1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Portion problem including Vrms

  1. Aug 23, 2016 #1
    1. The problem statement, all variables and given/known data
    The rms speed of molecules in a gas at 20.0C is to be increased by 2 percent. To what temperature must it be raised?

    2. Relevant equations
    Vrms = Squareroot((3kt)/m)

    3. The attempt at a solution
    vrms.02 + vrms = squareroot((3kt)/m)
    vrms(1.02) = squareroot((3kt)/m)

    Too many unknowns.

    In the solution manual it does
    (vrms1/vrms2) = 1.02 = (squareroot t1)/(squareroot t2)
    t2 1.02^2 = t1 then plug in 20 plus 275 for t2 to get t1

    My miss understanding is how does the solution manual know its 1.02 as a portion?
    They use proportionality or something. I can't see the logical steps.

    Thank you
     
  2. jcsd
  3. Aug 23, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You should have two equations here, one relating the unincreased vrms to the unraised temperature, and one relating the increased vrms to the new temperature.
     
  4. Aug 26, 2016 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    Once you've written an equation for each of the two different cases, in both of these equations the k and m have unchanged values for your gas sample (even though you may not know those values).
     
  5. Aug 27, 2016 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Equating k.e. to molecular temperature energy we get
    mv2/2 = 3kT/2 (kT/2 in each of the three directions)
    So we see that v2 ∝ T
    and so vRMS ∝ √T.
    Of course T is in Kelvin.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted