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Average speed of an atom

  1. Feb 5, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the ratio of the average speed of an atom of neon to another atom of neon at twice the temperature?
    (A) 1:1
    (B) 1:1.4
    (C) 1:2
    (D) 1:4

    2. Relevant equations
    KE = 3/2RT

    3. The attempt at a solution
    I used the above equation and substituted 2 for T for twice the temperature and 1 for half of that:

    KE=3/2RT = 3R
    KE=3/2RT = 3/2R

    My ratio is 3:1.5. The correct answer is B. So kinetic energy is proportional to temperature, but what is the relationship between speed and kinetic energy?
     
  2. jcsd
  3. Feb 5, 2015 #2

    DrClaude

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    Staff: Mentor

    Can't you figure it out? At least the proportionality between kinetic energy and speed? Write your ratio correctly (1 to something, 1:x), and then try to figure out the relation between x and 1.4.
     
  4. Feb 5, 2015 #3

    Borek

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    Is that really the first time you hear about kinetic energy? Of any object?

    Your relevant equation is only one of two required to solve the problem. There is another one, one so basic you will feel ashamed once you realize what it is. Or if you don't know it, you should feel ashamed for not knowing it ;)
     
  5. Feb 5, 2015 #4
    Ok. No, this is not the first time i heard about kinetic energy. For example, the average speed of an atom is defined in terms of KE, which is KE = 3/2RT. Also, another equation defines KE, which is KE = 1/2 mv^2. Putting these together I get v = sqrt. 3RT/m. Also what is known is that temperature is directly proportional to KE.

    I can eliminate A because it cannot be 1:1 if the temperature is doubled on one of them.

    When I use v = sqrt. 3RT/m, I get 0.024 (gas double temp) and 0.012 for the gas with standard temp. This still gives me a proportionality of 2:1.
     
  6. Feb 5, 2015 #5

    Quantum Defect

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    You are soooo very close!

    Do what Dr. Claude suggests (set this up as a ratio of speeds, using the equation for velocity that you have)

    v(T_2)/v(T_1) = SQRT (3RT_2/m)/SQRT (3RT_1/m) ==> all of the constants on the right side (3, R, T, m) cancel, and you get ...
     
  7. Feb 5, 2015 #6

    Borek

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    Show/check your math.
     
  8. Feb 5, 2015 #7
    Got it. So in order to relate T and avg. kinetic energy, we must set 1/2mv^2 = 3/2RT. We get sqrt. 3RT/m.

    Then, using Graham's law: v1/v2 = sqrt. ms/sqrt. m1, we get (sqrt. 10)/(sqrt. 5) = sqrt. 2 = 1.41.

    So, Graham's law only relates rms velocities to molar mass. In order to integrate temperature, we must set the above equations equal to each other and solve for v. Conceptually, we would not see a double in v since the relationship between v and KE is a sqrt. Am I finally understanding this now? Thanks in advance!
     
  9. Feb 5, 2015 #8

    Borek

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    You don't need Graham's law here.

    Let's start from the very beginning: please write formula relating average speed with the temperature (either use LaTeX or treat it with parentheses, to avoid any ambiguity).
     
  10. Feb 5, 2015 #9
    (1/2mv2)=(3/2RT)
    v = (sqrt. 3RT/m)
     
  11. Feb 5, 2015 #10

    Borek

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    (sqrt. 3RT/m) or sqrt(3RT/m)?
     
  12. Feb 5, 2015 #11
    sqrt (3RT/m); sorry.
     
  13. Feb 5, 2015 #12

    Borek

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    OK, so we have

    [tex]V_1 = \sqrt{\frac{3RT_1}{m}}[/tex]

    Assume [itex]T_2 = 2T_1[/itex] (twice higher), plug it into identical equation for V2, calculate ratio of [itex]\frac{V_1}{V_2}[/itex], what do you get?
     
  14. Feb 5, 2015 #13
    V1=sqrt (6R/20) and V2=sqrt. (3R/20). Since R is a constant, can I just use that value (0.08)?
     
  15. Feb 5, 2015 #14

    Borek

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    If R is a constant, it cancels out.
     
  16. Feb 5, 2015 #15
    Ok. Now that temperature is accounted for already, v1/v2 =0.38/0.70. I see how the math is done but for some reason I don't see the proportions.
     
  17. Feb 5, 2015 #16

    Borek

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    Why do you go for numbers instead of canceling everything out? For example

    [tex]\frac{\sqrt{40}}{\sqrt{60}}=\frac{\sqrt{2\times 2\times 2\times 5}}{\sqrt{3\times 2\times 2 \times 5}} = \frac {\sqrt {2}} {\sqrt{3}}[/tex]
     
  18. Feb 6, 2015 #17
    Yes, I see. But I get v1/v=sqrt. 6/sqrt.3 = sqrt. 2 = 1.4. I understand your example above; that was very helpful!
     
  19. Feb 6, 2015 #18

    Borek

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    You are looking for a ratio form, so it is best to leave it as a fraction:

    [tex]\frac{\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}}[/tex]

    and then it is obvious why 1:1.4 is the correct answer. Pure algebra.

    With some experience the answer is obvious. Energy is proportional to the temperature and to the velocity squared, 1.41... is a square root of 2 so it becomes a natural suspect ;)
     
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