11.2 Measurement of Temperature and Its Units 313
(11.8)
Equation (11.8) establishes a relationship between the temperature of the gas, its pressure, and
the reference pressure and temperature. If we were to proceed by lowering the surrounding
temperature, a lower gas pressure would result, and if we were to extrapolate the results of
our experiments, we would find that we eventually reach zero pressure at zero temperature.
This temperature is called theabsolute thermodynamic temperatureand is related to the Celsius
and Fahrenheit scales. The relationship between the Kelvin (K) and degree Celsius (C) in
SI units is(11.9)
The relationship between degree Rankine (R) and degree Fahrenheit (F) in U.S. Customary
units is(11.10)
Note that the experiment cannot be completely carried out, because as the temperature of
the gas decreases, it reaches a point where it will liquify and thus the ideal gas law will not be
valid. This is the reason for extrapolating the result to obtain a theoretical absolute zero tem-
perature. It is also important to note that while there is a limit as to how cold something can
be, there is no theoretical upper limit as to how hot something can be. Finally, we can also
establish a relationship between the degree Rankine and the Kelvin by(11.11)
In Equations (11.9) and (11.10), unless you are dealing with very precise experiments,
when converting from degree Celsius to Kelvin, you can round down the 273.15 to 273.
The same is true when converting from degree Fahrenheit to Rankine; round up the 459.67
to 460.Example 11.1 What is the equivalent value ofT 50 C in degrees Fahrenheit, Rankine, and Kelvin?
We can use Equations (11.2), (11.9), and (11.10):Note that we also could have converted the 582R to Kelvin directly using Equation (11.11):
T 1 K 2
5
9
T 1 °R 2 a
5
9
b1 5822 323 K
T 1 K 2 T 1 °C 2 273 50 273 323 K
T 1 °R 2 T 1 °F 2 460 122 460 582°R
T 1 °F 2
9
5
T 1 °C 2 32 a
9
5
b1 502 32 122°F
T 1 K 2
5
9
T 1 °R 2
T 1 °R 2 T 1 °F 2 459.67
T 1 K 2 T 1 °C 2 273.15
T 1 Tra
P 1
Pr
b
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