The ammonia molecule (NH3) has a dipole moment of 5.0×10?30C?m. Ammonia molecule

Question

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 1.1×106 N/C.

1- 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. (in J)

2- 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. (in K)

Explanation / Answer

Step1

Given data

dipole moment P = 5*10^-30 C.m

electric field is E = 1.1*10^6 N/C

1.

step2

change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E , dU = ?

here as the dipole is in electric field and torque will acts on it

torque acting T = P X E

T = P*E sin theta

theta is the angle between E and P = 90 degrees

T = 5*10^-30*1.1*10^6 = 5.5*10^-24 N.m

we know that the change in potential energy is equal to dU = P.E = P*E cos theta

Step3

The change in electric potential energy is  

dU = -P*E = P*E cos theta

dU = 5.5*10^-24 cJ

2. the average translational kinetic energy is k.e = (3/2)kT

here the energy is 5.5*10^-24 = (3/2)1.38*10^-23 *T

T = 0.2657 k

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