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Name: Walt A.
Status: educator
Age: 60s
Location: N/A
Country: N/A
Date: 4/26/2004

Air expands when heat is applied. By how much will the volume of air expand if the temperature is raise from 0 degrees F to 100 degrees F? Want to use this to instruct truck drivers on several issues:

1) How expanding air is used in a turbo/air blower in a diesel engine.

2) Why fuel mileage drops in winter due to the density of air. (3) Why tires can become over inflated when run on hot desert surfaces.


Other factors equal, the volume of the air in the tire will change by 1/273 for each degree (Celsius or Kelvin) change in temperature. A span of 0 degrees F (-17.8 C or 255 K) to 100 F (37.8 C or 311 K) represents an approximately 55 degree C or K difference. Thus, the volume of a tire (considered 1 unit) would expand to about 1.2 units. Likewise, the pressure inside would rise by a factor of about 1.2 as well.

For example: The pressure inside a tire containing air at 80 lb/in^2 (5.44 atm) at 0 F would rise to about 96 lb/in^2 (6.5 atm) at 100 F.

ProfHoff 844

Fine idea, Walt. Sounds simple enough and useful. To make this handy, one needs to memorize the value of Absolute Zero. For me, it is -273 degrees C. For your classes, it will be -459 dgrees F.

A trapped quantity of gas at constant pressure has Volume proportional to its Temperature-above-Absolute Zero. O degrees F + 459 F = 459 Absolute F (also called degrees R, for Rankine, if you want to teach that). 100 degrees F + 459 F = 559 Absolute F (absF?).

V2/V1 = T2/T1 = 559/459 = 1.22, or 22% expansion.

If the volume is constant, then the pressure is proportional instead. That would be 22% higher PSI Absolute.

Say a flacid but filled tire has 1psig at 0degF.

1 psig +15 psi = 16 psiA.

If that warms from 0F to 100F, and tries to expand 22%, the pressure inside will be

16 psiA x 1.22 = 19.5 psiA.

19.5 psiA -15 psi = 4.5 psig at 100 degF

I guess you cannot quite fill up a flat tire that way.

In scuba class they teach us "PV=NRT":

Pressure x Volume = Number_of_molecules x magic_number_R x Temperature.

Very general, always works. Maybe it would be helpful to buy a paperback scuba lesson text.

Magic_number_R depends on what units you will always remember to use:

Pressure: PSIA / Atmospheres / Torr / Pascals

Volume: Cubic Meters / Liters / cm3 / cuft / cubic inch

Temperature: degrees K (abs C) / degrees R (abs F)

Number: raw number / Moles (1 Mole = 6x10^23 each) / Liters @ STP / cuFt @ STP /

Pounds of Air ("STP" is Standard Temperature and Presssure, 25degC and 1 atmosphere absolute)

My memorized form of magic_number_R is "22.4 Liter-Atmospheres per Mole" at standard t emperature. Truck drivers would probably prefer units: { PSIA, cubic feet, abs_F, and StandardCuFt "scf" }, Then your magic number would be 14.7 psia-cuft per scf. If it helps you could rename the letters in the formula, N and R in particular. Like PV=aQT. "a" for atmosphere=14.7psi and "Q" for quantity of gas in scf. The "a" never changes. The four capitals are all up to you, fixed or variable, as needed in whatever problem you are solving.

Do your guys learn the number of pounds of air mass in a standard cubic foot?

Jim Swenson

Dear Walt,

I solved this problem by using Charles' Law, which says that if the pressure is constant and the number of grams of air does not change,

(V2/V1) = (T2/T1)

where V2 is the final volume, V1 is the initial volume, T2 is the final temperature (in degrees Kelvin) and T1 is the initial temperature (in degrees Kelvin). The relationship between degrees F and degrees K is

(deg. K) = (5/9)*(deg. F - 32) + 273.15

Since 0 deg. F = 255.4 deg. K and 100 deg. F = 310.9 K,

I get V2 / V1 = 310.9 / 255.4 = 1.22, or V2 = 1.22 V1.

This says that the final volume will be 1.22 times the initial volume. So, if you start off with a cylinder that is 100 cc's, the final volume would be 122 cc's. On a percentage basis, that's a 22% increase in the volume.

I hope this helps. I do not think I have made any errors but do not hate me if I slipped a digit somewhere!

Dr. Topper

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