Atmospheric Calculations 4/28/19

By Richard Bleil

Let’s have a little fun Bleil Style today, shall we. I asked this particularly nasty question on a test once.

At twenty degrees Celsius, at sea level, the pressure of the atmosphere is 14.696 pounds per square inch. What is the height of the atmosphere? We’ll talk about a few of the things we come across throughout the blog.

Okay, so, 14.696 psi is 1 atm. The ideal gas law is PV=nRT where P is pressure (1 atm), n is moles, R is the universal gas constant (0.08206 L.atm/mol.K) and T is temperature.

What temperature should we use? As it turns out, the average temperature of the atmosphere is somewhere around -30oC, and it fluctuates as you go higher.

The temperature is -30oC, but we have to convert this to an absolute temperature scale. Temperature is related to the speed at which molecules and atoms move. This implies that there must be a lower temperature, where molecular motion effectively stops. This temperature is called “absolute zero”, and has been determined to be -273.15oC. If we add 273.15 to degrees Celsius, we get a new temperature scale, called “Kelvin”, which can never have a value of zero. This is important in calculations since we sometimes divide by temperature, and division by zero is undefined. Thus, using absolute temperature scales is important.

By the way, there is an equivalent, but lesser known, absolute temperature in the Fahrenheit scale, called Rankine. In the Fahrenheit scale, absolute zero is at -459.67oF, so adding 459.67 converts degrees Fahrenheit to Rankine.

Okay, so, -30oC is 243.15 K. To use the ideal gas law, all we need now is n, or the number of moles of gas. A mole is just a number, equal to 6.02×10^23, or 602,000,000,000,000,000,000,000. It’s 6.02×10^23 of how many? Well, of anything that we are discussing. This is similar to a dozen, which is 12. Twelve of what? Twelve of whatever we are discussing. For our calculation, we need moles of a gas.

There are 453.592 grams in 1 pound. We know the pressure is 14.5 pounds per square inch, and using this conversion factor, 14.5 pounds is 6,316.084 grams. Air is roughly 70% nitrogen, 28% oxygen, 2% argon and additional trace gases. To convert to moles, the average molecular mass of air is 28.97 g/mol, so 14.5 lb, or 6,300 g, is 235 moles of gas.

The pressure we use is not constant. We can assume the pressure at the top of the atmosphere to be 0 atm, and at sea level it is at 1 atm. As such, the average pressure will be 0.5 atm.

Okay, now we know pressure (0.5 atm), temperature (which we are assuming to be constant at 243 K, which is actually a poor assumption), and moles (235 moles), so we can find volume. From PV=nRT and using R=0.082006, we find the volume of this column of air is 9,400 L. There are 2.54 cm in 1 inch, and 1 mL is 1 cm^3. So 9,400 L is 570,000 in^3.

We’re getting there. PSI means “pounds per square inch”, which means it is pressing on an area of 1 square inch. To have the volume of 570,000 in^3, we need a column with a height of 570,000 in. In other words, we have 570,000 inches of air pressing down on each inch of earth, which is 48,000 ft, or 9 miles. This is 15 km.

Does NASA agree? This calculation puts us at the top of the troposphere, just before the stratosphere where the ozone layer begins. We are actually far below the height of the atmosphere if it is defined as the top of the exosphere which is about 500 km. So I’ll call NASA and tell them that they’re WRONG! Well, maybe not. None the less, I find it interesting that the calculation puts us at the top of the troposphere.

So where could we have gone wrong? There are several assumptions that we have made that are not necessarily fair. For example, the temperature scale tends to fluctuate between -90 and 70oC, so my assumption of -30 may not have been very good. We also assumed that the density of the air decreased uniformly, which, again, may be a bad assumption.

Okay, maybe you thought this post was dry and tedious. But, with a few basic assumptions, we ran a calculation and ended up with a height corresponding to one of the critical divisions in the atmospheric makeup. I love these kinds of calculations. They aren’t meant to be accurate, just “back of the envelop” calculations, meaning rough estimates. This is the second such blog I’ve written along these lines. Next, I think we’ll estimate how many molecules of rubber we leave on the road per square inch that we drive. Or…the average number of aspirin molecules per cell in the human body for a regular dose.

It’s all fun!!!!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.