R-8.3145 JKmol K);ka-13806 x 105 J/K: h-66262 sovided. 1 atm = 760 torr = 101 32

Question

R-8.3145 JKmol K);ka-13806 x 105 J/K: h-66262 sovided. 1 atm = 760 torr = 101 325 Pa A list ofequations from the notes is 1. (20 points) A. In the following drawing showing the collision molecules is ur. [ 5, of hard sphere "molecules" A parameter is b and the angle of collision is y. The relative velocity of the dAB The component of relative velocity along the line of centers (along dax) at the point of collision is ur. The component of Kinetic Energy associated with Urs E. = (1/2)(w), Find an expression for E. in thefonmofE, = (1/2)(ur - ). (Hint: The final equation includes the 0 parameters b and das but not y.) A i A B AR use reacton:- A ressure and a temperature of 298 K. Assuming a molecular B. A vessel contains N2 at 7.0 torr p diameter of 0.316 nm, find the number of N2-N2 collisions per m per second. ith numbers and units. Indicate how you obtain the units of your answer The atomic mass of the nitrogen atom is 14.007 g/mol. this equation.

Explanation / Answer

According to Kinetic molecular theory, the collision frequency is equal to the root-mean-square velocity of the molecules divided by their mean free path.

V = vrms /

Step1 Calculate the Root-mean-square velocity for N2 at 298 K

vrms = 3RT/M

Where

R = the Universal Gas Constant =8.314 J/K-1 mol-1
T = the temperature = 298K
M = the molar mass of N2
Molar mass of N =14.007 g/mol

Molar mass of N2 = 14.007 g/mol x 2 = 28.014 g/mol

= 2.8014 x 10^-2 kg/mol

Putting the values into equation

vrms = square root of [(3 X 8.314 JK-1mol-1x 298 K) /2.8014 X 10^-kgmol-1) x (1 kg m2s-2/J)]

All the units cancel except m2s-2

vrms = square root of ( 3 X 8.314 X 298/2.8014 X 10^-3 m2 s-2)

vrms = square root of (265321.48 m2 s-2)

vrms = 515.1 m/s

Step2 Calculate mean free path for N2

= RT/ ()2 NA P

R= 0.083 14 barLK-1mol-1= 8.314×10-5barm3K-1mol-1
T= 298 K
NA=6.022×10^23mol-1

P = 7 torr = 9.33 x 10^-3 bar

Effective collision area = 2diameter = 0.316 x 2 = 0.632 nm x 6.32 x 10^-10 m

Putting the values into equation

= (8.314×10-5barm3K-1mol-1 x 298 K)/ 3.14 x (6.32 x 10^-10 m)^2 x 6.022×10^23mol-1 x9.33 x 10^-3 bar

= 2.48 x 10^-2/9935.856

= 2.496 x 10^-6 m

Step3 Calculate Collision frequency

V = vrms /

V = 515.1 m s-1/2.496 x 10^-6 m

V = 2.06 x 10^8 s-1

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