THE IDIAL GAS LAW
13.1 The ideal gas law:
¾
Ideal gas low is an equation relating p,V,n, and T that applies to many
gases at low density( high temperature and/or low pressure)
pV = nRT
where p = absolute pressure of the gas
V = total volume occupied by the gas
n = number of moles of the gas
R = ideal gas constant in appropriate units
T = absolute temperature of the gas
THE IDIAL GAS LAW
¾ Sometimes the ideal gas law is written as:
where in the equation
of the gas.
v̂ is the specific molar volume (volume per mole)
THE IDIAL GAS LAW
EXAMPLE: Calculate the volume, in cubic meters, occupied by 40 kg of CO2
at standard conditions assuming CO2 acts as an ideal gas.
Solution
Basis: 40 kg of CO2
At standard conditions:
= 22.415m3/kg mol
THE IDIAL GAS LAW
¾
Calculation of R Using the Standard Conditions:
Example: Find the value for the universal gas constant R to match the
following combination of units: For 1 g mol of ideal gas when the
pressure is in atm, the volume is in cm3, and the temperature is in K.
Solution
pV
(cm 3 )(atm)
1(atm) 22,415(cm 3 )
R=
=
= 82.06
T
273.15(K ) 1(g mol)
(K)(g mol)
THE IDIAL GAS LAW
¾
Application of the Ideal Gas Law: In processes going from initial state
to a final state:
or
where subscript 1 designates the initial state, and the subscript 2 designates the
final state)
Example: Calculate the volume occupied by 88 Ib of CO2 at 15°C and a
pressure of 32.2 ft of water.
Solution
Basis: 88 lb of CO2
THE IDIAL GAS LAW
¾
Volumetric flow rate of a gas through a pipe:
•
•
V = Aν
OR
ν = V/A
(m3/s or ft3/s)
THE IDIAL GAS LAW
¾
The density of a gas is defined as the mass per unit volume and can
be expressed in kilograms per cubic meter, pounds per cubic foot,
grams per liter,
¾
The specific gravity of a gas is usually defined as the ratio of the
density of the gas at a desired temperature and pressure to that of
air (or any specified reference gas) at a certain temperature and
pressure.
THE IDIAL GAS LAW
¾
For a gas and reference at the same temperature and pressure, the
specific gravity is just the ratio of the respective molecular weights
Where A is one gas and B is another.
THE IDIAL GAS LAW
EXAMPLE: What is the density of N2 at 27oC and 100 kPa in SI units?
Solution:
100(kPa ) × 28(kg/kg mol)
n.MW p.MW
3
ρ=
=
=
=
1
.
23
kg/m
8.314(kPa)(m 3 )(kg mol)(K) × (300)(K)
V
RT
THE IDIAL GAS LAW
13.2 ldeal Gas Mixtures and Partial Pressure
¾
The partial pressure of Dalton, pi, namely the pressure that would be
exerted by a single component in a gaseous mixture if it existed alone in
the same volume as that occupied by the mixture and at the same
temperature of the mixture, is defined by
where pi is the partial pressure of component i in the mixture.
THE IDIAL GAS LAW
Where yi is the mole fraction of component i.
Example: In air the percent of oxygen is
20.95, hence at the standard conditions of
one atmosphere, the partial pressure of
oxygen is:
pO = 0.21(1) = 0.21 atm
2
THE IDIAL GAS LAW
Example: Few organisms are able to grow in solution using organic
compounds that contain just one carbon atom such as methane or
methanol. However, the bacterium methylococcus capsulates can grow
under aerobic conditions (in the presence of air) on C-1 carbon
compounds. The resulting biomass is a good protein source that can be
used directly as feed for domestic animals or fish. In one process the offflue gas analyzes 14.0% CO2, 6.0% O2, and 80.0% N2. It is at 400°F and
765.0 mmHg pressure. Calculate the partial pressure of each
component.
Solution
Basis: 1 .00 kg (or Ib) mol flue gas
Use Equation: pi = ptotal yi
THE IDIAL GAS LAW
14
(
) × (765)
100
THE IDIAL GAS LAW
13.3 Material Balances Involving Ideal Gases
¾
The only difference between the subject matter of Chapters 6 through
12 and this chapter is that here the amount of material can be specified
in terms of p, V, and T rather than solely mass or moles.
Example: Gas at 15°C and 105 kPa is flowing through an irregular duct. To
determine the rate of flow of the gas, CO2 from a tank is passed into the
gas stream. The gas analyzes 1.2% CO2 by volume before and 3.4%
CO2 by volume after the addition. As the CO2 that was injected left the
tank, it was passed through a rotameter, and found to flow at the rate of
0.0917 m3/min at 7°C and 131 kPa. What was the rate of flow of the
entering gas in the duct in cubic meters per minute?
THE IDIAL GAS LAW
Solution
Steps 1,2,3, and 4
This is an open, steady-state system without reaction.
, 131 kPa
THE IDIAL GAS LAW
Step 5
Basis: 1 min (0.0917 m3 of CO2 at 7°C and 131 kPa)
Steps 6 and 7
Unknowns: F and P,
two independent component balances, CO2 and other.
Zero degree of freedom.
THE IDIAL GAS LAW
Steps 7,8, and 9
CO2 balance (in m3 at 15°C and 105 kPa):
F(0.012) + 0.1177 = P(0.034)
"Other" balance (in m3 at 15°C and 105 kPa):
F(0.988) = P(0.966)
F = 5.17 m3/min at 15oC and 105 kPa