When a tube, closed at one end an has trapped air is held horizontally, the length of the column of air is 50mm. When held vertically, closed end down, length of air column is 45mm. Mercury tread...

When a tube, closed at one end an has trapped air is held horizontally, the length of the column of air is 50mm. When held vertically, closed end down, length of air column is 45mm. Mercury tread that encloses air is 85mm. Find the atmospheric pressure. 

 

How do I solve this question? I know I have to use the formula p= constant over volume, but other then that, I'm stuck...

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valentin68 | College Teacher | (Level 3) Associate Educator

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See attached figure for two positions of the tube.

In the horizontal position 1) the Hg column does not press on the air inside the tube.

-the pressure of the air in the tube is `P1 =P0` (= atmospheric pressure).

-the volume of the air in the tube is `V1 =S*L1`

In the vertical position 2) the Hg column is pressing on the air inside the tube.

-the pressure of the air in the tube is

`P2 =P0 +ro_(Hg)*g*H_(Hg)`

-the volume of air in tube is

`V2 =S*L2`

Because the temperature is constant you have the equation of transformation of the gas

`P1*V1 =P2*V2`

`P0*S*L1 =(P0 +ro_(Hg)*g*H_(Hg))*S*L2`

`P0*50*10^-3 = (P0 +13534*9.81*85*10^-3)*45*10^-3`

`50*P0 =(P0 +11285.3)*45`

`P0 +11285.3 =50/45*P0`

`11285.3= 5/45*P0`

`P0 =101567.7 N/m^2`

The atmospheric pressure is P0 =101567.7 N/m^2

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