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A molecule of a substance has a permanent electric dipole moment of magnitude 10^(-29)C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 10^(6) V m^(-1) . The direction of the field is suddenly changed by an angle of 60^(@) . Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample. |
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Answer» Solution :Here dipole MOMENT of each molecules `=10^(-29)` C As 1 mole of the substance contains `6 xx 10^(23)` molecules. total dipole moment of all the molecules `p = 6 xx 10^(23) xx 10^(-29)Cm = 6 xx 10^(-6)` Cm INITIAL potential energy, `U_(1) =-pE cos theta =-6 xx 10^(-6) xx 10^(6) cos 0^(@) =-6 J` Final potential energy (when `theta = 60^(@)`), `U_(f) = -6 xx 10^(-6) xx 10^(6) cos 60^(@) = - 3J` CHANGE in potential energy `=-3 J -(-6J) = 3J` So, there is loss in potential energy. This must be the energy released by the substance in the form of HEAT in aligning its dipoles. |
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