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Absolute temperature for average translational kinetic energ

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    The ammonia molecule (NH3) has a dipole moment of 5.0×10−30C⋅m. Ammonia molecules in the gas phase are placed in a uniform electric field E⃗ with magnitude 1.3×106 N/C .

    Part A:
    What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E⃗ from parallel to perpendicular?

    Part B:
    At what absolute temperature T is the average translational kinetic energy 3/2 kT of a molecule equal to the change in potential energy calculated in part (a)? (Note: Above this temperature, thermal agitation prevents the dipoles from aligning with the electric field.)

    2. Relevant equations
    U = -p . E
    T = p X E

    3. The attempt at a solution

    I got Part A:
    delta U = p. E = p E cos theta = (5.0E-30)(1.3E6) = -6.5E-24 J,

    so delta U = 6.5E-24J

    Part B:

    My reasoning is as follows:
    KE + PE = 0
    3/2 kT + U = 0
    U = -3/2 kT
    (-2/3 U) / k = T
    T = (-2/3)(6.5E-24) / (8.99E9) = 482E-36 degrees K

    which is wrong.
    I don't know what I'm doing wrong :/
     
  2. jcsd
  3. Sep 18, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    What is k and what is its value?
     
  4. Sep 18, 2016 #3
    I thought it was maybe: k = 8.99E9 Nm^2/C^2
     
  5. Feb 4, 2017 #4
    Use Boltzmann constant for k, it is approximately 1.38*10^-23 J/K
     
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