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TOPIC 4 - THE COLLISION THEORY

Example of solved problems

1.   For molecular oxygen at 25 oC, (a) calculate the collision frequency z11 and (b)
     the collision density Z11 at a pressure of 1 atm. Given: doxy. = 0.361 nm.

     Answer:

     The collision diameter of O2 is 0.361 nm or 0.361×10-9 m

     (a) Collision frequency z11 = (πd2) × C rel ×ρ = (πd2) ×<v11> ×ρ

     Collision cross-section= σ = πd2 = π(0.361×10-9 m)2 = 4.094×10-19 m2

     Relative mean speed, C rel = <v11> = 2 < v1 >
                                              1/ 2
             8 × 8.314 kg m s −1 × 298 K 
     <v1> = 
             π × 32 × 10 −3 kg mol −1             = 444 ms-1
                                         

     <v11>= 2 < v1 >= 2 (444 ms-1) = 627.9 ms-1

                              N PN A (1 atm)(6.022 × 10 23 mol −1 )(10 3 L m −3 )
     Molecule density = ρ =     =    =
                              V   RT   (0.08206 atm L K −1 mol −1 )(298 K )

                             = 2.46×1025m-3

     So z11 = (πd2) <v11> ρ=(4.094×10-19 m2)( 627.9 ms-1)( 2.46×1025m-3)

            = 6.33×109 s-1

                                  1          1
     (b) Collision density Z11 = z11 × ρ = × (πd2) × C rel ×ρ = 7.78×1034 m-3s-1
                                  2          2
                          −3 −1      −3 3 −1
           (7.78 × 1034m s )(10 m L )
         =                                   = 1.29×108mol L-1s-1
                   6.022 × 10 23 mol −1

2.   Large vacuum chambers have been built for testing space vehicles at 10-7 Pa. Calculate
     the number of molecular impacts per square meter of wall per second for molecular
     CO2 (d = 0.40 nm) at 35 oC?

     Answer:
                                                      ρ<v>
     Use formula of collision with wall, JN =
                                                         4
     Calculate one item after another:

     i. Molecule density:
               N PN A           10 −7 Pa × 6.02 × 10 23 mol −1
           ρ= =           =                                    = 2.34 × 1013 m −3
               V    RT      8.3145 Pa m3 K −1mol −1 × 308.15 K

                                              16
ii. Mean speed :
                           1/ 2                                                      1/ 2
               8 RT               8 × 8.314 kg m s −1 K −1 mol −1 × 308 K 
       〈 v〉 =                   =                                                       = 385.07 ms −1
               πM                        π × 44 × 10 −3 kg mol −1         


     iii.       Put together:

                                    ρ<v>            (2.34 × 1013 m −3 )(385.07ms −1 )
                     Zw = J N =                 =                                     = 2.25×1015m-2s-1
                                          4                         4

3.     A 10 mL container with a hole1µm in diameter is filled with hydrogen. This
       container is placed in an evacuated chamber at 10 oC. How long will it take for
       80% of the hydrogen to effuse out?

       Answer:

                   1 dN ρ < v > N < v >
            JN = −      =      = ×                                            V = volume
                   A dt      4  V   4
              dN A < v >
            −    =        dt                                                  < v > = mean speed
               N     4V
                Nt                    t
                   dN A < v >
            −   ∫ N = 4V ∫ dt
                N0            0
                                                                              A = area , N = no. of molecules

                                        1 × 10 −6 2
                                                  m)   π(                     1/ 2
               No A < v >        100        2             8 × 8.3145 × 283 
            ln    =       t = ln     =                 ×                           ×t
               Nt   4V            20 4 × 10 × 10 −6 m 3  π × 2 × 10 −3 
                                                                           
            1.6094 = 1.963 × 10 −8 m × 1730.88ms −1 × t
            t = 47355 s ≈ 8 9 m i n
                         7                    ≈3 . 1 5 h
                                              1

4.     An effusion cell has a circular hole of diameter 2.50 mm. If the molar mass of
       the solid in the cell is 260 g mol-1 and its vapor pressure is 0.835 Pa at 400 K,
       by how much will the mass of the solid decrease in a period of 2 h?

       Answer:
                                                                       1/ 2
                          ∆w        ρ < v > 1  PN A  8 RT                        PN A
            J N = Zw =           =         =                              =
                        Ao m∆t         4      4  RT  πM                       (2πRTM )1 / 2
                  PN A Ao m ∆t       PMAo ∆t
            ∆w =            1/ 2
                                 =
                 (2πRTM )          (2πRTM )1 / 2
                                                       1
                 0.835 Pa × 260 × 10 −3 kg mol −1 × π ( × 2.50 × 10 − 4 m) 2 × 2 × 60 × 60s
                                                       2                                      kg m −1 s − 2
            ∆w =                                                                            ×
                        (2π × 8.3145 kg m 2 s −1 × 400 × 260 × 10 −3 kg mol −1 )1 / 2            Pa
            = 1.041 × 10 − 4 kg = . 0 1 0 4 g
                                0




                                                            17
Formula for collisions           Same molecule          Different molecules
       1 . Collision frequency, or              2                        2
                                       z11 = (πd ) × C rel ×ρ z12 = (πd 12 ) × C rel (12) × ρ 2
           collision per second, or
           collision of a single
           molecule.
       2. Collision density or                   1                           2
                                                                  Z12 = (πd 12 ) × C rel (12) × ρ1 ρ 2
           total number of coll. or    Z11 =       z11 ρ
                                                 2
           -coll. per volume per
           second.                          1
                                        =     × πd2× C rel ×ρ2
                                            2
       3. Collision flux or                                dN w ρ < v >   ∆wN A
          collision with wall, or                Zw=JN =       =        =
                                                            dt     4      MAo ∆t
          with surface, or
          coll. through a hole, or
          coll. per area s-1 or
          rate of effusion.



Exersice 4a

1.   What is the mean speed of nitrogen molecule relative to another nitrogen
     molecule at 290 K?(662 m s-1)

2.   What is the average relative speed of hydrogen molecules with respect to
     oxygen molecules at 300 K? (1837 m s-1)

3.   Calculate the average number of molecules per cubic centimeter for
     nitrogen gas at 1 bar pressure and 25 oC? (2.43×1019)

4.   The pressure in interplanetary space is estimated to be of the order of 10-14
     Pa. Calculate the collision frequency for an atom present in the space.
     Assume that only hydrogen atoms (d = 0.2 nm) are present and that the
     temperature is 1000 K and.(5.9×10-10s-1)

5.    How many collisions does a single Ar atom (d = 0.34 nm) make in 1.0 s
      when the temperature is 25 oC and the pressure is 1.0 µatm? Given: Ar:
      MR = 39.95 (4.98×103s-1)

6.   How many collisions per second does an N2 (d = 0.43 nm) molecule make
     at an altitude of 15 km where the temperature is 217 K and the pressure
     0.050 kPa. (4.1×106 s-1)

7.   Calculate the collision frequency for hydrogen molecule in hydrogen gas
     at 1.0 bar pressure and 20 oC? Given: dH2 = 2.47 Ǻ. (1.2×107s-1 )


                                            18
8.        Calculate the collision density for molecular chlorine at 35 oC and 770
          mmHg. The collision diameter is 5.44 ×10-6cm. (1.14×1035 s-1m3)

9.        What is the average time between collisions of an oxygen molecule in
          oxygen gas at 10-12 bar and 50 oC (d = 3.60 Ǻ)?

10. Calculate the mean free path for chlorine gas (d= 0.544 nm) at 0.1 Pa and
    25 oC? (0.0312 m)

11. How many molecules of O2 strike the wall per unit area per unit time at
    600 mmHg at 25 oC? (dO2 =3.60Å )( 2.1×1027 m-2s-1)

12. Large vacuum chambers have been built for testing space vehicles at 10-6 Pa.
    Calculate the number of molecules of nitrogen (d = 0.375 nm) impact per
    square meter of wall per second at 28 oC. (2.6×1021 m-2s-1)

13. A solid surface with dimensions 2.5 mm×3.0 mm is exposed to He gas at 90 Pa
    and 500 K. How many collision do the He atoms make with this surface in
    15s?(He: MR= 4.00) (5.9 ×1020 m-2s-1)



Exercise 4b

     1.     A box contains H2 (molecule b) and He (molecule c) at a total pressure of 1.4
            atm. If the mixture contains 18% H2 by weight, calculate the ratio of
            zb(b)/zc(c)? Given db =0.247 nm and dc= 2.20 nm.(6.97×10-3)

     2.     A gas mixture contains H2 at 2/3 atm and O2 at 1/3 atm at 27 oC. Calculate the
            number of collision per second of H2 with other H2, O2 with other O2, and H2
            with O2 (dH2 = 0.272 nm, dO2 = 0.361 nm) (9.5×109 s-1, 2.1×109 s-1 , 4.7×109) .

     3.     At 30 km above the Earth’s surface (roughly in the middle of the
            stratosphere) , the temperature is roughly 1000 K and the gas density is
            3.74×1023 molecules/m3. Assuming N2 (d = 0.38 nm) is representative of the
            stratosphere, determine z11; undergoes in this region of the stratosphere in 1 s.
            (2.09×108).

     4.     An equal number of moles of H2 (d1 =0.272 nm) and Cl2 (d2=0.544 nm) are
            mixed and held at 298 K and the total pressure of 1 kPa. (a) Calculate the
            collision frequencies z12 and z21, where hydrogen is component 1 and
            chlorine is component 2. (b) Calculate the collision density, Z12. (1.14×108s-1,
            1.14×108s-1, 2.30×104 mol L-1s-1)

     5.     For O2 (d = 0.361×10-9m) at 10-3 bar at 25 oC, (a) What is the collision
            density Z11 ? (b) What is the average time between collisions of a single
            molecule?(1.26×102 mol L-1s-1 ,1.6×10-7 s).




                                               19
6.    The average surface temperature of Mars is 220 K, and the surface pressure
      is 4.7 torr. The Martian atmosphere is mainly CO2 and N2 with smaller
      amounts of Ar, O2, CO, H2O, and Ne. Considering only the two main
      components, we approximate the Martian atmospheric composition as
      xCO2= 0.97 and xN2= 0.03. The collision diameters are dCO2=4.6 Ǻ and
      dN2= 3.7 Ǻ. For gas at 220 K at the Martian surface, calculate (a) the
      collision rate for one particular CO2 molecule with other CO2 molecules; (b)
      the collision rate for one particular N2 molecule with CO2 molecules;(c) the
      number of collisions per second made by one particular N2 molecule;(d) the
      number of CO2-N2 collisions per second in 1.0 cm3.

7.    A container of volume 1×10-5 cm3 holds three molecules of gas b, which we
      label b1, b2, and b3. In 1 s, there are two b1-b2 collisions, two b1-b3 collisions,
      and two b2-b3 collsions. Find zb(b) and Zbb.

      The vapor pressure of water at 25 oC is 3160 Pa. If every water molecule that
      strikes the surface of liquid water sticks, what is the rate of evaporation of
      molecules from a square meter of surface? (1.2×1026 m-2s-1)

8.   A 5- mL container with a hole10µm in diameter is filled with hydrogen. This
     container is placed in an evacuated chamber at 0 oC. How long will it take for
     90% of the hydrogen to effuse out? (347 min)

9.   Exactly 1 dm3 of nitrogen, under a pressure of 1 bar, will it takes 5.80 minutes
     to effuse through an orifice. How long will it take for Helium to effuse under
     the same conditions? (2.19 min)

10. The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At
    25 oC and assuming that air consists of N2 molecules with a collision diameter
    of 395 pm, calculate the mean free path of the gas. (4.46×104 m)

11. Dry air had a CO2 mole fraction of 0.004. Calculate the total mass of CO2 that
    strikes 0.5 cm2 of one side of a green leaf in 10 s in dry air at 39 o C and 1
    atm? (0.332 g )

12. A certain sample of a pure oxygen ( d = 3.61 Ǻ) has mean speed, <v> = 450 m
    s-1 and the average time between two successive collisions of a given
    molecule with other molecules is 4.0×10-10 s. Find the mean free path and
    molecule density in this gas. (1.8×10-7 m, 9.6×1024 m-3)




                                          20

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Topic 4 kft 131

  • 1. TOPIC 4 - THE COLLISION THEORY Example of solved problems 1. For molecular oxygen at 25 oC, (a) calculate the collision frequency z11 and (b) the collision density Z11 at a pressure of 1 atm. Given: doxy. = 0.361 nm. Answer: The collision diameter of O2 is 0.361 nm or 0.361×10-9 m (a) Collision frequency z11 = (πd2) × C rel ×ρ = (πd2) ×<v11> ×ρ Collision cross-section= σ = πd2 = π(0.361×10-9 m)2 = 4.094×10-19 m2 Relative mean speed, C rel = <v11> = 2 < v1 > 1/ 2  8 × 8.314 kg m s −1 × 298 K  <v1> =   π × 32 × 10 −3 kg mol −1   = 444 ms-1   <v11>= 2 < v1 >= 2 (444 ms-1) = 627.9 ms-1 N PN A (1 atm)(6.022 × 10 23 mol −1 )(10 3 L m −3 ) Molecule density = ρ = = = V RT (0.08206 atm L K −1 mol −1 )(298 K ) = 2.46×1025m-3 So z11 = (πd2) <v11> ρ=(4.094×10-19 m2)( 627.9 ms-1)( 2.46×1025m-3) = 6.33×109 s-1 1 1 (b) Collision density Z11 = z11 × ρ = × (πd2) × C rel ×ρ = 7.78×1034 m-3s-1 2 2 −3 −1 −3 3 −1 (7.78 × 1034m s )(10 m L ) = = 1.29×108mol L-1s-1 6.022 × 10 23 mol −1 2. Large vacuum chambers have been built for testing space vehicles at 10-7 Pa. Calculate the number of molecular impacts per square meter of wall per second for molecular CO2 (d = 0.40 nm) at 35 oC? Answer: ρ<v> Use formula of collision with wall, JN = 4 Calculate one item after another: i. Molecule density: N PN A 10 −7 Pa × 6.02 × 10 23 mol −1 ρ= = = = 2.34 × 1013 m −3 V RT 8.3145 Pa m3 K −1mol −1 × 308.15 K 16
  • 2. ii. Mean speed : 1/ 2 1/ 2  8 RT   8 × 8.314 kg m s −1 K −1 mol −1 × 308 K  〈 v〉 =   =  = 385.07 ms −1  πM   π × 44 × 10 −3 kg mol −1  iii. Put together: ρ<v> (2.34 × 1013 m −3 )(385.07ms −1 ) Zw = J N = = = 2.25×1015m-2s-1 4 4 3. A 10 mL container with a hole1µm in diameter is filled with hydrogen. This container is placed in an evacuated chamber at 10 oC. How long will it take for 80% of the hydrogen to effuse out? Answer: 1 dN ρ < v > N < v > JN = − = = × V = volume A dt 4 V 4 dN A < v > − = dt < v > = mean speed N 4V Nt t dN A < v > − ∫ N = 4V ∫ dt N0 0 A = area , N = no. of molecules 1 × 10 −6 2 m) π( 1/ 2 No A < v > 100 2  8 × 8.3145 × 283  ln = t = ln = × ×t Nt 4V 20 4 × 10 × 10 −6 m 3  π × 2 × 10 −3    1.6094 = 1.963 × 10 −8 m × 1730.88ms −1 × t t = 47355 s ≈ 8 9 m i n 7 ≈3 . 1 5 h 1 4. An effusion cell has a circular hole of diameter 2.50 mm. If the molar mass of the solid in the cell is 260 g mol-1 and its vapor pressure is 0.835 Pa at 400 K, by how much will the mass of the solid decrease in a period of 2 h? Answer: 1/ 2 ∆w ρ < v > 1  PN A  8 RT  PN A J N = Zw = = =    = Ao m∆t 4 4  RT  πM  (2πRTM )1 / 2 PN A Ao m ∆t PMAo ∆t ∆w = 1/ 2 = (2πRTM ) (2πRTM )1 / 2 1 0.835 Pa × 260 × 10 −3 kg mol −1 × π ( × 2.50 × 10 − 4 m) 2 × 2 × 60 × 60s 2 kg m −1 s − 2 ∆w = × (2π × 8.3145 kg m 2 s −1 × 400 × 260 × 10 −3 kg mol −1 )1 / 2 Pa = 1.041 × 10 − 4 kg = . 0 1 0 4 g 0 17
  • 3. Formula for collisions Same molecule Different molecules 1 . Collision frequency, or 2 2 z11 = (πd ) × C rel ×ρ z12 = (πd 12 ) × C rel (12) × ρ 2 collision per second, or collision of a single molecule. 2. Collision density or 1 2 Z12 = (πd 12 ) × C rel (12) × ρ1 ρ 2 total number of coll. or Z11 = z11 ρ 2 -coll. per volume per second. 1 = × πd2× C rel ×ρ2 2 3. Collision flux or dN w ρ < v > ∆wN A collision with wall, or Zw=JN = = = dt 4 MAo ∆t with surface, or coll. through a hole, or coll. per area s-1 or rate of effusion. Exersice 4a 1. What is the mean speed of nitrogen molecule relative to another nitrogen molecule at 290 K?(662 m s-1) 2. What is the average relative speed of hydrogen molecules with respect to oxygen molecules at 300 K? (1837 m s-1) 3. Calculate the average number of molecules per cubic centimeter for nitrogen gas at 1 bar pressure and 25 oC? (2.43×1019) 4. The pressure in interplanetary space is estimated to be of the order of 10-14 Pa. Calculate the collision frequency for an atom present in the space. Assume that only hydrogen atoms (d = 0.2 nm) are present and that the temperature is 1000 K and.(5.9×10-10s-1) 5. How many collisions does a single Ar atom (d = 0.34 nm) make in 1.0 s when the temperature is 25 oC and the pressure is 1.0 µatm? Given: Ar: MR = 39.95 (4.98×103s-1) 6. How many collisions per second does an N2 (d = 0.43 nm) molecule make at an altitude of 15 km where the temperature is 217 K and the pressure 0.050 kPa. (4.1×106 s-1) 7. Calculate the collision frequency for hydrogen molecule in hydrogen gas at 1.0 bar pressure and 20 oC? Given: dH2 = 2.47 Ǻ. (1.2×107s-1 ) 18
  • 4. 8. Calculate the collision density for molecular chlorine at 35 oC and 770 mmHg. The collision diameter is 5.44 ×10-6cm. (1.14×1035 s-1m3) 9. What is the average time between collisions of an oxygen molecule in oxygen gas at 10-12 bar and 50 oC (d = 3.60 Ǻ)? 10. Calculate the mean free path for chlorine gas (d= 0.544 nm) at 0.1 Pa and 25 oC? (0.0312 m) 11. How many molecules of O2 strike the wall per unit area per unit time at 600 mmHg at 25 oC? (dO2 =3.60Å )( 2.1×1027 m-2s-1) 12. Large vacuum chambers have been built for testing space vehicles at 10-6 Pa. Calculate the number of molecules of nitrogen (d = 0.375 nm) impact per square meter of wall per second at 28 oC. (2.6×1021 m-2s-1) 13. A solid surface with dimensions 2.5 mm×3.0 mm is exposed to He gas at 90 Pa and 500 K. How many collision do the He atoms make with this surface in 15s?(He: MR= 4.00) (5.9 ×1020 m-2s-1) Exercise 4b 1. A box contains H2 (molecule b) and He (molecule c) at a total pressure of 1.4 atm. If the mixture contains 18% H2 by weight, calculate the ratio of zb(b)/zc(c)? Given db =0.247 nm and dc= 2.20 nm.(6.97×10-3) 2. A gas mixture contains H2 at 2/3 atm and O2 at 1/3 atm at 27 oC. Calculate the number of collision per second of H2 with other H2, O2 with other O2, and H2 with O2 (dH2 = 0.272 nm, dO2 = 0.361 nm) (9.5×109 s-1, 2.1×109 s-1 , 4.7×109) . 3. At 30 km above the Earth’s surface (roughly in the middle of the stratosphere) , the temperature is roughly 1000 K and the gas density is 3.74×1023 molecules/m3. Assuming N2 (d = 0.38 nm) is representative of the stratosphere, determine z11; undergoes in this region of the stratosphere in 1 s. (2.09×108). 4. An equal number of moles of H2 (d1 =0.272 nm) and Cl2 (d2=0.544 nm) are mixed and held at 298 K and the total pressure of 1 kPa. (a) Calculate the collision frequencies z12 and z21, where hydrogen is component 1 and chlorine is component 2. (b) Calculate the collision density, Z12. (1.14×108s-1, 1.14×108s-1, 2.30×104 mol L-1s-1) 5. For O2 (d = 0.361×10-9m) at 10-3 bar at 25 oC, (a) What is the collision density Z11 ? (b) What is the average time between collisions of a single molecule?(1.26×102 mol L-1s-1 ,1.6×10-7 s). 19
  • 5. 6. The average surface temperature of Mars is 220 K, and the surface pressure is 4.7 torr. The Martian atmosphere is mainly CO2 and N2 with smaller amounts of Ar, O2, CO, H2O, and Ne. Considering only the two main components, we approximate the Martian atmospheric composition as xCO2= 0.97 and xN2= 0.03. The collision diameters are dCO2=4.6 Ǻ and dN2= 3.7 Ǻ. For gas at 220 K at the Martian surface, calculate (a) the collision rate for one particular CO2 molecule with other CO2 molecules; (b) the collision rate for one particular N2 molecule with CO2 molecules;(c) the number of collisions per second made by one particular N2 molecule;(d) the number of CO2-N2 collisions per second in 1.0 cm3. 7. A container of volume 1×10-5 cm3 holds three molecules of gas b, which we label b1, b2, and b3. In 1 s, there are two b1-b2 collisions, two b1-b3 collisions, and two b2-b3 collsions. Find zb(b) and Zbb. The vapor pressure of water at 25 oC is 3160 Pa. If every water molecule that strikes the surface of liquid water sticks, what is the rate of evaporation of molecules from a square meter of surface? (1.2×1026 m-2s-1) 8. A 5- mL container with a hole10µm in diameter is filled with hydrogen. This container is placed in an evacuated chamber at 0 oC. How long will it take for 90% of the hydrogen to effuse out? (347 min) 9. Exactly 1 dm3 of nitrogen, under a pressure of 1 bar, will it takes 5.80 minutes to effuse through an orifice. How long will it take for Helium to effuse under the same conditions? (2.19 min) 10. The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25 oC and assuming that air consists of N2 molecules with a collision diameter of 395 pm, calculate the mean free path of the gas. (4.46×104 m) 11. Dry air had a CO2 mole fraction of 0.004. Calculate the total mass of CO2 that strikes 0.5 cm2 of one side of a green leaf in 10 s in dry air at 39 o C and 1 atm? (0.332 g ) 12. A certain sample of a pure oxygen ( d = 3.61 Ǻ) has mean speed, <v> = 450 m s-1 and the average time between two successive collisions of a given molecule with other molecules is 4.0×10-10 s. Find the mean free path and molecule density in this gas. (1.8×10-7 m, 9.6×1024 m-3) 20