Solution We start from the usual relation between the molar polarization, P m, the polarizability, α, and the dipole moment, µ: N A.

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Justina Daniels
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1 Problem 1 P m at a dierent T and dipole moment from P m at a given T and pol. vol. The polarizability volume of bromoacetylene is cm 3 and its molar polarization, P m, in vapour phase, at 10 C is cm 3 mol 1. Assuming that the temperature dependence of P m on T 1 in vapour phase is linear, calculate P m of bromoacetylene in vapour phase at 110 C and its dipole moment. Why is the change in P m across 100 C so small for bromoacetylene? We start from the usual relation between the molar polarization, P m, the polarizability, α, and the dipole moment, µ: P m = N ( ) A α + µ2. 3ε 0 3kT By solving for µ we have and nally µ = ( µ 2 = P m α 3ε 0 P m = α + µ 2 3ε 0 9ε 0 kt, ) 9ε0 kt = 9P mε 0 kt 3αkT, 9Pm ε 0 kt 3αkT = C m = D. Once µ is known, the relation above can be used to obtain the value of P m at 110 C which is cm 3 mol 1. The change in P m across 100 C is so small for bromoacetylene because the dipole moment is quite small.

2 Problem 2 Contributions to the van der Waals interaction. Calculate the contributions to the van der Waals interaction for a pair of acetone molecules. The polarizability volume of acetone is cm 3, its dipole moment is 2.88 D and its ionization energy is 9.70 ev. The average intermolecular distance is 5.80 Å. Which of the contributions is the largest at 800 K? Which of the contributions is the largest at absolute zero? Discuss the results. There are three main contributions to the van der Waals interaction: the Keesom interaction energy, the dipoleinduced-dipole interaction and the instantaneous dipoleinduced dipole interaction. The expression for the Keesom interaction energy is V = 2 (µ 1µ 2 ) 2 3 (4πε 0 ) 2 kt r 6 = J. (1) The expression for the dipoleinduced-dipole interaction energy is V = µ2 1α 2 4πε 0 r 6 = J (2) The expression for the instantaneous dipoleinduced dipole interaction energy is V = 3 2 α 1α 2 I 1 I 2 (I 1 + I 2 ) r 6 = J (3) At absolute zero the assumption used to obtain the Keesom interaction, eq. 1, is no longer valid. Since the molecules are not moving, we simply use the dipole-dipole interaction energy for the case of antiparallel dipoles that are also perpendicular to the vector joining them, assuming that the two interacting acetone molecules will arrange the dipoles in the most favourable mutual orientation. The energy is V = µ 1µ 2 4πε 0 r 3 = J The question Which of the contributions is the largest refer to the contributions to the van der Waals interaction which is an attractive interaction. So, the largest contribution is the most attractive contribution. The most attractive contribution is the one with the most negative energy. At 800 K the largest contribution is therefore the instantaneous dipoleinduced dipole interaction with an energy of J. At absolute zero the largest contribution is the dipole-dipole interaction with an energy of J.

Problem 3 Water in a superhydrophobic capillary tube Calculate the displacement of the water meniscus in a capillary tube of internal diameter 0.

3 Problem 3 Water in a superhydrophobic capillary tube Calculate the displacement of the water meniscus in a capillary tube of internal diameter mm, with respect to the water level outside the tube. The internal surface of the tube is superhydrophobic with a contact angle of 160. The temperature is 20 C. Draw a schematic of the experiment. At 20 C, the density of water is g cm 3 and its surface tension is mn m 1. If the contact angle is θ c the equation for the capillary action is h = 2γ ρgr cos θ c = 23.3 cm. The schematic of the experiment should focus on the detail of the meniscus that is convex instead of the usual concave shape (tube on the right in the image below). This is due to the strong hydrophobicity of the internal surface of the tube. For this reason, the water level in the tube is 23.3 cm lower compared to the water level outside the tube.

4 Problem 4 Liquid crystal phases with positional order Describe the liquid crystal phases that have positional order. One-dimensional positional oder is present in the smectic A phase, smectic C phase, and columnar phase. Two-dimensional order is present in the smectic B phase.

5 Problem 5 Mean dipole moment Calculate the mean electric dipole moment, in units of debye, of a liquid phase formed by molecules with a permanent dipole of D along the direction of an applied uniform electric eld of intensity 1.50 MV cm 1 at a temperature of 15.0 K. Is it possible that the mean electric dipole moment of the phase is larger than the dipole moment of an individual molecule forming that phase? Discuss. The mean electric dipole moment is given by the following equation: µ z = µl(x) In many cases the argument, x = µe, of the Langevin function, L(x), is much smaller than unity and the kt expression for the mean electric dipole moment can be simplied to µ z = µ2 E 3kT But if the T is very low and the electric eld is very large, it is possible that µe kt fact, by doing the calculation, we have > 1, like in this case. In x = = In this case the approximation to the Langevin function is not valid and we have to use the full expression [ e x + e x µ z = µ e x e 1 ] x x By doing the calculations, we have [ e x + e x e x e 1 ] = 0.96 x x µ z = = 9.83 D Of course the mean electric dipole moment of a phase can not be larger than the dipole moment of an individual molecule. In fact, at the most, it can be equal to the dipole moment of an individual molecule, when the dipoles are all perfectly aligned.

Problem 1. Birefringence of a uniaxial material

Problem 1. Birefringence of a uniaxial material Problem 1 Birefringence of a uniaxial material Consider a uniaxial, birefringent material. The direction of propagation of unpolarized light is parallel to the optical axis. Draw a scheme of the experiment