How was the bottle crushed by air?

In [this](https://youtu.be/thSeYzcPrqs) video, a water bottle is shown to be crushed by the surrounding air.


My understanding: 


When the person pours cold water over the bottle which has hot water inside it, the hot water produces water vapor which increases the pressure inside in bottle. When the person places the bottle into the cold water bowl, the temperature of the bottle decreases, which causes the water vapor inside the bottle to get condensed, thus lowers the pressure inside the water bottle. When the pressure inside decreases, the pressure outside is greater which crushes the water bottle.


If my explanation of this is wrong in anyway, can someone correct it?




Your explanation is close, and I disagree with comments dismissing the water-vapor condensation effect. It's *not* primarily about heating the air but displacing it with water vapor. When air is cooled it contracts via the gas law as described in another answer, but when water vapor is cooled it condenses and is removed from the vapor phase altogether, just as if you had pulled the vapor out with a vacuum pump.


The experiment was not carried out very well in the video in the body of your question. They didn't provide the time and heating conditions to thoroughly displace the air with water vapor. You can tell that it happened to some degree though by the observation of steam in the bottle, condensing on the sides *before* sealing the bottle. Because of this the effect wasn't very dramatic.


The video in your comment shows the concept much more clearly. They are actively boiling the water such that steam is shooting out of the top and the vapor composition inside their container is almost entirely water vapor. This is an excellent chemical phase-change demonstration (the second video).


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You do not need the water vapour for this effect, the air that is contained in the bottle is sufficient.


You may use the ideal gas law to estimate the amount of air in the volume of the bottle of about $V=1\ \mathrm l=0.001\ \mathrm{m^3}$ when the air is heated to about $T=60\ \mathrm{^\circ C}=333.15\ \mathrm K$ (using hot water) at normal pressure $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ (since the bottle is still open):


$$\begin{align}
pV&=nRT\\[3pt]
n&=\frac{pV}{RT}\\[3pt]
&=\frac{100\,000\ \mathrm{Pa}\times0.001\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times333.15\ \mathrm K}\\[3pt]
&=0.036\ \mathrm{mol}
\end{align}$$


If you now close the bottle, the amount of air $n$ is trapped.


When this air is cooled to about $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$ (using ice water) at an approximately constant volume of $V=1\ \mathrm l=0.001\ \mathrm{m^3}$, the pressure in the bottle decreases according to the ideal gas law:


$$\begin{align}
pV&=nRT\\[3pt]
p&=\frac{nRT}{V}\\[3pt]
&=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{0.001\ \mathrm{m^3}}\\[3pt]
&=8.2\times10^4\ \mathrm{Pa}=0.82\ \mathrm{bar}
\end{align}$$


Therefore, the ambient air pressure of $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ compresses the bottle until the pressures are equalized (or the bottle fails). The volume in the bottle decreases according to the ideal gas law:


$$\begin{align}
pV&=nRT\\[3pt]
V&=\frac{nRT}{p}\\[3pt]
&=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{100\,000\ \mathrm{Pa}}\\[3pt]
&=8.2\times10^{-4}\ \mathrm{m^3}=0.82\ \mathrm l
\end{align}$$


Nevertheless, any condensation of water vapour would help to achieve this effect. The resulting changes in pressure and volume due to condensation of water can be even more extreme than the contraction of air. For example, if a volume of $V=1\ \mathrm l$ is filled with steam at $T=120\ \mathrm{^\circ C}$ and $p=1\ \mathrm{bar}$ (without any air) and then cooled until the steam is almost completely condensed to liquid water (at about $T=99.6\ \mathrm{^\circ C}$), the resulting liquid volume is only $V=0.00058\ \mathrm l=0.58\ \mathrm{ml}$.


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