**Question description**

**Lab 8 – Air Density, Pressure, Temperature, and Heat Transfer**

**Please make your answers highlighted in yellow.**

**Air Density and Air Pressure**

The atmosphere is made up of a mixture of several gases – though it is mostly nitrogen and oxygen – and is held to the Earth by the force of gravity. Because a gas is compressible, gravity pulls most of the gas molecules close to the surface of the Earth. This implies that the weight of the air overhead is greatest at the Earth’s surface, and this weight decreases with height. Another way of stating this fact is to say that the density of the atmosphere decreases with height where density in this context is defined as the mass (measured in grams or kilograms) of the air contained in a unit volume (measured in cubic centimeters or cubic meters). At sea level air has an average density of about 1.2 kg per cubic meter.

It should be pointed out that there is no clear “top” of the atmosphere since it would be difficult to find a level above the Earth where there are no gas molecules of any kind. Even in interstellar space, hydrogen gas molecules can be found. But for all practical purposes the “top” of the Earth’s atmosphere can be considered to be within the first 100 kilometers of the surface, which is about 62 miles.

There is a useful rule of thumb that states the following: for every 5.6 kilometers you rise into the atmosphere, there is **half** of the atmosphere above you than where you started. Thus, for example, if you started at 2 km above sea level and then went up to 7.6 km, then there would only be half as much atmosphere above your head at 7.6 km compared to how much was above your head when you started at 2 km above sea level.

- Complete the following table by indicating the percentage of atmosphere above each of the altitudes shown using the approximation that for every 5.6 km you go up into the atmosphere, the percentage of the atmosphere still above you is half of where you started.
- With this in mind go back to your original graph on the previous page, and now
**add a pressure scale directly beneath your percentage scale on the x-axis, and include the pressure readings from 0 mb to 1000 mb**. 0 mb would, of course, be the top of the atmosphere, and 1000 mb would be at sea level.

Height Above Sea Level (kilometers) |
% of Atmosphere Above |

22.4 | |

16.8 | |

11.2 | |

5.6 | |

0 (Sea Level) | 100% |

Use the data above, and graph the percentage of the atmosphere above each height level on the graph on the next page. On the horizontal x-axis, use the ten evenly-spaced intervals on the graph to represent the percentages from 0% to 100%, and label this axis, “*Percentage of Atmosphere Above At Different Heights*”. On the vertical y-axis, use the twelve evenly-spaced intervals on the graph to represent the different heights from sea level (which can be thought of as height level 0 kilometers) to 24 kilometers above sea level, and label this axis “*Height Above Sea Level*.” Title your graph at the top as, “*Percent Atmosphere Above vs. Height*.”

Finally, **connect your points with a smooth curve**!

**Air pressure** – also called **barometric pressure** – can be thought of as the weight of all of the air molecules contained in an imaginary column above any location on Earth. It turns out that an imaginary air column that is 1-inch by 1-inch square and starts at sea level and extends all the way to the top of the atmosphere, it would weigh, on average, about 14.7 pounds! It should come as no surprise that air pressure – like air density – also decreases with height simply because there are less air molecules overhead as you ascend higher into the atmosphere.

Air pressure is measured in units called **millibars** (**mb**). At sea level the air pressure averages about 1013.25 mb which is the same thing as 14.7 pounds per square inch. For this lab exercise, we will round this number down to 1000 mb for the sake of simplicity (and for making the math a little easier!).

Since the weight of the air is directly related to air density, this means that air density is directly related to air pressure. Therefore we can apply our general rule of thumb introduced earlier in this lab to air pressure. For example, if the air pressure at sea level is 1000 mb, then at 5.6 km above sea level, the pressure would drop by half, or 500 mb. And if we went from 5.6 km to 11.2 km, it would drop by half from where we started (which would be 500 mb), and thus it would drop to 250 mb by the time we reach 11.2 km above sea level.

This tells you, for example, that at sea level – where there is 100% of the atmosphere above your head – the air pressure is 1000 mb. Similarly, you can use your graph to determine what the air pressure would be at the various heights above sea level (and as already mentioned, you can see on the graph that the air pressure at 5.6 km above sea level would be 500 mb).

Using your graph, you can find the approximate average air pressure at any altitude above sea level. To be more accurate, we can use the following equation:

Average Air Pressure at Any Altitude = 1000 mb * (0.5)^{x/5.6}

where ” * ” is the symbol for multiplication, and the little “x” in the exponent is the height above sea level you are interested in, measured in kilometers.

**Do not let this equation with the exponent intimidate you. It’s not too difficult!** For example, let’s do a straightforward calculation. Say you had an equation such as the following:

y = 5 * 2^{x}

Let’s say that “y” equals 5 times 2 raised to the “x” power.

What would you get for “y” if you let “x” = 3. You would get:

y = 5 * 2^{3}

which is the same thing as

y = 5 * (2 * 2 * 2) = 5 * 8 = 40

So now let’s do an example using the equation for Average Air Pressure at Any Altitude “x.”. Let’s say you wanted to know the average air pressure in Amarillo, Texas. First you need to know how high above sea level Amarillo is. It turns out Amarillo is about 1,100 meters above sea level which is the same thing as 1.1 kilometers. Therefore, we would write:

**Average Air Pressure in Amarillo = 1000 mb * (0.5) ^{1.1/5.6}**

**Average Air Pressure in Amarillo = 1000 mb * (0.5) ^{.19642857}**

**Average Air Pressure in Amarillo = 1000 mb * .873**

**Average Air Pressure in Amarillo = 873 mb**

Therefore, on an average day in Amarillo, the air pressure would be about 873 mb.

What other information does this tell us? It tells us that at the elevation of Amarillo (i.e., 1.1 km above sea level) there is 87.3% of the atmosphere is above the city, and 12.7% of the atmosphere is below the elevation of Amarillo.

Remember, you could also use your graph to arrive at this same conclusion! Check it out. Go to 1.1 km on your graph, and draw a horizontal line across the graph until it intersects your curve. If you have drawn you curve accurately, you should get a pressure reading of about 873 mb for an altitude of 1.1 km.

Therefore, using information either from your graph or from the equation above, fill in the following blanks for #3 – 6. **You must “show your work” for each problem for credit (write out the multiplication and division that you did for EACH problem just like in the above example in bold text).**

*You need to use a math calculator for these problems to calculate the exponents OR do a “google” search for an online exponents calculator – one website I recommend is: https://www.calculatorsoup.com/calculators/algebra/exponent.php

3. A jet is flying at an altitude of 10 **km** above sea level. The air pressure at this level is approximately _________ mb, and there is _______% of the atmosphere above the jet.

Calculation: ____________________________________________________________________

4. Mountain climbers have climbed to the top of Mt. Everest at an altitude of 8.85 **km** above sea level. The air pressure at this level is approximately _________ mb, and there is _______% of the atmosphere above the mountain climbers.

Calculation: ____________________________________________________________________

5. Tornado chasers are driving across the High Plains of west Texas and are passing through Lubbock whose city elevation is 992 **meters** above sea level (don’t’ forget to convert m to km!). The air pressure at this level is approximately _________ mb, and there is _______% of the atmosphere above the storm chasers.

Calculation: ____________________________________________________________________

6. A daredevil skydiver has ascended in a helium-filled balloon to an altitude of 30 **km** above sea level. The air pressure at this level is approximately _________ mb, and there is _______% of the atmosphere above the skydiver before he/she jumps.

Calculation: ____________________________________________________________________

**Correcting Air Pressure Readings to Sea Level**

Since air pressure always decreases with height, it stands to reason that the air pressure, for example, in Amarillo (1099 meters above sea level) will always be less than the air pressure in Dallas (130 meters above sea level). But when we analyze the air pressure at several cities across the United States, we are interested in the changes in air pressure due to **weather systems** and not the changes in air pressure due to cities being located at different altitudes.

You see, air pressure can change from one location to the next not only because they have different altitudes, but the pressure can also be different because of the way temperature changes with height above adjacent locations. And it turns out that it is these changes of temperature of with height from one city to the next that determine how the winds will blow which in turn largely determines the type of weather any given location will have !

And so meteorologists are very interested in these changes in air pressure from one location to the next. But changes in air pressure from one location to the next due to changes in the vertical temperature profile can be very subtle, and these small but important changes can be easily masked by the overwhelming influence of altitude between these same two locations. So we must have a way to eliminate the affect of altitude on a city’s air pressure reading if we ever hope to find the more subtle influences to air pressure brought about by weather systems.

Perhaps now is a good time to point out that the **wind** is the result of air always trying to move from areas of higher pressure readings to areas of lower pressure readings. But because there is such a strong influence of altitude in any given location’s pressure reading, it would SEEM to stand to reason that air should always be blowing from low elevations to high elevations. And thus, for example, it would SEEM that the wind should always be blowing from the valley floor to the mountaintops since air pressure will always be lower at the top of mountains compared to the valley floor.

But experience tells us this is not the case. Only when we remove the influence of elevation on a location’s air pressure reading do we find it possible to have a “corrected” air pressure reading that is actually higher on the mountaintop than on the valley below! Thus, in this “corrected” atmospheric state, the wind would blow from the mountaintops to the valley below!

So how do we eliminate and “correct” the influence of altitude on a location’s air pressure reading and in so doing reveal the air pressure reading that is a result of weather influences instead? Air pressure readings are adjusted (i.e., corrected) to a common elevation so that when you compare the readings from one location to the next, it is as if they are all located at the same altitude. By international agreement, this altitude has been chosen to be sea level.

So we adjust a location’s air pressure reading so that it would reveal what the air pressure would be IF that location were at sea level! For locations above sea level we would have to ADD pressure units to their actual air pressure reading to correct it to sea level. For stations below sea level, we would have to SUBTRACT air pressure units to their actual air pressure reading to correct it to sea level. For stations at sea level, no correction needs to be applied. The number of pressure units used in these adjustments changes from day to day and from station to station, and the actual calculations need not concern us here. What is important is that you understand how and why the correction must be made.

Consider the following:

7. Which station – A, B, C, D, or E – would require the greatest adjustment to their actual air pressure reading in order to correct it to sea level? Why did you choose this answer?

8. Which stations would require little to no correction to their actual air pressure reading in order to correct it to sea level? Why did you choose this answer?

9. Which station or stations would require that you to add pressure units to their actual air pressure readings in order to correct it to sea level? Why did you choose this answer?

10. Which station or stations would require that you subtract pressure units to their actual air pressure reading to correct it to sea level? Why did you choose this answer?

**Change of Temperature with Height**

It probably comes as no surprise that temperatures usually get colder as you get higher into the atmosphere. After all it is usually much colder at the top of a mountain compared to the valley floor below. But the temperature does not always get colder with height! We will now examine how the temperature changes throughout the depth of the entire atmosphere in terms of the average change as well as by using actual temperatures observed on a single day at a given location, i.e., real data!

The following table shows the AVERAGE temperature at various heights above sea level. This is known as **Earth’s Standard Atmosphere**. Recall that 0 km is considered sea level.

Height Above Sea Level (km) |
Temperature (°C) |

0 | 15 |

5 | -17.5 |

10 | -50 |

10.9 | -55.8 |

11 | -56.5 |

20 | -56.5 |

25 | -51.6 |

30 | -46.6 |

32 | -44.7 |

35 | -36.6 |

40 | -22.8 |

45 | -9.0 |

47 | -3.5 |

48 | -2.5 |

51 | -2.5 |

53 | -6.9 |

55 | -12.4 |

60 | -26.1 |

65 | -39.9 |

70 | -53.6 |

75 | -64.8 |

80 | -74.5 |

82 | -78.4 |

85 | -84.3 |

90 | -86.3 |

95 | -86.3 |

100 | -77.6 |

105 | -63.5 |

107 | -53.0 |

110 | -29.5 |

115 | -25.8 |

11. Plot this information on the following graph. Label the horizontal x-axis, Temperature (°C), and the vertical y-axis, Height Above Sea Level (km).

12. Title your graph, “*Average Temperature vs. Height*.” Keep in mind that since most of your temperatures are below 0°C, **you will NOT have 0°C as the starting point of your graph on the left-hand side! It will be closer to the right-hand side**.

13. There are two layers in the atmosphere where temperatures actually increase (get warmer) with height. These layers are called **inversion layers**. The lowest inversion layer is called the **stratosphere**, and the highest inversion layer is called the **thermosphere**. **Label these layers on your graph: stratosphere and thermosphere.**

14. There are two layers in the atmosphere where temperatures decrease (get colder) with height. These layers are called **temperature lapse layers**. The lowest temperature lapse layer is called the **troposphere**, and the highest temperature lapse layer is called the **mesosphere** where the coldest temperatures in the Earth’s atmosphere can be found. **Label these layers on your graph: troposphere and mesosphere.**

15. Separating these different layers in the atmosphere are thin layers where the temperatures does not change with height. These are called **isothermal layers**. The lowest isothermal layer is called the **tropopause**, the next higher up one is called the **stratopause**, and finally the highest isothermal layer is called the **mesopause**. **Label these 3 “pauses” on your graph: tropopause, stratopause, and mesopause**