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 During the derivation of variation pressure with height in atmosphere why don't we take the weight of the air which is above the air slab which we are using?
If we take a slab of air with crosssectional area of [itex] A[/itex] and height [itex] dz[/itex] in our atmosphere. Now, what we do is make an argument like this :
Pressure from below must balance both the weight and Pressure from above to keep the slab at rest. ( I have added an attachment for clarification)
And from this are argument we do something like this :
[itex] P_{above} = P [/itex]
[itex] W_{slab} = mg [/itex] , Pressure from weight is [itex] \rho \times A \times dz \times / A = \rho ~dz~g [/itex] ( [itex] \rho [/itex] is the density of air)
Therefore, [itex] P_{below} = P + \rho~dz~g [/itex]
[itex] dP = P_{above}  P_{below} =  \rho ~dz~g [/itex]
[itex] \frac{dP} {dz} =  \rho ~g [/itex]
Now, my question is why we didn't not take into account the weight of air which is above the slab? I mean the air which lies above our sample slab must have weight this weight would have been pushing our slab downwards and therefore pressure from below must also balance this. The mass of air which lies above the slab must have number molecules and gravity would attract them and we should center of mass and assume the gravity to act over there and this force of gravity should cause a pressure on our slab.
I tried to argue that pressure from above is due to weight of air only, but after sometime I thought that this argument is not good as fluid causes pressure due to collision of it's molecules .
So, I want to restate my problem again : Why in our derivation we didn't take into account the pressure caused by the weight of the air that lies above our slab?
Thank you, any help will be much appreciated. I tried to be as clear as I could be.
Pressure from below must balance both the weight and Pressure from above to keep the slab at rest. ( I have added an attachment for clarification)
And from this are argument we do something like this :
[itex] P_{above} = P [/itex]
[itex] W_{slab} = mg [/itex] , Pressure from weight is [itex] \rho \times A \times dz \times / A = \rho ~dz~g [/itex] ( [itex] \rho [/itex] is the density of air)
Therefore, [itex] P_{below} = P + \rho~dz~g [/itex]
[itex] dP = P_{above}  P_{below} =  \rho ~dz~g [/itex]
[itex] \frac{dP} {dz} =  \rho ~g [/itex]
Now, my question is why we didn't not take into account the weight of air which is above the slab? I mean the air which lies above our sample slab must have weight this weight would have been pushing our slab downwards and therefore pressure from below must also balance this. The mass of air which lies above the slab must have number molecules and gravity would attract them and we should center of mass and assume the gravity to act over there and this force of gravity should cause a pressure on our slab.
I tried to argue that pressure from above is due to weight of air only, but after sometime I thought that this argument is not good as fluid causes pressure due to collision of it's molecules .
So, I want to restate my problem again : Why in our derivation we didn't take into account the pressure caused by the weight of the air that lies above our slab?
Thank you, any help will be much appreciated. I tried to be as clear as I could be.
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