
Equation
of State Answers
 To convert a Fahrenheit
temperature to an absolute temperature, add 459.69^{ }degrees.
(The absolute temperature scale in English units is known as the Rankine
scale and is expressed as ^{o}R.) Therefore our typical value
of 59 ^{o}F is 518.69^{o}R
 Now, let's algebraically
solve the equation of state, p = R r T, for R.
R = p/rT.
 We are ready to
calculate R, the gas constant for air, for typical values of pressure,
density, and temperature. Using the values with English units displayed
in the Air Properties Definitions slide, substitute into the equation
for R, remembering to use the temperature from #1 above.
R = ( ) / ( )( ). Compute R. R = 11.95 lbft^{2}/slugin^{2}^{o}R.
 As we go up in
altitude, the pressure, density, and temperature decrease. At an altitude
of 36,089 ft, p = 472.68 lb/ft^{2}, r = 7.06 x 10^{4}
slugft^{3}, and T = 390 ^{o}R.
How many miles high are we? 6.84 miles high.
 What is the air
temperature in ^{o}F at this altitude? 70
^{o}F
That's cold! (That is why the walls of an airplane are cold.)
 Notice that the
units of pressure at the 36,089 ft data set are in lb/ft^{2}.
To be consistent with our first calculation of R, we must convert to
lb/in^{2}. Therefore 472.68 lb/ft^{2}
= 3.28 lb/in^{2} Comparing this to the
typical value of 14.7 lb/in^{2}, we notice that we have only
about one fifth as much air! That would make it difficult to breathe;
therefore, airplanes are pressurized inside.
 Calculate R at
36,089 ft (using pressure from #6 above, and r and T in ^{o}R
from #4. R ~ 11.91. Compare
this to the R in #3. WOW! R is constant!
 Now that we have
found (and checked ) R, let's use it to determine the density, r, at
an altitude of 65,620 ft where p = 0.80 lb/in^{2} and T = 390^{o}R.
Solve the equation of state algebraically for r. r = ( ) / ( )( ). Then
substitute in R and the other parameters and compute r.
r
= 1.72 x 10^{4} slug/ft^{3}
