A simple barometer tube contains some air in it. The length of the tube above the mercury level in the trough is 80 cm. The hieght of the mercury in the the tube is 71 cm at normal atmospheric pressure. What is the actual decrease in the Atmospheric pressure if the barometer reads 65 cm.
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Answers
Answer:
Find the length of the tube above the nercury level in the trough and height of Hg in the tube normal atmosphere pressure (Pa)
Is the given barometer tube contains air in it ?
Then, Is Pa= Pressure due to entrapped air (P1) and the height of Hg column ? (1)
Take area of cross section of the tube as 'a'.
Get the value of 'P1' from (1)
Find the volume of entrapped air in the tube. Take in the tube. Take it as 'V1'.
If barometer reads 65 cm of Hg column, then the atmosphere pressure (Pa1) is equal to 65 cm of Hg + pressure due to air entrapped in the tube (P2).
i.e., P1a=65cm+P2 (2)
Now, find the volume of the air entrapped in the barometer tube.
Take it as 'V2'.
Find the value of P2 by using Boyle's law.
i.e., P1V1=P2V2
⇒P2=P1V1V2 (3) Substitute the value of 'P2' in (2) and find the value of P1a.
Then, actual decrease in atmospheric pressure will be equal to (Pa−P1a).
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Answer:
Let A be the area of the cross-section
V
1
=(90−74.8)A=15.2Acm
3
P
1
=76−74.8=1.2 cm of Hg
P
2
=(P−75.4) cm of Hg
V
2
=(90−75.4)A=14.6Acm
3
T
1
P
1
V
1
=
T
2
P
2
V
2
=
303×14.6
1.2×15.2×283
=75.4+1.17
=76.57 cm of Hg
Explanation:
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