The ammonia molecule (NH3) has a dipole moment of5.0×1030Cm. Ammonia molecules i

Question

The ammonia molecule (NH3) has a dipole moment of5.0×1030Cm. Ammonia molecules in the gas phase are placed in a uniform electric field E  with magnitude 1.6×106 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 E  from parallel to perpendicular?

Express your answer using two significant figures.

Part B

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.)

Express your answer using two significant figures.

Explanation / Answer

Given,

Dipole moment, p = 5 x 10-30 C m

Electric field, E = 1.6 x 106 N/C

Part A:

The change in the potential energy of the ammonia molecule is given by, U = p E

U = (5 × 1030 C m)( 1.6 × 106 N/C) = 8 x 10-24 J

Therefore, the change in the potential energy of the ammonia molecule is 8 x 10-24 J.

Part B:

The average translational kinetic energy of the ammonia molecule is given by, E = 3/2 kT

Since, the average translational kinetic energy of a molecule equal to the change in potential energy calculated in part (a). So, E = U

3/2 kT = U

        T = (2/3k)U

Here, k is Boltzmann constant = 1.38× 10-23

        T = (2/3k)U

= {2/ 3 (1.38× 10-23 J/ K)}( 8 x 10-24 J)

T = 0.386 K

So, the absolute temperature at which the average translational kinetic energy of a molecule equal to the change in potential energy is 0.39 K.

(Please rate my answer if you find it helpful, good luck...)

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