| ### Problem | | ### Problem |
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| $6.6.3.$ | | $6.6.3.$ |
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| The saturated vapor pressure of water at $18^{\circ}C$ is $2 \cdot 10^{3} \ Pa$, and its dielectric constant is 1.0078. From these data, find the average dipole moment of a water molecule in an electric field of strength $10^{3}$ V/m. Reference books give the dipole moment of water as $-0.61 \cdot 10^{-29} C \cdot m$. How can the discrepancy in the results be explained? | | The saturated vapor pressure of water at $18^{\circ}C$ is $2 \cdot 10^{3} \ Pa$, and its dielectric constant is 1.0078. From these data, find the average dipole moment of a water molecule in an electric field of strength $10^{3}$ V/m. Reference books give the dipole moment of water as $-0.61 \cdot 10^{-29} C \cdot m$. How can the discrepancy in the results be explained? |
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| ### Solution | | ### Solution |
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| Before looking at the solution, you can review the first solution method (the second one is also possible, for broadening your knowledge) of problem $6.6.2$ and then try to solve this one. | | Before looking at the solution, you can review the first solution method (the second one is also possible, for broadening your knowledge) of problem $6.6.2$ and then try to solve this one. |
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| #### 0. Write down the given data (a purely formal step to make the solution clear). | | #### 0. Write down the given data (a purely formal step to make the solution clear). |
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| $\varepsilon = 1.0078$ | | $\varepsilon = 1.0078$ |
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| $E = 10^{3} \ V/m$ | | $E = 10^{3} \ V/m$ |
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| $T = 18^{\circ}C = 291 \ K$ | | $T = 18^{\circ}C = 291 \ K$ |
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| $\varepsilon_{0} = 8.85 \cdot 10^{-12} \ F/m$ | | $\varepsilon_{0} = 8.85 \cdot 10^{-12} \ F/m$ |
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| $P = 2 \cdot 10^{3} \ Pa$ | | $P = 2 \cdot 10^{3} \ Pa$ |
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| $p' = -0.61 \cdot 10^{-29} \ C \cdot m$ | | $p' = -0.61 \cdot 10^{-29} \ C \cdot m$ |
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| $K = 1.38 \cdot 10^{-23} \ J/K$ | | $K = 1.38 \cdot 10^{-23} \ J/K$ |
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| #### 1. Find the answer to the first question. | | #### 1. Find the answer to the first question. |
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| We will find the answer by working backwards, that is, we write the formula for the dipole moment and start from there. | | We will find the answer by working backwards, that is, we write the formula for the dipole moment and start from there. |
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| \[ | | \[ |
| \boxed{p=\frac{P_{pol}}{n}} | | \boxed{p=\frac{P_{pol}}{n}} |
| \] | | \] |
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| where $P_{pol}$ is the polarization vector, but since our dielectric is isotropic (the same in all directions), the vector concept does not play a role here, and $n$ is the concentration of molecules; we find it from the basic MKT equation. | | where $P_{pol}$ is the polarization vector, but since our dielectric is isotropic (the same in all directions), the vector concept does not play a role here, and $n$ is the concentration of molecules; we find it from the basic MKT equation. |
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| $P = nKT \Rightarrow n = \frac{P}{KT}$ | | $P = nKT \Rightarrow n = \frac{P}{KT}$ |
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| $P_{pol}$ is found from the polarization vector formula: | | $P_{pol}$ is found from the polarization vector formula: |
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| $P_{pol} = \varepsilon_{0}(\varepsilon - 1)E$ | | $P_{pol} = \varepsilon_{0}(\varepsilon - 1)E$ |
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| Now we find the dipole moment: | | Now we find the dipole moment: |
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| \[ | | \[ |
| \boxed{p = \frac{\varepsilon_{0}(\varepsilon - 1)EKT}{P}} | | \boxed{p = \frac{\varepsilon_{0}(\varepsilon - 1)EKT}{P}} |
| \] | | \] |
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| $p = \frac{8.85 \cdot 10^{-12} \cdot (1.0078 - 1) \cdot 10^{3} \cdot 1.38 \cdot 10^{-23} \cdot 291}{2 \cdot 10^{3}} \approx 1.4 \cdot 10^{-34} \ C \cdot m$ | | $p = \frac{8.85 \cdot 10^{-12} \cdot (1.0078 - 1) \cdot 10^{3} \cdot 1.38 \cdot 10^{-23} \cdot 291}{2 \cdot 10^{3}} \approx 1.4 \cdot 10^{-34} \ C \cdot m$ |
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| #### 2. Reasoning for the answer to the second question. | | #### 2. Reasoning for the answer to the second question. |
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| First of all, we need to understand what we found and what is given in reference materials. | | First of all, we need to understand what we found and what is given in reference materials. |
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| We found the induced dipole moment, while reference books give the permanent dipole moment of the molecule. | | We found the induced dipole moment, while reference books give the permanent dipole moment of the molecule. |
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| What is the difference between the induced and the permanent dipole moment? | | What is the difference between the induced and the permanent dipole moment? |
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| The induced dipole moment is a temporary dipole moment that arises in a particle under the influence of an external electric field. | | The induced dipole moment is a temporary dipole moment that arises in a particle under the influence of an external electric field. |
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| The permanent dipole moment remains with the molecule forever (as the name suggests), even at $E = 0$, in our equations. | | The permanent dipole moment remains with the molecule forever (as the name suggests), even at $E = 0$, in our equations. |
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| If we speak quantitatively about the difference, the Langevin-Debye formula helps, which has the form: | | If we speak quantitatively about the difference, the Langevin-Debye formula helps, which has the form: |
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| \[ | | \[ |
| \boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n}{3\varepsilon_{0}}\left(\alpha + \frac{p_{0}^{2}}{3KT}\right)} | | \boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n}{3\varepsilon_{0}}\left(\alpha + \frac{p_{0}^{2}}{3KT}\right)} |
| \] | | \] |
