PV
45 cm
dry air
capillary
tube
15 cm
mercury
thread
15 cm
30 cm
dry air
form cross-sectional area in two
arrangements. Using the data in the
7
Fig. 10.17
The set-up illustrated above shows a capillary
tube of uniform cross-sectional area in two
different arrangement. using the
diagrams, calculate the pressure of the
atmosphere.
A 76 cm Hg B 75 cm Hg
C 74 cm Hg D 60 cm Hg
E 45 cm Hg

The dry air pressure is 75 cm of Hg.

Explanation:

From the set up it can be understood that the mercury thread is present in the middle of the capillary tube. So from the bottom, the top of the mercury thread will exhibit a pressure of P1 as

P1 = p + 15. Here p is the air pressure. Similarly, if the tube is made upside down, then the top of mercury thread will be P2 = p-15 as p is the air pressure.

In both the cases, the product of volume and pressure will be constant as per kinetic theory of gases.

P1 V1 = P2 V2

(p+15) V1 = (p-15) V2

(p+15) 30 = (p-15) 45

Since the volume is given as  30 and 45 respectively.

30 p + 450 = 45 p - 675

45p-30 p = 450 + 675

15 p = 1125

p = 1125/15 = 75 cm of Hg.

So the dry air pressure is 75 cm of Hg.

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

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