An open ended U-tube with a uniform cross section equal to 1 cm2 and height 100

Question

An open ended U-tube with a uniform cross section equal to 1 cm2 and height 100 cm on both sides is mounted on a stand in the lab. The tube is filled with mercury to a height of 50 cm on both sides when the barometric pressure in the lab is 750 Torr. The left side of the tube is then sealed and the right side is connected to a vacuum pump and evacuated to a pressure of essentially zero Torr. (a) What distance in cm does the mercury level fall from its original position on the left side of the tube? (b) What is the final pressure in Torr of the air trapped in the left side of the tube? An open ended U-tube with a uniform cross section equal to 1 cm2 and height 100 cm on both sides is mounted on a stand in the lab. The tube is filled with mercury to a height of 50 cm on both sides when the barometric pressure in the lab is 750 Torr. The left side of the tube is then sealed and the right side is connected to a vacuum pump and evacuated to a pressure of essentially zero Torr. (a) What distance in cm does the mercury level fall from its original position on the left side of the tube? (b) What is the final pressure in Torr of the air trapped in the left side of the tube? An open ended U-tube with a uniform cross section equal to 1 cm2 and height 100 cm on both sides is mounted on a stand in the lab. The tube is filled with mercury to a height of 50 cm on both sides when the barometric pressure in the lab is 750 Torr. The left side of the tube is then sealed and the right side is connected to a vacuum pump and evacuated to a pressure of essentially zero Torr. (a) What distance in cm does the mercury level fall from its original position on the left side of the tube? (b) What is the final pressure in Torr of the air trapped in the left side of the tube?

Explanation / Answer

a) when the right side is connected to vacuum

as pressure at the bottom of tube = 750 torr

therefore , as 1 torr = 1 mm of mercury

hr - hl = 75 -----(1)

hl + hr = 100 ----(2)

hr = 87.5 cm

inital height in right side = 50 cm

as rise in right side = fall inleft side

fall in left tube = 87.5 - 50 = 37.5 cm

the fall in height of mercury in left side is 37.5 cm

b)

air pressure in left side ,

Pl = 750 torr

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