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/cm^{3}) was used in place of mercury?

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/c

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/cm^{3}) was used in place of mercury?

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/cm^{3}) was 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/s^{2}

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:

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

Hence, we have:

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