Правка разделов «Problem», «Solution», «0. Write down the given data (a purely f»

Adler правка от
Посмотреть решение История
правка #18991 предыдущая #18990 ← раньше позже →
en/6.6.3.md +70 −5
@@ -1,11 +1,76 @@
−### Statement
+### Problem
−$6.6.3.$ [Insert the problem statement]
+$6.6.3.$
+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?
+
### Solution
−12
+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.
−#### Answer
+#### 0. Write down the given data (a purely formal step to make the solution clear).
−[Insert a concise answer or boxed result]
+$\varepsilon = 1.0078$
+
+$E = 10^{3} \ V/m$
+
+$T = 18^{\circ}C = 291 \ K$
+
+$\varepsilon_{0} = 8.85 \cdot 10^{-12} \ F/m$
+
+$P = 2 \cdot 10^{3} \ Pa$
+
+$p' = -0.61 \cdot 10^{-29} \ C \cdot m$
+
+$K = 1.38 \cdot 10^{-23} \ J/K$
+
+#### 1. Find the answer to the first question.
+
+We will find the answer by working backwards, that is, we write the formula for the dipole moment and start from there.
+
+\[
+\boxed{p=\frac{P_{pol}}{n}}
+\]
+
+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.
+
+$P = nKT \Rightarrow n = \frac{P}{KT}$
+
+$P_{pol}$ is found from the polarization vector formula:
+
+$P_{pol} = \varepsilon_{0}(\varepsilon - 1)E$
+
+Now we find the dipole moment:
+
+\[
+\boxed{p = \frac{\varepsilon_{0}(\varepsilon - 1)EKT}{P}}
+\]
+
+$p = \frac{8.85 \cdot 10^{-12} \cdot (1.0078 - 1) \cdot 10^{3} \cdot 1.38 \cdot 10^{-23} \cdot 291}{2 \cdot 10^{3}} \approx 1.4 \cdot 10^{-34} \ C \cdot m$
+
+#### 2. Reasoning for the answer to the second question.
+
+First of all, we need to understand what we found and what is given in reference materials.
+
+We found the induced dipole moment, while reference books give the permanent dipole moment of the molecule.
+
+What is the difference between the induced and the permanent dipole moment?
+
+The induced dipole moment is a temporary dipole moment that arises in a particle under the influence of an external electric field.
+
+The permanent dipole moment remains with the molecule forever (as the name suggests), even at $E = 0$, in our equations.
+
+If we speak quantitatively about the difference, the Langevin-Debye formula helps, which has the form:
+
+\[
+\boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n}{3\varepsilon_{0}}\left(\alpha + \frac{p_{0}^{2}}{3KT}\right)}
+\]
+
+Here $p_{0}^{2}$ is precisely that permanent dipole moment, and $\alpha$ is the electronic polarizability.
+
+More detailed information about this formula can be read here: https://www.sciencedirect.com/topics/pharmacology-toxicology-and-pharmaceutical-science/lachesine
+### Answer
+
+\[
+\boxed{p = 1.4 \cdot 10^{-34} \ C \cdot m}
+\]