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13 July, 18:14

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 E? with magnitude 1.8*106 N/C.

At what absolute temperature T is the average translational kinetic energy 32kT 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.)

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  1. 13 July, 19:23
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    Question (continuation)

    (a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E S from parallel to perpendicular?

    (b) At what absolute temperature T is the average translational kinetic energy 3/2kT of a molecule equal to the change in potential energy calculated in part (a) ?

    Answer:

    a. 9.0 * 10^-24 Joules

    b. 0.44K

    Explanation:

    Given

    Let p = dipole moment = 5.0 * 10^-30 Cm

    Let E = Magnitude = 1.8 * 10^6 N/m

    a.

    The charge in electric potential = Final Charge - Initial Charge

    Initial Charge = Potential Energy

    Initial Energy = - pE cosФ where Ф = 0

    So, initial Energy = - 5.0 * 10^-30 * 1.8 * 10^6

    Initial Energy = - 9 * 10^-24 Joules

    Final Energy = 0

    Charge = 0 - (-9.0 * 10^-24)

    Charge = 9.0 * 10^-24 Joules

    b.

    Absolute Temperature

    Change in Kinetic Energy = Change in Potential Energy = 9.0 * 10^-24

    Change in Kinetic Energy = 3/2kT where k is Steven-Boltzmann constant = 1.38 * 10^-23

    So,

    9.0 * 10^-24 = 3/2 * 1.38 * 10^-23 * T

    T = (9.0 * 10^-24 * 2) / (3 * 1.38 * 10^-23)

    T = (18 * 10^-24) / (4.14 * 10^-23)

    T = 0.44K
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