q1-How many collisions per second does an N2 (= 0.43 nm) molecule make at an altitude of 15 km where the temperature is 217 K and the pressure 0.050 kPa.  Please enter a number with format 1.01E9

Q2-Calculate the mean free path (in m) of Ar at 25oC and 1atm. the cross section area is 0.36nm2. Enter a number with format 2.55E-10

Q3-Assuming the diameter of argon atom is 0.34nm, from the result of the previous question, calculate the ratio of the mean free path to the diameter. use format 5.1E2.  Is intermolecular distance large compared to its own size at ambient condition?

Explanation / Answer

Here =

Temperature = 217 K, Pressure = 0.050 kPa or 50 Pa

And = 0.43 nm^2 or 0.43*10^-18 m^2

We know that mass of N2 = 28.02 amu *1.66*10^-27 kg/1 amu

= 4.65*10^-26 kg

The collision number z is given as follows:

z = (16/ mkT)^1/2 p

= (16/ 3.14* 4.65*10^-26 * 1.381*10^-23*217 )^1/2 *0.43*10^-18 m^2*50 Pa

= (16/4.38*10^-46)^1/2 0.43*10^-18 *50

=(3.6566548e+46)^1/2 0.43*10^-18 *50      

= 1.9131126e+23*0.43*10^-18 *50

=4.1131921e6

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