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Boyles Law And Atmospheric Pressure

It's probably too easy, but it's causing some confusion to me.

First of all why do we plot h vs 1/L in order to find the atmospheric pressure? and specifically, why ONE/L ?

I can't figure out how we will get to P(a) if we extrapolate the line to when 1/L = 0?





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Anyone?
Original post by Daniel Atieh
Anyone?


You are more likely to get useful replies if you supply all the required info. in your question. Here, you haven't told us what h,L,p(a)h, L, p(a) are. Regardless of that:

1. hh is the displacement (+ve or -ve) of the column of mercury from its position in figure (a)

2. LL is the height of the gas column in the left bend of the tube. LL is proportional to the volume VV of the gas, so 1/L∝1/V∝p1/L \propto 1/V \propto p, the gas pressure, according to Boyle's law. That just says that 1/L1/L gives us a measurement of the gas pressure, if we calibrate it correctly.

3. In figure (a), the trapped gas is at 1 atm since the pressure of the gas perfectly balances the mercury against the atmospheric pressure.

4. hh measures the excess pressure due to the column of mercury over atmospheric pressure. So when h=0h=0 then 1/L∝p1/L \propto p represents a gas pressure of 1 atm. If we add enough mercury to halve the volume of gas, experimentally we find that we added 76 cm of mercury i.e. h=76h=76, which is equal to another 1 atm.

5. If we take figure (a) and imagine removing some trapped gas, then the gas pressure will fall, and atmospheric pressure will push mercury down below its initial level (i.e. h<0h <0) until the weight of the mercury + gas pressure in the trapped gas tube balances atmospheric pressure. If we continue to remove gas until there is almost none there, then all of the atmospheric pressure is balanced solely by the weight of the mercury, and it fills almost all of the tube. At this point, the -ve value of hh corresponds to 1 atm. That is the situation illustrated by the extrapolation of the straight line into -ve values of hh

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