Edits to “Statement”, “Answer”
en/5.3.9.md
+4 −2
| @@ -1,6 +1,8 @@ | |||
| ### Statement | |||
| − | $5.3.9.$ | ||
| + | $5.3.9.$ The diffusion coefficient of molecules $A$ in gases $B_1$ and $B_2$ is equal to $D_1$ and $D_2$, respectively, if the unit | ||
| + | volume of these gases contains $n$ particles. Find the diffusion coefficient of molecules $A$ in a mixture of | ||
| + | gases, the unit volume of which contains $n_1$ gas molecules $B_1$ and $n_2$ gas molecules $B_2$. | ||
| ### Solution | |||
| We'll consider molecules $A$ as impurities, meaning that the number of collisions of molecules $A$ with each other is much less than the number of collisions with molecules of gases $B_{1}$ and $B_{2}$. We'll also assume that the temperature is the same in all cases. This means we can assume that: | |||
| $$\frac{D_{1}}{D_{2}}=\frac{(R_{A} + R_{B_{2}})^{2}}{(R_{A} + R_{B_{1}})^{2}}$$ | |||
| Number of collisions in the mixture: | |||
| $$\nu_{12}=\pi\cdot (R_{A} + R_{B_{1}})^{2}\cdot n_{1}\cdot <\upsilon_{sq_{A}}> + \pi\cdot (R_{A} + R_{B_{2}})^{2}\cdot n_{2}\cdot <\upsilon_{sq_{A}}>$$ | |||
| Let's write down the ratio of $D_{1}$ to $D_{12}$, after some transformations we get: | |||
| $$\frac{D_{1}}{D_{12}}=\frac{(R_{A} + R_{B_{1}})^{2}\cdot n_{1} + (R_{A} + R_{B_{2}})^{2}\cdot n_{2}}{(R_{A} + R_{B_{1}})^{2}\cdot n}$$ | |||
| Let's use the relation for $D_{1}$ and $D_{2}$: | |||
| $$\frac{D_{1}}{D_{12}}=\frac{n_{1} + \frac{D_{1}}{D_{2}}\cdot n_{2}}{n}$$ | |||
| Expressing $D_{12}$ we get: | |||
| $$D_{12}=\frac{n\cdot D_{1}\cdot D_{2}}{n_{1}\cdot D_{2} + n_{2}\cdot D_{1}}$$ | |||
| @@ -31,4 +33,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$D_{12}=\frac{n\cdot D_{1}\cdot D_{2}}{n_{1}\cdot D_{2} + n_{2}\cdot D_{1}}$$ | ||
| @@ -1,6 +1,8 @@ | |||
| ### Statement | ### Statement | ||
| $5.3.9.$ |
$5.3.9.$ The diffusion coefficient of molecules $A$ in gases $B_1$ and $B_2$ is equal to $D_1$ and $D_2$, respectively, if the unit | ||
| volume of these gases contains $n$ particles. Find the diffusion coefficient of molecules $A$ in a mixture of | |||
| gases, the unit volume of which contains $n_1$ gas molecules $B_1$ and $n_2$ gas molecules $B_2$. | |||
| ### Solution | ### Solution | ||
| We'll consider molecules $A$ as impurities, meaning that the number of collisions of molecules $A$ with each other is much less than the number of collisions with molecules of gases $B_{1}$ and $B_{2}$. We'll also assume that the temperature is the same in all cases. This means we can assume that: | We'll consider molecules $A$ as impurities, meaning that the number of collisions of molecules $A$ with each other is much less than the number of collisions with molecules of gases $B_{1}$ and $B_{2}$. We'll also assume that the temperature is the same in all cases. This means we can assume that: | ||
| $$\frac{D_{1}}{D_{2}}=\frac{(R_{A} + R_{B_{2}})^{2}}{(R_{A} + R_{B_{1}})^{2}}$$ | $$\frac{D_{1}}{D_{2}}=\frac{(R_{A} + R_{B_{2}})^{2}}{(R_{A} + R_{B_{1}})^{2}}$$ | ||
| Number of collisions in the mixture: | Number of collisions in the mixture: | ||
| $$\nu_{12}=\pi\cdot (R_{A} + R_{B_{1}})^{2}\cdot n_{1}\cdot <\upsilon_{sq_{A}}> + \pi\cdot (R_{A} + R_{B_{2}})^{2}\cdot n_{2}\cdot <\upsilon_{sq_{A}}>$$ | $$\nu_{12}=\pi\cdot (R_{A} + R_{B_{1}})^{2}\cdot n_{1}\cdot <\upsilon_{sq_{A}}> + \pi\cdot (R_{A} + R_{B_{2}})^{2}\cdot n_{2}\cdot <\upsilon_{sq_{A}}>$$ | ||
| Let's write down the ratio of $D_{1}$ to $D_{12}$, after some transformations we get: | Let's write down the ratio of $D_{1}$ to $D_{12}$, after some transformations we get: | ||
| $$\frac{D_{1}}{D_{12}}=\frac{(R_{A} + R_{B_{1}})^{2}\cdot n_{1} + (R_{A} + R_{B_{2}})^{2}\cdot n_{2}}{(R_{A} + R_{B_{1}})^{2}\cdot n}$$ | $$\frac{D_{1}}{D_{12}}=\frac{(R_{A} + R_{B_{1}})^{2}\cdot n_{1} + (R_{A} + R_{B_{2}})^{2}\cdot n_{2}}{(R_{A} + R_{B_{1}})^{2}\cdot n}$$ | ||
| Let's use the relation for $D_{1}$ and $D_{2}$: | Let's use the relation for $D_{1}$ and $D_{2}$: | ||
| $$\frac{D_{1}}{D_{12}}=\frac{n_{1} + \frac{D_{1}}{D_{2}}\cdot n_{2}}{n}$$ | $$\frac{D_{1}}{D_{12}}=\frac{n_{1} + \frac{D_{1}}{D_{2}}\cdot n_{2}}{n}$$ | ||
| Expressing $D_{12}$ we get: | Expressing $D_{12}$ we get: | ||
| $$D_{12}=\frac{n\cdot D_{1}\cdot D_{2}}{n_{1}\cdot D_{2} + n_{2}\cdot D_{1}}$$ | $$D_{12}=\frac{n\cdot D_{1}\cdot D_{2}}{n_{1}\cdot D_{2} + n_{2}\cdot D_{1}}$$ | ||
| @@ -31,4 +33,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$D_{12}=\frac{n\cdot D_{1}\cdot D_{2}}{n_{1}\cdot D_{2} + n_{2}\cdot D_{1}}$$ | ||