The ammonia molecule (NH3) has a dipole moment of 5.0 X 10^-30 C. m. Ammonia mol
ID: 1399258 • Letter: T
Question
The ammonia molecule (NH3) has a dipole moment of 5.0 X 10^-30 C. m. Ammonia molecules in the gas phase are placed in a uniform electric field E with magnitude 1 .6 X 10^6 N/C. (a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E 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)? (Note: Above this temperature, thermal agitation prevents the dipoles from aligning with the electric field.)Explanation / Answer
a)
when dipole moment is parallel to E
potential energy, U1 = -P*E*cos(0)
= -5*10^-30*1.6*10^6
= -8*10^-24 J
when dipole moment is perpendicular to E
potential energy, U2 = -P*E*cos(90)
= 0 J
delta U = U2 - U1
= 0 - (-8*10^-24)
= 8*10^-24 J <<<<<------Answer
b) Apply, U = (3/2)k*T
==> T = (2/3)*U/k
= (2/3)*8*10^-24/(1.38*10^-23)
= 0.386 K <<<<<------Answer
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