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Why does collision frequency decrease with increasing T?

  1. Apr 23, 2015 #1
    From the equation for Z from this website https://books.google.com.sg/books?i...page&q=collision frequency Z equation&f=false

    It shows that as T increases Z decreases. However this seems quite counter intuitive to me as I had always been taught that as temperature increases so does the frequency of effective collisions/collisions. The only reason that I could think of was due to the mean free path increasing as temperature increases causing the frequency of collision to decrease. But I feel that a problem with that statement is that even though the mean free path increases, the velocity of the gas increases so wouldn't that cancel out the effect of having a greater distance to cover? So why does an increase in T decrease Z?

    Secondly, in the equation for Z the average speed of the molecules, c is in the numerator of the equation while temperature, T is in the denominator. This again seems weird to me because I always thought that the speed of molecules and temperature should be proportional to each other so they shouldn't be on opposite sides of the fraction. Why aren't they both on one side of the equation?
     
  2. jcsd
  3. Apr 23, 2015 #2

    Quantum Defect

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    P/RT on the right is the same thing as 1/Vm, where Vm is the molar volume.

    What you have is essentially a volume swept out per second ( c * sigma) divided by the molar volume.

    The collision cross section has some small temperature dependence, but you can assume that it is a constant.

    c is proportional to T^(1/2), and Vm is proportional to T (Charles' Law -- assumes pressure is constant)

    As a result, Z is proportional to T^(-1/2) if pressure is constant. The molecules are moving faster, which should increase the collision rate, but the gas is expanding (P is a constant, so Vm increases) and the space between molecules is increasing which will decrease the collision rate. The net is a reduction in collision rate.

    If you assume that the molar volume is fixed (a closed container) then the collision rate will be proportional to T^(1/2) -- an increasing T will result in a larger collision frequency.
     
  4. Apr 23, 2015 #3
    Thanks for the reply. How would I combine the c is proportional to T^(1/2), and Vm is proportional to T to get Z is proportional to T^(-1/2) if pressure is constant? Sorry my math is not very strong so I'm not sure how to combine them together.

    If we were to assume that molar volume is fixed that would mean that the 1/Vm term is constant right? So the only temperature dependent term is the c term and so it would increase the collision frequency? Is this in time with your second paragraph?

    Thank you for the quick response.
     
  5. Apr 24, 2015 #4

    Quantum Defect

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    Z = 2^(1/2) Na * c_bar * sigma / Vm

    c_bar = c1*T^(1/2) , Vm = c2 * T (constant p)

    Z = c3 * T^(1/2)/ T = c3 * T^(-1/2), where c1, c2, and c3 are constants.

    If Vm is a constant, then Z = c4 * T^(1/2)
     
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