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Homework Help: Why is the Velocity equal to joules per kilogram

  1. Feb 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculating the rms speed of an N2 molecule. The units I end up with are J/kg and this is equivalent to Velocity. my question is why?

    2. Relevant equations
    v = J / kg
    KE = .5mv^2
    3. The attempt at a solution
    I tried to derive it myself using the kinetic energy formula. What I end up with is that ##v=\sqrt{\frac{2*J}{kg}}##

    The units are right, J/kg, but it isn't a direct conversion like the solution on Chegg says. it looks like to convert, my 517 J/kg doesn't equal 517 m/s. It would seem I would have to take the square root of 1034
     
  2. jcsd
  3. Feb 8, 2017 #2
    J/ kg has units of velocity square.
     
  4. Feb 8, 2017 #3

    fresh_42

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    I'm not sure that I understand what you mean.
    ##1J = \frac{kg\cdot m^2}{s^2}## so Joule per kilogram is ##\frac{m^2}{s^2}## which is the unity of the squared velocity: ##[v^2] = \frac{m^2}{s^2}##
    The reason is, that all these units base on the metric SI system, i.e. ##m\, , \,kg\, , \,s\, , \,A\,##.
     
  5. Feb 8, 2017 #4
    @Chestermiller @fresh_42

    you are absolutely right. the ##V_{rms} = \sqrt{\frac{3kT}{m}}## is the square root of J/kg or v^2. But what about the 2? I guess that is already factored into the formula? What I am seeing now, based on your responses, is that velocity is equal to ##v=\sqrt{2*v^2}## it is just not working out in my mind. I am missing something.

    the kinetic energy formula is ##KE={\frac{1}{2}*m*v^2}## or in units ##J=\frac{1}{2}*kg*\frac{m^2}{s^2}##
    solved for ##v## ##v=\sqrt{\frac{2*KE}{m}}## the units being ##\frac{m}{s} = \sqrt{\frac{2*J}{kg}}##
    and you are saying that ##\frac{J}{kg} = \frac{m^2}{s^2}## or ##\frac{J}{kg} = v^2##
    so that would be ##\frac{m}{s} = \sqrt{2*\frac{m^2}{s^2}}## or ##v = \sqrt{2*v^2}##

    that 2 is throwing me off. what am I missing here?
     
    Last edited: Feb 8, 2017
  6. Feb 8, 2017 #5

    TSny

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    The factor of ##\frac{1}{2}## in the KE expression has no dimensions. So, ##\frac{1}{2}## doesn't contribute any units.

    Another example: The SI units for an area are m2. The formula for the area of a circle is ##A = \pi r^2##. When analyzing the units on the right side of ##A##, you would not say that the units for the area are ##\pi##m2.
     
  7. Feb 8, 2017 #6

    fresh_42

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    You shouldn't mix notations of physical quantities like ##v## with units. ##\frac{J}{kg} = v^2## isn't a good way to write it. That's why I wrote ##[v^2] = \frac{J}{kg}## with brackets to indicate "unit of". The two doesn't play a role for the units. If we have three yards in each of three downs, we get ##3 \cdot (3 \cdot yd) = (3 \cdot 3) \cdot yd = 9 \, yd## in total. So the unit of the drive ##dr## was ##[dr] = yd## and it doesn't matter, that it was the result of ##3 \cdot 3\, yd## or ##0\,yd +0\,yd +9\,yd\;##.
     
  8. Feb 8, 2017 #7
    What you are missing is that just because something has units of velocity doesn't mean that it is velocity.
     
  9. Feb 8, 2017 #8
    I'm not arguing, I'm just trying to understand a well-accepted conversion. I know it is right, I just really want to understand why.
    What about the 1/2. What happened to it during conversion.
    Right but if I want to convert 9 yd to downs, I don't just say 9 yards equals 9 downs, I have to divide by that 3.
    this is the way I am thinking of it. 1 yd = 3 x 1ft,3*1*ft the 3 doesn't play a role for the units. except that it does. I am not seeing why 1 yd must equal 3 ft, but with a joule we can ignore the two. I'm just not seeing it. Perhaps it is something that needs to be explained with pencil and paper.
    If the two doesn't matter, why is it present in the kinetic energy formula. It doesn't have a unit so it doesn't change the units, but according to that formula, Joules equal ##\frac{1}{2}*kg*v^2##, not ##kg*v^2##

    I don't know what that means. It doesn't have a direction, but why does that dump the sqrt(2).

    at this point, the only explanation that seems to ease my mind is the the root mean squared speed formula as that two factored in already. that it would be ##v=\sqrt{\frac{2*1.5*k*T}{m}}## that the 2 has already been factored in.
    That in any other conversion from Joules and kg to speed (or velocity), I would need that 2.
     
    Last edited: Feb 8, 2017
  10. Feb 8, 2017 #9
    it's ok if you all want to give up on me. I'm sorry if I am a pain. You can go ahead and delete the thread. I'm thinking maybe I should just ask my professor. Maybe he can get through to me. It's a lot easier when you can talk in person and the other person can yell at you and hammer it in. ;)

    I mean it is not really "important". can we say that anytime I am presented with Joules/kg, I can just take the square root and get the speed? That it's always a

    edit: lol a ##\sqrt{9}:3## conversion.
     
  11. Feb 9, 2017 #10

    fresh_42

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    A factor in front of a unit doesn't affect the unit, only the quantity. ##cm## and ##m## both measure the physical quantity length ##L##. If this length is achieved by a velocity ##V## during a time ##T##, say ##V=10\frac{cm}{h}## and ##T=\frac{1}{2}\,h##, probably a snail, then the snail moved ##L=V \cdot T = 10\frac{cm}{h} \cdot \frac{1}{2}\,h = 5 \, cm## during this half hour. If we do the same calculation in the units meter and seconds, we get ##L = 10 \cdot 10^{-2} \cdot m \cdot \frac{1}{3600\,s} \cdot \frac{1}{2} \cdot 3600 \,s = 5 \cdot 10^{-2} \,m = 0.05 \, m = 5\, cm##. Now the only difference between ## 0.05 \, m## and ##5\, cm## is whether the ##\frac{1}{100}## is swallowed by the ##5## making it ##0.05## or by the unit, making it centi.

    You can always handle units as "times something", i.e. a factor. This is the reason why we can add ##1\,m + 2\,m = 1 \cdot m + 2 \cdot m = (1+2) \cdot m = 3\, m## and can not add ##1\,m + 2##. And it is the reason, why we can multiply both: ##1\,m \cdot 2\,m= 2\,m^2## and ##1\,m \cdot 2 = 2\, m##.
    Therefore any conversion factors can be treated as such, as factors, and you can likewise multiply the amount of quantity or change the unit, but you must not do both for this would square the factor because it would be applied twice. But it's only available once.
     
  12. Feb 9, 2017 #11
    According to your rationale, if s is distance and a is acceleration, then s/a is time squared:$$\frac{s}{a}=t^2$$and $$t=\sqrt{\frac{s}{a}}$$But wait! The SUVAT equations tell us that $$s=\frac{1}{2}at^2$$and$$t=\sqrt{\frac{2s}{a}}$$So,$$t=\sqrt{2t^2}$$There's that pesky little 2 again. How do you explain this?
     
  13. Feb 9, 2017 #12

    gneill

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    Yes, it's already factored into the formula. When you use the formula ##v = \sqrt{\frac{2 \cdot KE}{m}}## the constant "2" arrived via the derivation of ##KE = \frac{1}{2}m v^2## from the work-energy theorem, Newton's second law, and basic kinematics (you can find the derivation easily enough via a google search).

    The expression for Vrms took a different route in its derivation, involving statistical mechanics. If you take the KE to be an average KE for a particle in an ensemble of particles of mass m then if you equate the two velocity formulae:

    ##\sqrt{\frac{3 k T}{m}} = \sqrt{\frac{2 KE}{m}}##

    so

    ##3 k T = 2 KE##

    ##\frac{3}{2}k T = KE##

    The left hand side of the last expression should look familiar from chemistry classes (or statistical mechanics) :smile:
     
  14. Feb 9, 2017 #13
    I don't explain it. That's what I am asking. I am not arguing that the 2 belongs. Obviously it doesn't because it is a well accepted conversion.

    I am asking why. How do you explain the pesky little 2
     
  15. Feb 9, 2017 #14

    this is exactly the issue that I am having. Thankyou.


    ##\frac{3}{2}k T = KE##

    and if I want to convert that kinetic energy to v, I have to multiply it by 2, divide by the mass, and take the square root.

    ##\sqrt{\frac{2*\frac{3}{2}kT}{m}}##
     
  16. Feb 9, 2017 #15
    The 2 obviously does belong.
     
  17. Feb 9, 2017 #16
    ... so then how does ##t=\sqrt{2}t##
     
  18. Feb 9, 2017 #17
    It doesn't. That's the result of your derivation, not the SUVAT result.
     
  19. Feb 9, 2017 #18
    I don't think you understand the question I am asking. I already know that a constant doesn't change the units.
     
  20. Feb 9, 2017 #19
    so then the sqrt(2) doesnt belong. I'm going to backtrack a bit for a second
     
  21. Feb 9, 2017 #19
    I think @gnell has answered my question completely. that kinetic energy is equal to ##KE = \frac{3}{2}kT## it looks like my problem is solved.
     
  22. Feb 9, 2017 #19
    The 2 belongs in the SUVAT equations and the equation for the kinetic energy. The equations that you wrote are the ones that are incorrect.
     
  23. Feb 9, 2017 #20
    hold up
     
  24. Feb 9, 2017 #21
    That formula is familiar to me now. This is all I needed. Is for someone to say "gramps, you Dummy! 3kT equal 2*KE not KE" and so it fits nicely in the ##v=\sqrt{\frac{2KE}{kg}}##

    that brings me back to units.
    so there is a difference between a Kinetic Energy Joule (1/2 Mass x Velocity squared) and a Joule (Mass x Velocity squared)
     
    Last edited: Feb 9, 2017
  25. Feb 9, 2017 #22
    so there is a difference between a Kinetic Energy Joule (1/2 Mass x Velocity squared) and a Joule (Mass x Velocity squared)

    it seems clear that a joule in potential energy is different from a Joule of KE. etc.
     
    Last edited: Feb 9, 2017
  26. Feb 9, 2017 #23
    Nope. A Joule is a unit of energy (any kind).
     
  27. Feb 9, 2017 #24

    fresh_42

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    There is a difference between the kinetic energy ##E=\frac{1}{2}mv^2## and the energy ##E=mc^2##. It's the energy of a different thing, not a different unit.
     
  28. Feb 9, 2017 #25
    I remember we had to be able to derive the Kinetic Energy (via integration). So I know how to do it. I guess I just wasn't really thinking about what I was doing at the time because I was coming from a different angle. And so, now I am confused when it is appearing to me that a joule is defined two different ways.
     
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