Chemistry Practice Problems Manometer Practice Problems Solution: You may want to reference (Pages 397 - 400)Section...

# Solution: You may want to reference (Pages 397 - 400)Section 10.2 while completing this problem.What would be the height of the column in a barometer if the external pressure was 101 kPa and water (d = 1.00 g/cm3) was used in place of mercury?

###### Problem

You may want to reference (Pages 397 - 400)Section 10.2 while completing this problem.

What would be the height of the column in a barometer if the external pressure was 101 kPa and water (d = 1.00 g/cm3) was used in place of mercury?

###### Solution

We are asked to determine the height of the column in a barometer if the external pressure was 101 kPa and water (d = 1.00 g/cm3was used in place of mercury.

Recall:

The pressure can be calculated using the equation:

$\overline{){\mathbf{P}}{\mathbf{=}}\frac{\mathbf{F}}{\mathbf{A}}}$

The force, F, due to the air acting on the column is given by its mass times the acceleration due to gravity,

F = mg

where g = 9.8 m/s2

The pressure caused by the air is:

$\overline{){\mathbf{P}}{\mathbf{=}}\frac{\mathbf{F}}{\mathbf{A}}{\mathbf{=}}\frac{\mathbf{mg}}{\mathbf{A}}}$

Recall that mass is related to density:

$\mathbf{density}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{volume}}$

We can treat the water as column whose volume equals its cross-sectional area times its height:

$\mathbf{V}\mathbf{=}\mathbf{A}\mathbf{×}\mathbf{h}$

Hence, we have:

$\mathbf{P}\mathbf{=}\frac{\mathbf{mg}}{\mathbf{A}}\mathbf{=}\frac{\mathbf{dVg}}{\mathbf{A}}\mathbf{=}\frac{\mathbf{d}\left(\mathrm{Ah}\right)\mathbf{g}}{\mathbf{A}}\mathbf{=}\mathbf{dhg}$

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