A neutral water molecule (H 2 O) in its vapor state has an electric dipole momen

ID: 1477033 • Letter: A

Question

A neutral water molecule (H2O) in its vapor state has an electric dipole moment of magnitude 6.2 x 10-30 C.m.

a) How far apart are the molecule's centers of positive and negative charge?

b)If the molecules is placed in an electric field of 1.5 x 104 N/C, what maximum torque can the field exert on it?

c) How much work must an external agent do to rotate this molecule by 180 in this field, starting from its fully aligned position, for which theta=0?

d) Compare this energy to a typical thermal energy kT when the molecule is part of a gas at a temperature 400K

There are 10 electrons and 10 protons in a neutral water molecule so the magnitude of its dipole moment is

d = p/10e = 6.20 × 10^-30/10 × 1.6 × 10^-19

d = 3.9 × 10^-12 m.

b) torque = pE * sin(theta)

torque = (6.2 × 10^-30 )*(1.5 × 10^4)*sin90°

Torque = 9.3 × 10^-26 N-m

C) Wa = U180 - U0

Wa = (-pEcos180) - (-pEcos0)

Wa = 2pE

Wa = 2*6.2*10^-30*1.5*10^4

Wa = 1.9 × 10^-25 J

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