Volume of Air Bubble at Atmospheric Pressure

[kriegera]

Homework Statement

My teache recently assigned this problem but its nothing like the atmospheric problems we've done in class. Any suggestions on where to start?

The cross section of the tube of a mercury barometer is 1 square centimeter, and when the barometric height is 760 millimeters, 6 centimeters of the upper end of the tube is empty. When a bubble of air is passed up into the tube it is found that the mercury surface in the tube drops 2 centimeters. Calculate the volume of the bubble of air at atmospheric pressure.

Homework Equations

The Attempt at a Solution

The volume of mercury would be V=ha = (760)(1) = 760
but that is as far as I've gotten.

 

[ehild]

What is the pressure of the air inside the volume above the mercury column?

ehild

 

[kriegera]

Not sure how to calculate this. Pressure is:
P=F/A

but I don't know what the force would be?

 

[ehild]

You see a mercury barometer in the attachment. At the level of the mercury surface outside the the glass tube, the pressure should be th esame inside and outside: The hydrostatic pressure of the mercury column balances the atmospheric pressure. This pressure is

$$p_0=\rho g h_0$$.

($\rho$ is the density of mercury, g=9.8 m/s^2, ho=76 cm).

When some air gets inside the glass tube, it fills the empty space above the mercury column, and exerts some pressure on the mercury. To reach balance between the pressure outside the tube and inside, the mercury column falls, so its hydrostatic pressure + the pressure of the air inside the tube equals the atmospheric pressure. The new height is 2 cm lower as before, that is 74 cm.

$$76 \rho g =74 \rho g +P \rightarrow P=2 \rho g$$

The volume of the air inside the glass tube is V=(6+2) cm^3 and its pressure is P. At the atmospheric pressure, this amount of air would occupy the volume Vo=(P/Po )*V=2/76 *8 cm^3.

 

[kriegera]

so is this "Vo=(P/Po )*V=2/76 *8 cm^3 = 0.000513 cm or 0.000513 mm

the volume of all the air in the barometer? Do I need to find for 2cm:
2*(0.000513/8)

or is that what the P=2 in the equation accounts for?

Thanks for this-very helpful.

 

[ehild]

The pressure is equal to high of the mercury column * density of mercury *g. So the pressure is proportional to the equivalent column height. As we use the ratio of pressures, it is the same as the ratio of heights.

If you want the value of pressures in pascals, they are :

Po=0.76 (m) *13600 (kg/m^3) * 9.8 (m/s^2 )= 1.013 *10^5 Pa
P=0.02 (m) * 13600 (kg/m^3)*9.8 (m/s^2) = 2.666*10^3 Pa

P/Po=0.026

ehild

 

[kriegera]

The pressure is equal to high of the mercury column * density of mercury *g. So the pressure is proportional to the equivalent column height. As we use the ratio of pressures, it is the same as the ratio of heights.

If you want the value of pressures in pascals, they are :

Po=0.76 (m) *13600 (kg/m^3) * 9.8 (m/s^2 )= 1.013 *10^5 Pa
P=0.02 (m) * 13600 (kg/m^3)*9.8 (m/s^2) = 2.666*10^3 Pa

P/Po=0.026

ehild

so we have to set the equations equal to each other to find the volume of the air bubble?
0.026*V=2/76 *8 cm^3.
0.026*V =0.0000513
Solve for V=0.001973 cm??

also, what is the "13600" in the pressure equation?

 

[ehild]

13600 is the density of mercury in kg/m^3

You know the pressure of the air in the glass tube above the mercury column and you know its volume: 8cm^3= 8.*10^-6 m^3.

Boyle' law states for ideal gases that PV = constant at constant temperature.

So P (above mercury)* V(above mercury) = Po (atmospheric pressure) *Vo (volume at atmospheric pressure)

You need Vo:

$$V_0= P*V/P_0=\frac{2.666*10^3}{1.013*10^5} \cdot8*10^{-6}=0.21*10^{-6} m^3 = 0.21 cm^3$$

ehild

 

[kriegera]

You know the pressure of the air in the glass tube above the mercury column and you know its volume: 8cm^3= 8.*10^-6 m^3.

ehild

So 0.21cm^3 is the volume of the air bubble.

I just don't get where you got the equation above. I know 8cm^3 is the volume of the air bubble but where did you get the 8*10^-6?