2: Thermodynamics Review

Example using the perfect gas relation:

Calculate the density of air at a pressure of 1 atmosphere and a temperature of 25 deg. C.

 

Solution:

Pressure: 1 atmosphere = 101300 Newtons per square meter

Temperature: 25 deg. Celsius=273.15 + 25 =298.15Kelvin

The thermal equation of state relates the pressure P, absolute temperature T, and density of a gas:

P = (rho) RT, where

R=Ru / MW,

Ru being the Universal Gas Constant (8314.3 in SI units ),and MW the molecular weight of the gas.Air is composed of 79% Nitrogen, and 21% Oxygen. The molecular weight of Nitrogen (N2) is 28, and that of Oxygen (O2) is 32. Thus the mean molecular weight is

MW= 0.79 * 28 + 0.21*32= 28.84

Thus, the gas constant for air, R = 8314.3 / 28.84 = 288.29 mK-1s-2

rho = 101300 / (298.15 * 288.29)=1.1785 kg/m3.

 

( Note: A more exact representation of air at sea level is: 79% nitrogen, 20% oxygen, and 1% argon (MW=44), giving a molecular weight of 28.96. This makes R = 287.04 mK-1s-2)

 

It is useful to remember that in SI units, atmospheric pressure at sea-level is approximately 100,000, temperature is 300, and density is 1.2.


 

Caloric Equations of State

For a calorically perfect gas, internal energy per unit mass depends only on temperature.

e = e(T)

 

Enthalpyper unit mass

h = e + p/r,

 

so that h = h(T)as well.

 

Specific Heats

Specific heat at constant volume

or

Specific heat at constant pressure

Now,

,and h = e + p/r,

so that

At constant pressure, dp = 0, so that dq = dh

or


 

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