| ### Solution | | ### Solution |
| | | |
| | | |
| | | |
| 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:$$ n=N\cdot \frac{V}{V_0} \approx 10^{15}$$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 $p= \frac{V}{V_0}.$ |
| | | |
| 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 $P=10^{-10^{15}}$, 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. |
| | | |
| 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 |
| $V=10^{-17} mm^3$ | | $V=10^{-17} mm^3$ |
|  | |  |
| | | |
| Intereating to notice, that with decreasing of $x$, $P$ goes to $1$. | | Intereating to notice, that with decreasing of $x$, $P$ goes to $1$. |
| | | |
| 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. |
| | | |
| But much more truthful is the next fact: | | But much more truthful is the next fact: |
| | | |
| 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. |
| | | |
| #### Answer | | #### Answer |
| | | |
| $P=10^{-10^{15}}$, | | $P=10^{-10^{15}}$, |
| | | |
| $P=10^{-4}% $ if $V=10^{-17} mm^3$, | | $P=10^{-4}% $ if $V=10^{-17} mm^3$, |
| | | |
| $P=(\frac{V}{V_0})^{N \cdot \frac{V}{V_0}}$ | | $P=(\frac{V}{V_0})^{N \cdot \frac{V}{V_0}}$ |
