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B Atmospheric pressure etc.

  1. Jul 26, 2017 #1
    Hi,

    The air pressure at the surface is about 1,00,000 N/m2.

    It means that on average, a virtual column of air one square meter [m2] in cross-section, measured from sea level to the top of the earth's atmosphere, has a mass of about 10,000 kilograms and weight of about 1,00,000 newtons. That weight (across one square meter) is a pressure of 1,00,000 N/m2.

    Let's look at the air molecules in that virtual column of air. Let's imagine that the column is closed at one end by earth's surface where we also have a pressure sensor and the other end at the top of earth's atmosphere by some other material. At any instant only a fraction of molecules will be hitting the bottom of earth's surface downward, some of them will be hitting the other end upward, some of them will be hitting the column laterally and others will be in the space between the column of box.

    Although at any instant only a fraction of molecules will be hitting the bottom of earth's surface downward, the pressure sensor will still read 1,00,000 N/m because the molecules strike the bottom of the column more frequently and harder than the top of the column. The sum total of the greater number of collisions and the greater strength of the collisions will be that the average force against the bottom will be 1,00,000 N.

    Please have a look on the attachment.

    The cylinder is filled with air having the mass of 1.2 kg. Please also notice that at 25 C° the density of air at sea level could also be taken to be 1.2 kg/m3. The valves 1 and 2 are closed.

    When valves 1 and 2 are opened simultaneously what would happen? I think that outside air pressure will press the mercury downward because outside air pressure is greater than that of the cylinder. What do you think? Thank you for your help.
     

    Attached Files:

  2. jcsd
  3. Jul 26, 2017 #2

    jbriggs444

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    As I understand it, you have arranged for a cylinder with a volume of 1m3 containing 1.2 kg air at a reasonable temperature. You note that the density of air in the cylinder matches the density of air at sea level. Why would you not then conclude that the pressure of air in the cylinder will match the pressure of air at sea level?

    Have you ever been exposed to the ideal gas law: PV=nRT?
     
  4. Jul 26, 2017 #3
    Thank you for the reply.

    I was expecting this question but was having hard time to phrase my question in my previous post. I'll try my best what is really confusing me and why I think that the pressure should be different. I'll repeat my original post with some addition and changes. Please have a look on the both attachments. We can assume the temperature to be 25 C°.

    The air pressure at the surface is about 1,00,000 N/m2.

    It means that on average a virtual cylinder of air one square meter [m2] in cross-section, measured from sea level to the top of the earth's atmosphere, has a mass of about 10,000 kilograms and weight of about 1,00,000 newtons. That weight (across one square meter) constitutes pressure of 1,00,000 N/m2 at the surface of earth.

    Please note that the atmosphere becomes thinner and thinner with increasing altitude, with no definite boundary between the atmosphere and outer space. The Kármán line at 100 km is often used as the border between the atmosphere and outer space. Therefore, we can say that the volume of that virtual cylinder is 1,00,000 m x 1 m2 = 1,00,000 m3.

    Let's look at the air molecules in that virtual cylinder containing atmospheric air. Let's imagine that the cylinder is closed at both ends - at one end by earth's surface where we also have a pressure sensor and at the other end, at the top of earth's atmosphere, by some other material. At any instant only a fraction of molecules will be hitting the bottom of earth's surface downward, some of them will be hitting the other end where the atmosphere ends, some of them will be hitting the cylinder laterally and others will be in the space between the cylinder.

    Although at any instant only a fraction of molecules will be hitting the bottom of earth's surface downward, the pressure sensor will still read 1,00,000 N/m2. This is interesting because although only a fraction hit the surface, the force still measures 1,00,000 N as if all the molecules are striking the surface. In my opinion, this is what is happening. The molecules strike the bottom of the cylinder, i.e. earth's surface, more frequently and harder than the top of the cylinder. The sum total of the greater number of collisions and the greater strength of the collisions will still be that the average force against the bottom will be 1,00,000 N.

    There are greater number of collisions with the surface of earth compared to the other end for several reasons. The earth tries to pull all the air molecules in that cylinder downward toward the ground but as the force of gravity becomes weaker as the distance increases from the surface therefore its pull on the molecules far from the surface is weaker and on the other hand the air molecules also resist this compression due to gravity due to their inter-molecular repulsion and they tend to keep maximum distance from each other, and moreover the molecules have their own independent motion due to their kinetic energy and the molecules which have greater kinetic energy tend to escape the gravity pull toward the surface more easily. But overall the gravity keeps comparatively more air molecules closer to its surface and hence greater number of collisions with the surface of earth.

    The striking impact of the individual air molecules with the surface of earth is more powerful for two reasons. Firstly, each molecule has its own kinetic energy and moreover it's getting pulled by gravity vertically downward which constitutes more force at the instance of impact with the surface. It's just like when you fire a bullet from your roof into a table sitting on the ground - the bullet will have forceful impact with the table compared to a bullet which hits the table along the ground. Secondly, the most important reason is cumulative effect of gained momentum. Let me explain. The bottom of the cylinder, i.e. earth's surface, is pulling all the air molecules vertically downward. A molecule which is at the top end of the cylinder, where the atmosphere ends, might strike another molecule which is below it and transfer its vertical momentum to that molecule which in turn can transfer its vertical momentum toward the ground to another molecule which is below it and it becomes a domino effect. As a result the molecules closer to the surface of earth which lie below the majority of the molecules end up with far greater individual momenta. So, the impact of each molecule would be the result of contribution of momentum from many other molecules.

    I assume that what I have said above is almost correct.

    Question:
    How can a cylinder which measures only 1 m3, let's call it Cylinder #1, compared to a cylinder, let's call it Cylinder #2, which has far greater volume of 1,00,000 m3 exert an equal amount of pressure at the surface of earth?


    Let me try to elaborate more. If you wee to liquefy the air in both cylinders, you would end up with 1.3 kg of liquefied air in one cylinder and in the other virtual cylinder you would end up with 10,000 kilograms. How could 1.3 kg of liquefied air exert same pressure as 10,000 kg?


    Yes, I'm familiar with PV=nRT gas law but don't know how to convert 1.3 kg or 10,000 kg into moles.

    Thank you for your help and time.
     

    Attached Files:

  5. Jul 26, 2017 #4

    jbriggs444

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    Mostly correct, but almost completely irrelevant. Pressure is what it is, regardless of whether the air is being pushed down by a 100 mile column of air above or the sealed top of a cylinder above.
     
  6. Jul 27, 2017 #5
    Thank you for the reply.

    Let me ask you few questions to clarify it.

    Suppose you have a cylinder with base area 1 m2 and height 1 m which gives it a volume of 1 m3. The mass of the cylinder is 5 kg. You fill up this cylinder with 1 kg of gas X. How much the cylinder plus gas weigh? 6 kg?

    Now let's assume that the height of cylinder is 2 m which gives it volume of 2 m3 and mass of 8 kg. You fill the cylinder up with 2 kg of gas X. How much the cylinder plus gas weigh? 10 kg or 9 kg?

    Thanks a lot for your help.
     
    Last edited: Jul 27, 2017
  7. Jul 27, 2017 #6

    jbriggs444

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    6 kg.

    But... are you measuring this weight with a scale under the cylinder? If so, the scale will read less than 6 kg. The cylinder has buoyancy because it is in air. If the air inside and the air outside are at the same density then the weight of the air inside is exactly offset by the buoyancy from the air outside.
    Its mass is 10 kg. You tell me what its down force on the scale will be.
     
  8. Jul 27, 2017 #7
    Thank you for the reply.

    I also thought that it should measure 6 kg.

    Yes, the weight is being measured with a scale under the cylinder. The density is same inside and out.

    The force should be 100 N assuming gravitational pull of 10 N/kg.

    So, as the base area for both cases is similar, i.e. 1 m2, therefore pressure on the base for 6 kg cylinder would be 60 N/m2, and for the 10 kg cylinder it would be 100 N/m2. I hope that you agree with this.

    Assuming you agree with my statement above then how come 1 m high cylinder and 1,00,000 m high cylinder from one of my previous posts, i.e. post #3, would exert the same pressure.

    Thank you.
     
  9. Jul 28, 2017 #8

    jbriggs444

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    Because of the 99,999 m of air pressing down on the top of the cylinder.
     
  10. Jul 28, 2017 #9

    russ_watters

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    Because pressure isn't weight, so you are looking at apples and asking why they aren't oranges. In the special case of the atmosphere, the weight of the column of air causes it to have the pressure we see at the bottom, but for gases in containers, the weight has essentially nothing to do with the pressure (though density does). But in neither case do you measure the pressure with a scale: you measure it with a pressure gauge.
     
  11. Jul 29, 2017 #10
    Thank you.

    I understand your point to some extent but I'm still confused so let me ask you few questions.

    Before I ask any question I tried to calculate air pressure using PV=nRT as suggested above. Please have a look on this attachment. The molar mass of dry air is 29 g/mol so the total number of moles is (1/29 x 1300)=44.83. The temperature in Kelvin is 25 C = 298.15 K. The value of R=8.314 J/mol.K. The volume is 1 m3. The pressure would be 111125 N/m2. The value is pretty close to atmospheric pressure at sea level.

    Question 1:
    A 1 m3 cylinder containing 1.3 kg of air at 25 C will have same pressure as 'weight' of 1,00,000 m column of air with base 1 m2. The density of atmospheric air and the air in cylinder is same. How is so that both cylinder and column happen to have the same pressure? Just a coincidence?

    Question 2:
    Suppose that you have a thermally insulated 1 m3 rectangular cylinder in vacuum containing 1.3 kg of air at 25 C. The pressure would be 111125 N/m2. This pressure is result of collisions with the faces of cylinder and the molecules move around due to their 'intrinsic' motion.

    Now you place the cylinder on surface of earth. You weigh it and total weight would be "weight of cylinder" plus "weight of air". The weight of air is a result of collisions of air molecules with the base of cylinder although the molecules strike the top face of cylinder too but those collisions are not that forceful compared to collisions with the base because of gravity; as a matter of fact the collisions with the bottom face of cylinder are stronger by almost 13 N.

    The collision of each molecule with the bottom face of cylinder consist of two components: one component due to intrinsic thermal motion and the other component by virtue of gravity. Therefore, don't you think that the bottom face should experience more pressure compared to all other faces?

    Thank you for the guidance.
     
  12. Jul 30, 2017 #11

    jbriggs444

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    No coincidence. You chose the numbers so that it would come out that way.

    The second question is all over the map. It is not clear what the point is. You start with a 1 m^3 cylinder in vacuum. But you never use the fact that it is in vacuum. Instead, I suspect that you want us to consider that cylinder in an environment free from gravity. Fine. In zero g, both "top" and "bottom" inside faces are under identical pressure equal to 111125 N/m2.. The outside faces are, of course under a pressure of 0 N/m2 -- vacuum as specified.

    Now you put the cylinder on earth. The air inside sags a very tiny bit under gravity, arranging itself so that there is a pressure differential between top and bottom. By no coincidence, this pressure differential is just enough to support the air inside. The pressure at the inside top will be low by about 6.5 N/m2 and the pressure at the bottom will be high by about 6.5N/m2. The pressure at the top would be, perhaps, 111,119 N/m2 and the pressure at the bottom 111,131 N/m2.

    You want us to "weigh" the cylinder. Its true weight would reflect the mass of the cylinder plus the mass of the contained air. Its measured weight will have a small discrepancy due to atmospheric buoyancy. The cylinder displaces 1 cubic meter of air massing 1.3 kg. So the measured weight will be low by 13N.

    You ask whether the bottom will be under more pressure than the top. Yes, of course it will. So?
     
  13. Aug 5, 2017 #12
    Thank you for the help.

    I'm sorry that you might be thinking that I'm going round in circles but seriously I'm finding it hard to explain my confusion. I will give it a last try.

    You already confirmed that in an environment free from gravity a 1 m3 rectangular cylinder containing 1.3 kg air will exert a pressure of 111125 N/m2 at 25 C. This pressure is due to intrinsic motion of air molecules. And when such a cylinder is placed on earth, it creates a pressure differential between top and bottom faces to account for air weight. I believe that the side faces of cylinder still experience the same average pressure of 111,125 N/m2 although the pressure will vary along the side faces with height because the density differs from top to bottom with being maximum at bottom. So, it can be concluded that two factors come into play - pressure due to intrinsic motion of air molecules and pressure differential created due to weight.

    When it comes to atmospheric pressure, it is said that the pressure is due to the weight of air as if the motion of molecules of air play no part at all. Let's suppose an ideal situation by assuming that the atmosphere is enclosed within in a sphere so that the atmosphere is enclosed between earth and that outer sphere. The distance between this sphere and earth being 100 km because the atmosphere also ends around this distance from the surface. It's just like containing the air in a container. Even if the gravity of earth disappears the pressure due to motion of molecules will still be there and it should be equal to 111,125 N/m2 but the pressure differential between top and bottom will disappear. I'm assuming ideal conditions like uniform density of air, i.e. 1.3 kg/m3, 25 C temperature, earth being perfect sphere with smooth surface etc.

    I understand that in reality earth is like a one sided container - the surface of earth being the only side.

    Thank you for your patience.
     
  14. Aug 5, 2017 #13

    jbriggs444

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    The pressure in air is entirely due to the motion of the molecules, regardless of whether those molecules are constrained by the top of a cylinder or by the weight of all the other molecules higher up.
     
  15. Aug 6, 2017 #14
    Thank you.

    If you don't mind let me ask you few questions about what I wrote in my last post.

    You already confirmed that in an environment free from gravity a 1 m3 rectangular cylinder containing 1.3 kg air will exert a pressure of 111125 N/m2 at 25 C. This pressure is due to intrinsic motion of air molecules. And when such a cylinder is placed on earth, it creates a pressure differential between top and bottom faces to account for air weight. I believe that the side faces of cylinder still experience the same average pressure of 111,125 N/m2 although the pressure will vary along the side faces with height because the density differs from top to bottom with being maximum at bottom. So, it can be concluded that two factors come into play - pressure due to intrinsic motion of air molecules and pressure differential created due to weight. Question: Do you agree with this? A short answer like "yes" or "no" would suffice.

    When it comes to atmospheric pressure, it is said that the pressure is due to the weight of air as if the motion of air molecules plays no part at all. You have already directly answered this by saying "The pressure in air is entirely due to the motion of the molecules, regardless of whether those molecules are constrained by the top of a cylinder or by the weight of all the other molecules higher up". So, you are essentially saying that the atmosphere would still exert the same pressure, i.e. 111,125 N/m2 even if there was no gravity assuming that the atmosphere doesn't just disappear into space. Question: Do I have it correct?

    I assume that your answer to the question above is "yes". Question: Then, I would say that what is the effect of gravity on atmospheric pressure?

    Thanks a lot for your help and patience.
     
  16. Aug 6, 2017 #15

    jbriggs444

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    No.

    There is no such thing as "intrinsic motion" as distinct from just plain old "motion". There is no such thing as a pressure differential due to weight that is not also a pressure differential due to motion and density.
     
  17. Aug 7, 2017 #16
    Thank you.

    I have few more questions and I would request you to help me.

    There is a 1000 m3 rectangular cylinder (1 m2 base x 1000 m length) in vacuum free from gravity. It contains 1300 kg air (density 1.3 kg/m3). In zero gravity, both "top" and "bottom" inside faces are under identical pressure equal to 111,125 N/m2. The pressure will be same along all other faces.

    Now the cylinder is put on a planet free from atmosphere with gravitational pull of 10 kg/N. The density of air is adjusted due to gravity in the sense that the air is more dense closer to bottom than at the top. Will the pressure increase only because more number of molecules are striking the bottom per unit time, or is the second reason also being that the molecules striking the bottom with more force? And are both factors equally important?

    It is said that pressure acts equally in all directions. I tend to agree with this statement when it comes to gases or liquids in containers ignoring gravity. But when I think of the effect of gravity, it confuses me. Please have a look on the attachment. There are pressure sensors at locations labelled A, A', B, B' and so on. For example, "A" pressure sensor measures the pressure vertically downward and " A' " sensor measures lateral pressure; both A and A' sensors are almost at the same level. The gravity provides an extra momentum to all the molecules only vertically downward and not laterally, therefore sensors A, A' and other sensor pairs should give different readings. What do you say? Where am I going wrong?

    Thanks a lot.
     

    Attached Files:

  18. Aug 7, 2017 #17
    Let's consider one air molecule in earth's atmosphere at 1 km altitude. The molecule must carry the weight of trillion air molecules above it. Here "carry the weight" means to transmit the weight to the next molecule below.

    So our molecule punches the molecule below downwards and the molecule above upwards.

    Now let's assume the force of the down-punches is the same as the force of the up-punches. The net force on our molecule from other molecules is zero, so the molecule free falls downwards.

    Now let's assume the molecule does not fall but floats. We can see that in this case the force of the down-punches must be larger than the force of the up-punches. About one trillionth larger.
     
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