A hermetically closed absolutely rigid vessel of height $H$ is filled to the top with water. The water pressure at the upper wall of the vessel is zero. At the bottom of the vessel there is a small air bubble. How will the pressure of the water in the vessel change if the bubble floats up? Assume that water is incompressible. Neglect the solubility of air in water.
Decision:
Pressure at the bottom of the vessel (at depth $H$)
$p(H) = \rho gH$,
where $\rho$ is the density of water. Pressure inside the bubble near the bottom,
$p_{0}=\rho gH + \frac{2 \sigma }{r}$,
where $\sigma$ is the surface tension of water; $r$ is the radius of the bubble. Since the bubble floats at constant temperature and the solubility of air in water can be neglected, the pressure $p_{0}$ inside the bubble and its volume $V$ are related by the formula $p_{0}V = const$. When surfacing, the bubble radius $r$ and its volume $V$ remain unchanged, since water is incompressible according to the problem condition. Consequently, $p_{0}$ does not change either. Thus, for the pressure p in the water at the upper wall of the vessel after the bubble pops, we can write:
$p(0)=p_{0}- \frac{2 \sigma }{r} = \rho gH$;
the pressure at depth $h$ will be equal to $p(h) = p(0) + \rho gh = \rho g(H + h)$, i.e. by $\rho gH$ more than before the bubble surfaced.
Decision:
Pressure at the bottom of the vessel (at depth $H$)
$p(H) = \rho gH$,
where $\rho$ is the density of water. Pressure inside the bubble near the bottom,
$p_{0}=\rho gH + \frac{2 \sigma }{r}$,
where $\sigma$ is the surface tension of water; $r$ is the radius of the bubble. Since the bubble floats at constant temperature and the solubility of air in water can be neglected, the pressure $p_{0}$ inside the bubble and its volume $V$ are related by the formula $p_{0}V = const$. When surfacing, the bubble radius $r$ and its volume $V$ remain unchanged, since water is incompressible according to the problem condition. Consequently, $p_{0}$ does not change either. Thus, for the pressure p in the water at the upper wall of the vessel after the bubble pops, we can write:
$p(0)=p_{0}- \frac{2 \sigma }{r} = \rho gH$;
the pressure at depth $h$ will be equal to $p(h) = p(0) + \rho gh = \rho g(H + h)$, i.e. by $\rho gH$ more than before the bubble surfaced.