Question

A thin tube closed at the top and open to the atmosphere at the bottom contains a 12 cm high column of air trapped above a 20 cm column of mercury. If the tube is flipped so that the closed end is now at the bottom, what is the new height of the trapped column of air

Answer 1

Answer:

The column height of air after the inversion is 7 cm.

Step-by-step explanation:

The initial pressure balance is given as

P_1+20 cm=76 cm of Hg

P_1=76-20 cm of Hg

P_1=56 cm of Hg

The initial volume of the air with cross sectional area as 12 cm2 and the length of air column as 12 is given as V_1=12 cm *1 =12 cm3

After the inversion

P_2=20 cm+76 cm of Hg

P_2=96 cm of Hg

The volume of the air after the inversion with cross sectional area as 1 cm2 and the length of air column as x is given as V_2=x *1 =x cm3

Now as temperature is constant and the cross sectional area is also constant so

$P_1V_1=P_2V_2\\56\times 12=96\times x\\x=\dfrac{56\times 12}{96}\\x=7 cm$

So the column height of air after the inversion is 7 cm.

Answer 2

Answer:

7cm

Step-by-step explanation:

P1 =H-h =76-20=56; P2 =H+h =96

P1V1 =P2V2

56x12=96xV2

V2 =56x12/96 = 7 cm