4
$\begingroup$

Say I had a plastic water bottle that was slightly crushed, I could uncrush it (mostly) by covering the opening with my mouth and blowing air into it hard.

Would it be easier to do this if the bottle was full of water? or while empty (full of air)? Why?

My thoughts are that it would be easier if it was full of water because water doesn't compress as easily as air and so more force will be applied to the walls of the bottle. Is this correct thinking? Are there other factors involved?

$\endgroup$
3
$\begingroup$

Change in interior volume of the bottle when it is uncrumpled, is the same in both cases, call it $\Delta V$. If $\Delta p$ is the pressure difference between inside and outside of the bottle, then assuming this pressure difference changes negligibly during uncrumpling, work done to uncrumple the bottle is $\Delta p\times \Delta V$. Say you need to apply a pressure difference greater than some threshold, $\Delta p_{threshold}$, to uncrumple the bottle, i.e. we must have $\Delta p>\Delta p_{threshold}$. Now part of the pressure difference required is provided by fluid already present inside the bottle, $\Delta p_{fluid}$, and the rest has to be provided by you, $\Delta p_{you}$. When the fluid inside the bottle is water, hydrostatic pressure is approximately thousand times greater than would be the case if the fluid were to be air (at atmospheric pressure). Therefore if the same total pressure difference is to be applied then comparing the two cases we have \begin{align} \Delta p = & \Delta p_{water}+\Delta p_{you,water}=\Delta p_{air}+\Delta p_{you,air} \\ \Rightarrow & \Delta p_{water}-\Delta p_{air}=\Delta p_{you,air}-\Delta p_{you,water}\gg 0 \end{align} Therefore if water were to be present in bottle, you need to apply much less pressure to uncrumple the bottle, than would be case if bottle contained air. It also follows that work contribution from you towards uncrumpling is much less when the bottle is filled with water.

$\endgroup$
2
$\begingroup$

The pressure required will be the same in both cases. The force on the walls of the bottle is the pressure multiplied by the area, and assuming some uncrumple force $F_u$ is required to push the walls back into shape that means there will be some uncrumple pressure $P_u$ required. Whether that pressure is due to air or water makes no difference.

However the work required will be greater for air than for water. The work done in the process consists of:

  1. the work needed to raise the pressure of the air or water to $P_u$

  2. the work needed to uncrumple the bottle

The work required for part (2) will be the same for both air and water because the pressure is the same in both cases. In both cases the work will be:

$$ W_2 = P_u\,\Delta V $$

where $\Delta V$ is the volume increase of the bottle when you uncrumple it.

However because water is largely incompressible it takes negligable work to raise the water pressure to $P_u$ i.e.

$$W_1(\text{water})=0$$

By contrast air is compressible and it will take work for you to raise the pressure of the air from atmospheric to the uncrumple pressure $P$. Assuming air can be treated as an ideal gas, and ignoring changes in temperature the work done for air is:

$$ W_1(\text{air}) = nRT\ln\frac{V_0}{V_\text{bottle}} $$

where $V_0$ is the original volume of all the air you pushed into the bottle. That means the work for step (1) is greater for air than for water.

$\endgroup$
1
$\begingroup$

It is equally "easy" in either scenario, if by "easy" we mean work done on the system.

We could model this scenario as a container with a piston that has some friction with the cavity.

In either scenario, the gas/liquid can be adiabatically pressurized until the piston moves to the final position, and then depressurized to the original pressure.

Since the only "change" is in the position of the piston (which took energy $W = F_{fric}*d$) the work done is equivalent regardless of the compressibility of the gas/liquid -- in the end it just gets dissipated by friction.

This of course assumes that you have some much larger reservoir which can provide for the difference in volume caused by the displacement of the piston.

$\endgroup$
0
$\begingroup$

You would still require the same amount of pressure. Think of it like blowing up a balloon with water vs air. If you blow them both up to the same size, then, regardless of the medium, the balloon pushes back with the same amount of force per unit area (pressure) for both cases. The difference is that you would've used a larger volume of air (at STP) than water since the air is much more compressible, which is also why hydraulic systems are typically used over pneumatic ones.

On a side note, there is a slight exception. The water's weight could add to the pressure of you blowing into the bottle. However, this additional pressure would be pretty insignificant and would be a function of position within the bottle, among other things.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.