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| ### Statement |
| ### Statement |
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| $5.8.4.$ [Insert the problem statement] |
| $5.8.4.$ Estimate the probability that the air density in the volume of 0.1 mm3 of any |
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| part of your room will be twice as high as the usual density. What should be |
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| the volume of this section so that this probability is large enough? |
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| ### Solution |
| ### Solution |
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| What should happen is the same number of molecules that are already in $V=1 mm^3$ should appear in $V.$ |
| What should happen is the same number of molecules that are already in $V=1 mm^3$ should appear in $V.$ |
| Total of molecules in the room (my bathroom after i take a shower $:)$ ): |
| Total of molecules in the room (my bathroom after i take a shower $:)$ ): |
| $$N=N_{A} \cdot \frac{PV_0}{RT}\approx N_{A} \cdot \frac{10^5 \cdot 10}{8.3 \cdot 300}\approx10^{26}$$ |
| $$N=N_{A} \cdot \frac{PV_0}{RT}\approx N_{A} \cdot \frac{10^5 \cdot 10}{8.3 \cdot 300}\approx10^{26}$$ |
| We need:$$ n=N\cdot \frac{V}{V_0} \approx 10^{16}$$Probability of existing of $n$ molecules is $1$. |
| We need:$$ n=N\cdot \frac{V}{V_0} \approx 10^{15}$$Probability of existing of $n$ molecules is $1$. |
| Probability of existing in $V$ for $1$ molecule in $p= \frac{V}{V_0}.$ |
| Probability of existing in $V$ for $1$ molecule in $p= \frac{V}{V_0}.$ |
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| So final answer is |
| So final answer is |
| $$ P= (\frac{V}{V_0})^n=(\frac{V}{V_0})^{N \cdot \frac{V}{V_0}}=x^{Nx}, x=\frac{V}{V_0}$$In our case $\frac{V}{V_0}=10^{-10}$ and $P=10^{-10^{16}}$, which is close to Savchenko's estimation. The upper power depends strongly on temperature and volume. |
| $$ P= (\frac{V}{V_0})^n=(\frac{V}{V_0})^{N \cdot \frac{V}{V_0}}=x^{Nx}, x=\frac{V}{V_0}$$In our case $\frac{V}{V_0}=10^{-10}$ and $P=10^{-10^{15}}$, which is close to Savchenko's estimation. The upper power depends strongly on temperature and volume. |
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| We can plot $P(x)$ and see that $P=10^{-4}% $ if $x=\frac{V}{V_0}\approx10^{-27}$, so |
| We can plot $P(x)$ and see that $P=10^{-4}% $ if $x=\frac{V}{V_0}\approx10^{-27}$, so |
| $V=10^{-17} mm^3$ |
| $V=10^{-17} mm^3$ |
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| Intereating to notice, that with decreasing of $x$, $P$ goes to $1$. |
| Intereating to notice, that with decreasing of $x$, $P$ goes to $1$. |
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| Explanation in terms of our model is that if we consider a volume less than length of free fly of molecule, there should be all pairs of colliding molesules, so double density. | | Explanation in terms of our model is that if we consider a volume less than length of free fly of molecule, there should be all pairs of colliding molesules, so double density. |
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| But much more truthful is the next fact: | | But much more truthful is the next fact: |
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| The key postulate of thermodynamix say that we consider a system that has many enough partickles, so fluctuations of $n$ are negligible. But with decreasing of volume, we can no longer have this fact true. | | The key postulate of thermodynamix say that we consider a system that has many enough partickles, so fluctuations of $n$ are negligible. But with decreasing of volume, we can no longer have this fact true. |
