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Gas Laws
M. L. Watson
Ideal Gas Law
An ideal gas is defined as one in which all collisions between atoms or
molecules are perfectly elastic and in which there are no intermolecular
attractive forces. One can visualize it as a collection of perfectly hard
spheres which collide but which otherwise do not interact with each other.
In such a gas, all the internal energy is in the form of kinetic energy and
any change in internal energy is accompanied by a change in temperature.
An ideal gas can be characterized by three state variables: absolute
pressure (P), volume (V), and absolute temperature (T). The relationship
between them may be deduced from kinetic theory and is called the
Where:
n = number of moles
R = universal gas constant = 8.3145 J/mol K
N = number of molecules
k = Boltzmann constant = 1.38066 x 10-23 J/K = 8.617385 x 10-5 eV/K
k = R/NA
NA = Avogadro's number = 6.0221 x 1023
Ideal Gas Law
PV=nRTZ
n = Mass
R = Universal gas constant
T = Temperature
Z = Supercompressability
(P1V1/T1)Z1=(P2V2/T2)Z2
Ideal Gas Law
The ideal gas law can be viewed as arising from
the kinetic pressure of gas molecules colliding
with the walls of a container in accordance with
Newton's laws. But there is also a statistical
element in the determination of the average
kinetic energy of those molecules. The
temperature is taken to be proportional to this
average kinetic energy; this invokes the idea of
kinetic temperature. One mole of an ideal gas at
STP occupies 22.4 liters.
Ideal Gas Law
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An Ideal Gas (perfect gas)is one which obeys Boyle's Law and Charles' Law exactly.
An Ideal Gas obeys the Ideal Gas Law (General gas equation):
PV = nRT
where P=pressure, V=volume, n=moles of gas, T=temperature, R is the gas constant which
is dependent on the units of pressure, temperature and volume
R = 8.314 J K-1 mol-1 if Pressure is in kilopascals(kPa), Volume is in litres(L), Temperature
is in Kelvin(K)
R = 0.0821 L atm K-1 mol-1 if Pressure is in atmospheres(atm), Volume is in litres(L),
Temperature is in Kelvin(K)
An Ideal Gas is modelled on the Kinetic Theory of Gases which has 4 basic postulates
Gases consist of small particles (molecules) which are in continuous random motion
The volume of the molecules present is negligible compared to the total volume occupied by
the gas
Intermolecular forces are negligible
Pressure is due to the gas molecules colliding with the walls of the container
Real Gases deviate from Ideal Gas Behaviour because
at low temperatures the gas molecules have less kinetic energy (move around less) so they
do attract each other
at high pressures the gas molecules are forced closer together so that the volume of the gas
molecules becomes significant compared to the volume the gas occupies
Under ordinary conditions, deviations from Ideal Gas behaviour are so slight that they can
be neglected
A gas which deviates from Ideal Gas behaviour is called a non-ideal gas.
There are many applications to the
Ideal Gas Law Equation
How can the ideal gas law be applied in dealing with how gases behave?
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PV = nRT
Used to derive the individual ideal gas laws:
For two sets of conditions: initial and final set of conditions:
P1V1 = n1RT1 and P2V2 = n2RT2
Solving for R in both equations gives:
R = P1V1 / n1T1 and R = P2V2 / n2T2
Since they are equal to the same constant, R, they are equal to each other:
P1V1 / n1T1 = P2V2 / n2T2
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For the Volume Pressure relationship (ie: Boyle's Law):
n1 = n2 and T1 = T2 therefore the n's and T's cancel in the above expression resulting in the following simplification:
P1V1 = P2V2 (mathematical expression of Boyle's Law)
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For the Volume Temperature relationship (ie: Charles's Law):
n1 = n2 and P1 = P2 therefore the n's and the P's cancel in the original expression resulting in the following simplification:
V1 / T1 = V2 / T2 (mathematical expression of Charles's Law)
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For the Pressure Temperature Relationship (ie: Gay-Lussac's Law):
n1 = n2 and V1 = V2 therefore the n's and the V's cancel in the above original expression:
P1 / T1 = P2 / T2 (math expression of Gay Lussac's Law)
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For the Volume mole relationship (Avagadro's Law)
P1 = P2 and T1 = T2 therefore the P's and T's cancel in the above original expression:
V1 / n1 = V2 / n2 (math expression for Avagadro's Law)
Used to solve single set of conditions type of gas problems where there is no observable change in the four variables of a gas sample.
Knowing three of the four variables allows you to determine the fourth variable. Since the universal Gas Law constant, R, is involved in the
computation of these kinds of problems, then the value of R will set the units for the variables.
Ideal Gas Law w/Constraints
For the purpose of calculations, it is convenient to place the ideal gas
law in the form:
where the subscripts i and f refer to the initial and final states of some
process. If the temperature is constrained to be constant, this
becomes:
which is referred to as Boyle's Law.
If the pressure is constant, then the ideal gas law takes the form
which has been historically called Charles' Law. It is appropriate for
experiments performed in the presence of a constant atmospheric
pressure.
Boyle’s Law
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At constant temperature, the volume of a given quantity of gas is inversely
proportional to its pressure : V 1/P
So at constant temperature, if the volume of a gas is doubled, its pressure is
halved.
OR
At constant temperature for a given quantity of gas, the product of its
volume and its pressure is a constant : PV = constant, PV = k
At constant temperature for a given quantity of gas : PiVi = PfVf
where Pi is the initial (original) pressure, Vi is its initial (original) volume, Pf
is its final pressure, Vf is its final volume
Pi and Pf must be in the same units of measurement (eg, both in
atmospheres), Vi and Vf must be in the same units of measurement (eg,
both in litres).
All gases approximate Boyle's Law at high temperatures and low pressures.
A hypothetical gas which obeys Boyle's Law at all temperatures and
pressures is called an Ideal Gas. A Real Gas is one which approaches
Boyle's Law behaviour as the temperature is raised or the pressure lowered.
Boyle’s Law
P1V1=P2V2
Charles Law
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At constant pressure, the volume of a given quantity of gas is directly proportional to the absolute
temperature : V T (in Kelvin)
So at constant pressure, if the temperature (K) is doubled, the volume of gas is also doubled.
OR
At constant pressure for a given quantity of gas, the ratio of its volume and the absolute
temperature is a constant : V/T = constant, V/T = k
At constant pressure for a given quantity of gas : Vi/Ti = Vf/Tf
where Vi is the initial (original) volume, Ti is its initial (original) temperature (in Kelvin), Vf is its final
volme, Tf is its final tempeature (in Kelvin)
Vi and Vf must be in the same units of measurement (eg, both in litres), Ti and Tf must be in
Kelvin NOT celsius.
temperature in kelvin = temperature in celsius + 273 (approximately)
All gases approximate Charles' Law at high temperatures and low pressures. A hypothetical gas
which obeys Charles' Law at all temperatures and pressures is called an Ideal Gas. A Real Gas is
one which approaches Charles' Law as the temperature is raised or the pressure lowered.
As a Real Gas is cooled at constant pressure from a point well above its condensation point, its
volume begins to increase linearly. As the temperature approaches the gases condensation point,
the line begins to curve (usually downward) so there is a marked deviation from Ideal Gas
behaviour close to the condensation point. Once the gas condenses to a liquid it is no longer a
gas and so does not obey Charles' Law at all.
Absolute zero (0K, -273oC approximately) is the temperature at which the volume of a gas would
become zero if it did not condense and if it behaved ideally down to that temperature.
Charles Law
V1/V2=T1/T2
P1V1/T1=P2V2/T2
Supercompressability
Why? Molecular Squish
Theoretical
Actual
- under 1% for pressures under 1500 PSI
- different for each gas mixture
- similar to alcohol & water or sand & gravel
Isentropic Expansion
Thermodynamics!
P1V1k=P2V2k
k=(Cp/Cv)
k=Ratio of specific heat at a constant pressure
to the specific heat at a constant volume
K=1.3 for natural gas
Specific Heat = kJ/kg
All the possible states of an
ideal gas can be represented by
a PvT surface as illustrated. The
behavior when any one of the
three state variables is held
constant is also shown.
Molecular Constants
In the kinetic theory of gases, there are certain constants which
constrain the ceaseless molecular activity. A given volume V of any
ideal gas will have the same number of molecules. The mass of the
gas will then be proportional to the molecular mass. A convenient
standard quantity is the mole, the mass of gas in grams equal to the
molecular mass in amu. Avogadro's number is the number of
molecules in a mole of any molecular substance.
The average translational kinetic energy of any kind of
molecule in an ideal gas is given by
State Variables
A state variable is a precisely measurable physical property which
characterizes the state of a system, independently of how the system was
brought to that state. It must be inherently single-valued to characterize a
state. For example in the heat-work example, the final state is characterized
by a specific temperature (a state variable) regardless of whether it was
brought to that state by heating, or by having work done on it, or both.
Common examples of state variables are the pressure P, volume V, and
temperature T. In the ideal gas law, the state of n moles of gas is precisely
determined by these three state variables. If a property, e.g., enthalpy H, is
defined as a combination of other state variables, then it too is a state
variable. Enthalpy is one of the four "thermodynamic potentials", and the
other three, internal energy U, Helmholtz free energy F and Gibbs free
energy G are also state variables. The entropy S is also a state variable.
Some texts just use the term "thermodynamic variable" instead of the
description "state variable".
The Mole
A mole (abbreviated mol) of a pure substance is a mass of the material
in grams that is numerically equal to the molecular mass in atomic
mass units (amu). A mole of any material will contain Avogadro's
number of molecules. For example, carbon has an atomic mass of
exactly 12.0 atomic mass units -- a mole of carbon is therefore 12
grams. For an isotope of a pure element, the mass number A is
approximately equal to the mass in amu. The accurate masses of
pure elements with their normal isotopic concentrations can be
obtained from the periodic table.
One mole of an ideal gas will occupy a volume of 22.4 liters at STP
(Standard Temperature and Pressure, 0°C and one atmosphere
pressure).
Avogadro's number
Standard Temperature and
Pressure
STP is used widely as a standard reference point for
expression of the properties and processes of ideal
gases. The standard temperature is the freezing point of
water and the standard pressure is one standard
atmosphere. These can be quantified as follows:
Pressure ( P ) is the ratio of the force applied to a surface (F) to the surface area ( A ).
P=F/A
Standard temperature: 0°C = 273.15 K = 32 F
Standard pressure = 1 atmosphere = 760 mmHg = 101.3 kPa =14.696 psi
Standard volume of 1 mole of an ideal gas at STP: 22.4 liters
Gauge Pressure
Does the flat tire on your automobile have zero air pressure? If it is completely
flat, it still has the atmospheric pressure air in it. To be sure, it has zero
useful pressure in it, and your tire gauge would read zero pounds per
square inch. Most gauges read the excess of pressure over atmospheric
pressure and this excess is called "gauge pressure". While a useful
measurement for many practical purposes, it must be converted to absolute
pressure for applications like the ideal gas law.
Since a partial vacuum will be below atmospheric pressure, the phrase
"negative pressure" is often used. Certainly there is no such thing as a
negative absolute pressure, but small decreases in pressure are commonly
used to entrain fluids in sprayers, in carburetors for automobiles, and many
other applications. In the case of respiration, we say that the lungs produce
a negative pressure of about -4 mmHg to take in air, which of course means
a 4 mmHg decrease from the surrounding atmospheric pressure.
Example
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Here is a sample problem involving a single set of conditions:
A sample of chlorine gas weighs 1.31 g at STP. Calculate the volume this
sample of chlorine would occupy under the following new conditions.
3.20atm and 0.00 C
Here are the steps that will lead to an answer:
– Calculate the moles of Cl2 from 1.31 grams Cl2 using the formula weight of 2 x
at weight of Cl
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1.31 grams Cl2 X 1 mole Cl2 / 71 grams Cl2 = 0.0184 moles Cl2
– Check the temperature and convert to Kelvin if necessary:
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K = C + 273 = 0.00 + 273 = 273 K
– Check the pressure given and convert to atmospheres unit
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Pressure is already in atmospheres, 3.20 atm
– Use the value of R = 0.0821 liter-atm/mole-K
– Using the PV = nRT plug in the moles, temperature, pressure, and R and solve
for the Volume in liters
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V = nRT / P = (0.0184 moles) ( 0.0821 liter-atm / mol-K) (273 K) / 3.20 atm
= 0.129 liters
Example
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Let's let you try one:
A sample of chlorine gas weighs 1.31 g at STP. Calculate the volume this sample of
chlorine would occupy under the following new conditions.
760 torr and -23.0*C
Use the Ideal Gas Law Equation to solve Gas stoichiometry problems involving
conversion of moles of gas to volume or the conversion of a given volume of a gas
sample under a stated temperature and pressure to moles or mass of another
component in the process under a stated temperature and pressure.
Use of Ideal Gas Law Equation to determine the density of a gas.
PV = nRT if n = mass of gas / molecular mass
then PV = (mass of gas/molecular mass) R T Solving for molecular mass
molecular mass = (mass of gas) R T / P V
Since Density = mass / V Then:
molecular mass = D R T / P
and solving for Density: D = (molecular mass) (P) / R T
Using the equation to solve for Partial Pressure of a known amount of gas in a gas
mixture:
P1 = n1 R T / V and P2 = n2 R T / V
Example
• Ideal Gas Law Calculations
• Calculating Volume of Ideal Gas: V = (nRT) ÷ P
• What volume is needed to store 0.050 moles of
helium gas at 202.6kPa and 400K? PV = nRT
• P = 202.6 kPa
n = 0.050 mol
T = 400K
V=?L
R = 8.314 J K-1 mol-1 202.6V=0.050x8.314x400
202.6 V = 166.28
V = 166.28 ÷ 202.6
V = 0.821 L (821mL)
Notes
Notes
Notes