Is the atmospheric pressure in a closed container the same as that of the surroundings (1 bar at sea level)?

Consider a tube with both ends open with one end dipped into water (like a pipette in chem lab). Now if we close the other end with a thumb and draw it out of the water, the water level in the tube will be higher than the surrounding water level.

If we say it is because the atmosphere that pushes up on the water in the tube is same as that of remaining air in tube pushing down on the water, won't the water fall out due to its own weight as the upward and downward pressure is balanced? I would like an explanation of the whole process which compares the weight of the water with up and down pressure by the atmosphere.

  • $\begingroup$ Will someone please help me figure this stupid question out..it's burning me since last 2 days.. :-/ $\endgroup$ – Shashank Apr 13 '14 at 6:37
  • $\begingroup$ When the undipped end is closed, the pressure due to the air inside it will change, i.e. decrease. Thus, since the external pressure is more, it would remain "hanging" inside. $\endgroup$ – Shubham Apr 14 '14 at 14:12
  • $\begingroup$ That right but how?? I mean how the pressure of air in the top 'll decrease.. $\endgroup$ – Shashank Apr 14 '14 at 14:28
  • $\begingroup$ I mean if i'll close a empty bottle the pressure inside it will decrease!!that's not reasonable $\endgroup$ – Shashank Apr 14 '14 at 14:30
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    $\begingroup$ It is reasonable. When the bottle is open, the whole column of air above acts on it, thus more weight of air acts, which implies more pressure. When closed, the amount(and hence the weight of air) in the bottle decreases, thus meaning less weight and hence less pressure. The air that is outside the bottle is balanced by the cap. $\endgroup$ – Shubham Apr 14 '14 at 16:21

Lets define some variables first. Lets say that the length of the column of air that you trap in the tube between the water level and your thumb is $h_0$ and it is initially at the same pressure as the surrounding air which we will call $P_0$. Lets also define the cross-sectional area of the tube to be $A$. As you draw the tube out of the water, the water level in the tube will rise above the surrounding water level, and the pressure at the bottom of this column of water will be given by $$ P_0=P_t+\rho g h_w, $$ where $P_t$ is the pressure of the volume of air trapped by your thumb, $\rho$ is the mass density of the water, $g$ is the gravitational acceleration at the surface of the Earth, and $h_w$ is the height difference between the water in the tube and the surrounding water. The second term on the right hand side is a standard equation from fluid statics. You can see from this already that the pressure can not be the same at the top of the tube anymore, as you suspected.

We can use the ideal gas law to rewrite the first term on the right hand side $$ P_tV_t=P_0V_0\qquad\Rightarrow\qquad P_t=P_0\frac{V_0}{V_t} $$ Now the volume of air at the top of the tube before you start to draw it up is given by $V_0=Ah_0$, but as you draw it up the pressure and volume will change to $V_t=Ah_t$. So, putting all of this information into the first equation we get $$ P_0=P_0\frac{h_0}{h_t}+\rho g h_w $$ The final thing we need to take care of is replacing $h_t$ with something we actually know. Namely, the height we draw the tube up to, which we will call $h_s$ and is given by $h_s=h_t+h_w$. Sticking this into the above equation yields $$ P_0=P_0\frac{h_0}{(h_s-h_w)}+\rho g h_w $$

Solving this equation for $h_w$ gives $$ h_w=\frac{1}{2\rho g}\left(P_0+\rho g h_s+\sqrt{P_0^2+(\rho gh_s)^2+P_0\rho g(4h_0-h_s)}\right). $$ I've plotted this equation below for differing values of $h_0$. Notice that there is a maximum height to which you can draw water with this method, this is the height at which the pressure in the tube reaches zero.

enter image description here

Finally, since your question was actually about the pressure in the tube, we can rearrange the above equations to solve for the pressure in the tube under your thumb.
$$ P_t=P_0-\rho gh_w=\frac{1}{2}\left(P_0-\rho g h_s-\sqrt{P_0^2+(\rho gh_s)^2+P_0\rho g(4h_0-h_s)}\right). $$ This equation is plotted below for a number of different values of $h_0$.

enter image description here

  • $\begingroup$ Note that the implementation of the ideal gas law here assumes constant temperature. The gas expansion will cause the temperature to fall. After a sufficient period of time, it will equalize to the environment's temperature, and the model here will become relevant. $\endgroup$ – Alan Rominger Apr 14 '14 at 16:54
  • $\begingroup$ @AlanSE You are correct. I didn't bother to mention it because the post was already starting to get rather long. $\endgroup$ – Chris Mueller Apr 14 '14 at 16:56
  • $\begingroup$ Thank you very much for your utter generosity and taking my silly question into consideration..explanation was very clear and satisfactory..thanks a lot.. One more question if that doesn't bother you, what is the philosophy behind the fact that the pressure at the bottom of a open and closed container are both same i.e equal to atmospheric pressure,when in closed container weight of the air column is doesn't affect the air inside... What if i similarly consider a water container in the bottom of ocean,what 'll be the pressure at the bottom of the container..what's physical concept govern these $\endgroup$ – Shashank Apr 14 '14 at 17:16
  • $\begingroup$ @shashank I'm glad I could help. The weight of the air column does exert a pressure, but it is approximately constant within tens of meters of the ground. You can see the effect if you go higher up in the atmosphere though, have a look at this portion of the Wikipedia article on atmospheric pressure to see the scale of the effect. $\endgroup$ – Chris Mueller Apr 14 '14 at 18:50

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