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Question

(B) Given,dipole moment of one molecule,$p_{molecule} = 10^{-29} \; C \; m$.Number of molecules in $1$ mole is $N = 6.022 	imes 10^{23} \approx 6 	i

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given,dipole moment of one molecule,$p_{molecule} = 10^{-29} \; C \; m$.Number of molecules in $1$ mole is $N = 6.022 	imes 10^{23} \approx 6 	imes 10^{23}$.Total dipole moment of the sample,$p = N 	imes p_{molecule} = 6 	imes 10^{23} 	imes 10^{-29} = 6 	imes 10^{-6} \; C \; m$.The potential energy of a dipole in an electric field is given by $U = -p E \cos 	heta$.Initial potential energy $(U_i)$ at $	heta = 0^{\circ}$:$U_i = -p E \cos 0^{\circ} = -(6 	imes 10^{-6}) 	imes 10^{6} 	imes 1 = -6 \; J$.Final potential energy $(U_f)$ at $	heta = 60^{\circ}$:$U_f = -p E \cos 60^{\circ} = -(6 	imes 10^{-6}) 	imes 10^{6} 	imes 0.5 = -3 \; J$.The change in potential energy is $\Delta U = U_f - U_i = -3 \; J - (-6 \; J) = 3 \; J$.Since the system releases energy to align with the new field,the heat released is equal to the magnitude of the decrease in potential energy,which is $3 \; J$.

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