**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

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

P
= (rho) RT,
where

R=R_{u }/ MW,

R_{u
}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 (N_{2})
is 28, and that of Oxygen (O_{2}) 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^{-1}s^{-2}

rho = 101300 / (298.15 *
288.29)=**1.1785 kg/m ^{3}.
**

** **

*( 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 ^{-1}s^{-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

_{}