Pressure inside a sealed container

Click For Summary
SUMMARY

The discussion centers on the behavior of gas pressure in a sealed container, specifically addressing the application of Boyle's Law and the ideal gas law (pV = RNT). Participants clarify that sealing the container does not change the number of air molecules present, and thus the pressure remains at 1 atm, as the concentration of gas remains constant. The pressure exerted by the gas is a result of the number of molecules per volume, and sealing the container simply alters the interaction between the gas and the external atmosphere without affecting the internal pressure.

PREREQUISITES
  • Understanding of Boyle's Law and its application in gas behavior.
  • Familiarity with the ideal gas law (pV = RNT).
  • Basic knowledge of pressure concepts in physics.
  • Concept of molecular concentration and its effect on pressure.
NEXT STEPS
  • Study the implications of Boyle's Law in real-world applications.
  • Explore the ideal gas law and its derivations in various conditions.
  • Investigate the effects of temperature changes on gas pressure and volume.
  • Learn about the behavior of gases in different states and under varying pressures.
USEFUL FOR

Students of physics, engineers working with gas systems, and anyone interested in thermodynamics and fluid mechanics will benefit from this discussion.

brochesspro
Messages
155
Reaction score
22
Homework Statement
A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice.
Relevant Equations
Theoretical doubt.
1704015650052.png

I do know how to solve this question:
We find the equilibrium pressure due to air molecules inside the container (after water has stopped flowing). Then, assuming initial pressure of gas inside the container to be atmospheric pressure, we use Boyle's law to find the new volume of the gas and then after a bit of manipulation we find the depression in water level to be about 6 mm.
Now, my doubt is, how can we take the initial pressure of he gas to be 1 atm after we seal the container? Should the pressure not decrease since the no. of air molecules inside the container decreases?
 
Physics news on Phys.org
brochesspro said:
Now, my doubt is, how can we take the initial pressure of he gas to be 1 atm after we seal the container? Should the pressure not decrease since the no. of air molecules inside the container decreases?
The gas pressure before the container is sealed is 1 atm because it is in direct contact with the atmosphere. Why would sealing the container change the number of molecules inside it?
 
Orodruin said:
Why would sealing the container change the number of molecules inside it?
I meant the no. of molecules that can apply the pressure on the surface of the water, sorry for the poor wording. Initially, all the air above the water surface will apply 1 atm on it. But after the container is sealed, the air molecules above the seal do not apply any pressure on the water, right? Only the molecules inside the container will exert any pressure, so why is the pressure still 1 atm?
 
brochesspro said:
I meant the no. of molecules that can apply the pressure on the surface of the water, sorry for the poor wording.
That’s not what gives pressure. What gives pressure is number of molecules per volume.

Think of pressure as the molecules bouncing off the surface. If there are the same number of molecules in a larger volume, fewer will bounce on the surface every unit time, therefore less pressure. The ideal gas law is pV = RNT so ##p \propto N/V##.

brochesspro said:
the air molecules above the seal do not apply any pressure on the water, right?
The sealing procedure does not change N/V.
 
To add to that: Without the seal there is of course the atmosphere outside applying pressure to the gas inside the cylinder. If it were not for this pressure the air would simply escape. When you seal the container this pressure is exchanged for a pressure from the seal. The pressure inside the cylinder remains the same (until you open the orifice).
 
Orodruin said:
When you seal the container this pressure is exchanged for a pressure from the seal.
Can we instead say that the atmosphere applies a pressure through the seal? And, is the same thing applicable in this question?
1704020611630.png


After we seal the tube, the tube, the pressure above the mercury due to the gas is 1 atm cuz the concentration of gas remains the same and we do the same steps as in the previous question, Boyle's law and all.
 
brochesspro said:
Can we instead say that the atmosphere applies a pressure through the seal?
No. If this were the case the pressure in the container could never be less than on the outside.
brochesspro said:
And, is the same thing applicable in this question?
Yes.
 
Orodruin said:
No. If this were the case the pressure in the container could never be less than on the outside.
Then what does the pressure from the seal mean?
 
The seal just removes the contact between the gas in the container and the outside atmosphere. It does not change the pressure of the gas in the container, which is now pressing on the seal rather than the outside atmosphere. Do you think that you always need to have a column of gas to cause of pressure to be present, and without gravity, you can't have pressure?
 
  • Like
Likes   Reactions: Orodruin
  • #10
brochesspro said:
Can we instead say that the atmosphere applies a pressure through the seal? And, is the same thing applicable in this question?

After we seal the tube, the tube, the pressure above the mercury due to the gas is 1 atm cuz the concentration of gas remains the same and we do the same steps as in the previous question, Boyle's law and all.
Rather than water or mercury, just imagine an initially supported piston of mass m inside a vertical cylinder.
Pressure of air above and below the faces of the mechanically supported piston in place (unable to fall due to its own weight) are initially equal (the atmosphere reaches the piston both ways).

Then, we solidly close the top open face of the cylinder, and remove the support of the piston.
The piston will then tend to descend under the force equal to mg.
As that happens, the pressure on the bottom face of the piston remains the same atmospheric one.

Nevertheless, the descending movement of the piston makes the volume of that top sealed chamber (fixed number of molecules of air) increase, which induces lower pressure acting on the top face of the piston (increased area inside the chamber against which that fixed number of molecules of air collide).

The weight of our piston is the only thing compensating for that difference of pressures acting on top and bottom equal surfaces of our piston, which should stop its descending trajectory at a unique height of balance.

original_439552.png
 
Last edited:

Similar threads

Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
972
Replies
19
Views
2K
Replies
18
Views
2K
Replies
2
Views
1K
Replies
8
Views
2K
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
1K