12.2: Measuring Pressure
Objectives
The student will know:
- How atmospheric pressure is measured.
- How gauge pressure is defined.
Vocabulary
- atmospheric pressure: The pressure from the Earth's atmosphere.
- barometer: The oldest design for scientifically measuring pressure.
- gauge pressure: The pressure above the current atmospheric pressure.
Introduction
We live at the bottom of an immense ocean of air. The air high up is pulled down by gravity, pressing on the air below it, which also presses on the air below it. The weight of all the air above above you combines to exert atmospheric pressure, the pressure from the Earth's atmosphere. This constantly presses on all sides of things in contact with open air, and is what makes suction cups stick to walls, lets vacuum cleaners suck up dirt, and lets people drink through straws. At sea level such as a coastal city like New Orleans or New York, the typical atmospheric pressure is 101,325 pascals – the equivalent of a three-story high (10.3 meter) column of water. This is also known as one standard atmosphere of pressure, abbreviated \begin{align*}1 ~ \text{atm}\end{align*}.
\begin{align*}1 ~ \text{atm} = 101,325 ~ \text{Pa} = 1.013 \times 10^5 \frac{\text{N}}{\text{m}^2} = 101.3 ~ \text{kPa}\end{align*}
Calculating pressure from gas is different than pressure from fluids, because gases change their density easily. The Earth's atmosphere is denser close to sea level, and gets less dense the higher you go. Thus, there is no simple relationship between the height of air and pressure. Indeed, the actual pressure at sea level often changes with the location, climate, and current weather. Human beings can detect the differences in air pressure when our altitude changes a few hundred feet. This is most noticeable when your ears pop when driving up or down a mountain, or taking a plane flight.
The oldest design for scientifically measuring pressure is the barometer, invented in 1643. The main part of it is just a glass tube closed at one end and open on the other, filled with liquid and put open side down in a dish. The closed end can hold liquid up, just like when you put your finger on the end of a straw, it lets you hold water in the straw. The open end has full atmospheric pressure pressing on it, while the sealed side has no pressure. The Figure below shows a mercury-filled barometer.
If the tube is filled with mercury and has no air, then the atmospheric pressure pushes on the top of the dish, lifting up the mercury in the tube that has no air pushing down on it from above. All suction works this way – it is really atmospheric pressure pushing, rather than the vacuum pulling. The mercury is stable when the atmospheric pressure on the mercury in the dish is equal to the hydrostatic pressure which the column of mercury exerts at the bottom of the tube.
Illustrative Example 1
We can calculate the height of mercury lifted up by one standard atmosphere of pressure, \begin{align*}P_{atm}\end{align*}. Mercury has a density of \begin{align*}\rho = 13.6 \times 10^3 ~ \text{kg/m}^3\end{align*}. So, if the hydrostatic pressure from this mercury is equal to \begin{align*}P_{atm}\end{align*}, we have:
\begin{align*}& P_{atm} = P = \rho gh ~~ \rightarrow ~~ h = \frac{P_{atm}}{\rho g}\\ & h = \frac { \left( 101,396 \frac{\text{N}}{\text{m}^2} \right) } {\left(13.6 \times 10^3 \frac{kg}{m^3} \right) \left(9.81 ~ \frac{\text{m}}{\text{s}^2} \right) } = 0.760 ~ \text{m}\end{align*}.
So the pressure of one standard atmosphere hold a column of mercury with a height of 0.76 meters. As the atmospheric pressure changes, the height of the column of mercury changes.
Check Your Understanding
In the above experiment, if water had been used rather than mercury what would the height of the column of water have been?
Answer: Mercury is 13.6 times denser than water. So the atmosphere can support a column of water 13.6 times higher than a column of mercury.
Height of water column: \begin{align*}(0.76 \ m)(13.6) = 10.34 \rightarrow 10.3 \ m\end{align*}
A column of water 10.3 m high has the same weight as a column of mercury 0.76 m. Mercury barometers are therefore much more practical than water barometers.
Gauge Pressure
In practice, most examples of pressure happen inside atmospheric pressure. When a faucet pushes out water, that water is pressing out against atmospheric pressure. The gauge pressure is the pressure above the current atmospheric pressure. For example, a completely deflated tire still has pressure on it. Atmospheric pressure is pushing on both the inside and outside of the tire. What matters is how much additional pressure we put in when the tire is inflated.
When dealing with liquids, we typically refer to the pressure \begin{align*}P = \rho gh\end{align*} as the gauge pressure \begin{align*}P_g\end{align*}. The total pressure \begin{align*}P_t\end{align*} is therefore \begin{align*}P_t = P_{atm} + P_g = P_{atm} + \rho gh\end{align*}.
Check Your Understanding
1. A tire pressure gauge reads 32 pounds per square inch (psi) when measuring tire pressure. What is the total pressure in pounds per square inch?
Answer: Atmospheric pressure is 101,325 Newtons per square meter. To convert, we use:
- 1 pound = 4.45 newtons
- 1 inch = 0.0254 meters
\begin{align*}& P_{atm} = 101,325 ~ \frac{\text{N}}{\text{m}^2} \times \left( \frac{1 ~ \text{lbs}}{4.45 ~ \text{N}} \right) \times \left( \frac{0.0254 ~ \text{m}}{1 ~ \text{in}} \right)^2 = 14.7 ~ \frac{\text{lbs}}{\text{in}^2} \\ & P_t = P_{atm} + P_g = 14.7 ~ \frac{\text{lbs}}{\text{in}^2} + 32 ~ \frac{\text{lbs}}{\text{in}^2} = 46.7 ~ \frac{\text{lbs}}{\text{in}^2}\end{align*}.
2. The bottom of Lake Pontchartrain has a gauge pressure of \begin{align*}3.9 \times 10^4 ~ \frac{\text{N}}{\text{m}^2}\end{align*}. What is the total pressure in units of standard atmospheres of pressure, \begin{align*}\text{atm}\end{align*}?
Answer: To find the pressure in the units of \begin{align*}\text{atm}\end{align*}, we convert from \begin{align*}\frac{\text{N}}{\text{m}^2}\end{align*} to \begin{align*}\text{atm}\end{align*}, then adding in the atmospheric pressure of 1.0 \begin{align*}\text{atm}\end{align*}.
\begin{align*}& P_g = \left( 3.9 \times 10^4 ~ \frac{ \text{N} }{ \text{m}^2} \right) \times \left( \frac{1 ~ \text{atm} }{ 1.01 \times 10^5 ~ \frac{ \text{N} }{\text{m}^2} } \right) = 0.39 ~ \text{atm} \\ & P_t = P_{atm} + P_g = 1.0 ~ \text{atm} + 0.39 ~ \text{atm} = 1.39 ~ \text{atm}\end{align*}.
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