Gases Chapter 9
What parameters do we use to describe gases?
pressure: force/unit area 1 atm = 101 kPa;
volume: liters (L)
Temperature: K
Amount of gas: moles
A gas consists of small particles that move rapidly in straight lines
until they collide;
they have enough kinetic energy to overcome any attractive forces;
gas molecules are very far apart;
gas molecules have very small volumes compared to the volumes of
the containers they occupy;
have kinetic energies that increase with an increase in temperature;
collisions of the gas cause pressure (force /unit area)
What is meant by % volume? In principle, the actual molecular
volume of a gas is so small in comparison to the volume it will
occupy that we treat gases at mathematical points.
What is air pressure due to?
A column of air 1 m2 has a mass
of 10,300 kg, producing a
pressure of 101 kPa due to
gravity (14.7 pounds/in2)
1 atm = 76 cm Hg; 101 kPa
Atmospheric pressure
is the pressure exerted by a column of air from
the top of the atmosphere to the surface of the
Earth;
Pressure is:
about 1 atmosphere at sea level;
depends on the altitude and the weather;
is lower at high altitudes where the density of
air is less;
How do we measure pressure?
A barometer
measures the pressure exerted by the gases in
the atmosphere
indicates atmospheric pressure as the height in
mm of the mercury column
A water barometer would be 13.6 times taller
than a mercury barometer because the density
of Hg is 13.6 times as dense as water
• What is the relationship between the
variables used to describe gases?
Boyle’s Law is concerned with the relationship of pressure and
volume using a fixed amount of gas ( a fixed number of
mols of gas)
P*V = constant at constant temperature
Initial
Note that the initial product between
pressure and volume is 4L*1atm = 4 L atm
In the final diagram: 2L*2atm = 4 Latm
Decreasing the volume to ½ results in a
doubling of the pressure
We find that the product of volume and
pressure is equal to a constant
PV = a constant
Final
Temperature and Volume
Suppose we had a frictionless piston and a gas
enclosed within the piston. What could we say
about the pressure inside and outside of the piston if
the piston was notionless?
pressure P (inside) = P (outside)
temperature (inside) = temperature (outside)
Supppose we now heat the gas inside the piston
What happens to the number of collisions a
molecule makes against the wall if we heat
a gas?
the number of collisions increase causing
an increase in the pressure inside
What will happen to the volume of the gas
if the pressure outside the piston does not
change and the pressure inside increases?
V/T(K) = constant
Charles’ Law is concerned with the relationship of temperature and
volume when dealing with a constant amount of gas (mols)
V ∝ T when T is expressed in K. The K temperature scale is derived
from the behavior of gases
if V ∝ T then V = kT where k is a constant at constant pressure
Suppose now that we double the
number of molecules in the same
volume and at the same
temperature. What will that do to
the number of collisions with the
walls?
the pressure?
What will it do to the volume if we
allow the pressure to return to its
orginal value?
If we allow the
pressure
to return to itwith
Avogadro’s
Law
is concerned
value, what
will happen
theoriginal
relationship
between
theto the
volume?
number of molecules or mols (n)
and the volume of a gas under
n is directly
to pressure and
conditions
of proportional
constant pressure
temperature
Pressure and the amount of gas , n
mols is proportional to
volume if T and P is
kept constant
Summary
Ideal gas law: PV = nRT where R is a constant
R = 0.0821 L.atm/K.mol
Note that at constant n and T, PV = constant
Boyle’s Law
Note that at constant P and T, V/n = constant
Avogadro’s Law
Note that at constant P and n, V/T = constant
Charles’s Law
Standard conditions of pressure and temperature
T = 0 °C (273 K)
Pressure: 1 atm
What volume does a mol of any ideal gas occupy at STP?
PV = nRT
V = 1mol(0.0821 L*atm/K*mol)(273 K)/(1 atm)
V = 22.4 L
This means that equal volumes of gases under identical
conditions of temperature and pressure contain equal number
of molecules
What is the difference between an ideal gas and a real gas?
The ideal gas equation was generated from the kinetic theory of gases
making the following assumptions or approximations:
1. The molecules could be treated as points (ie molecular volume = 0)
2. There are no attractive interactions between molecules.
3. Gas particles move around at random
4. Collision of gas molecules with the wall are totally elastic
5. The kinetic energy of the gas particle is ∝ to temperature (K)
In general, the ideal gas law works best at low pressures and high
temperatures
Real Gases: van der Waal’s equation
(P + an2/V2)(V-nb) = n RT
an2/V2 corrects for intermolecular attractions
nb corrects for the real volume of molecules
Dalton’s Law of partial pressures:
Total atmospheric pressure = 1 atm;
How much of the pressure is contributed by N2?
Pressure is a consequence of molecules colliding with each
other and the walls of the container
*062
For air if
If PTV = nTRT
and nT = (nO2 + n N2 + ...)
at constant T, PTV = (no2 + n N2 + ...)RT
Since the actual volume of the molecules is small in comparison
to the volume occupied by the gas, all molecule occupy the same
volume V.
The contribution to the total pressure is dependent on the number
of collision of each gas with the wall and this is dependent on the
number of molecule of each gas. Hence:
P = (PN2 + PO2 + ...)
PO2V = nO2RT ; PN2V = nN2RT ...
Temperature: a measure of the average kinetic energy of
molecules
Distribution of molecular speeds as a function of temperature
What are some of the consequences associated with the fact that
molecules at the same temperature have different speeds?
Diffusion: mixing of gases at
the same pressure
Effusion: escape
through a small opening
The size of the pinhole needs to be small for effusion
*07
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Two molecules of different mass at the same
temperature effusing through an opening
If M > m, then :
1/2Mv2 = 1/2mV2
V>v
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Two molecules of different mass at the same
temperature effusing through an opening
From the kinetic theory of gases
speed of a molecule v = (3RT/M)1/2
For two gases at the same temperature
1/2mava2 = 1/2mbvb2
va = average speed of molecule a
vb = average speed of molecule b
ma /mb = vb2 /va2
or
vb /va = (ma /mb )1/2
The rate at which molecule a hits the pinhole ∝ v if the comparisons
are made at the same concentration and temperature. If you wait
long enough ...
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Two molecules of different mass at the same
temperature effusing through an opening at
equilibrium
Solving some problems involving gases
1. A sample of gas at 25 °C and 2 atm pressure in a 5 L vessel was
found to have a mass of 18 g. What is its molecular weight?
PV = n RT
2 atm*5 L = n*0.0821 (Latm/K mol)*298 K
n = 10/(0.0821*298) mol; 0.4087
n = wt/ mw;
mw = 44 g/mol
0.4087 = 18g/mw
Suppose the gas at the right
exerted a pressure of 15 cm as
shown.
Would the pressure of the gas
be greater or less than 1 atm?
15.2 cm
Hg
How many atm of pressure is
the gas exerting?
1 atm = 76 cm
76 - 15.2 = 60.8
Suppose the gas at the right
exerted a pressure of 15 cm as
shown.
Would the pressure of the gas
be greater or less than 1 atm?
15.2 cm
How many atm of pressure is
the gas exerting?
1 atm = 76 cm
76 - 15.2 = 60.8
Suppose we have a sample of equal amounts of H2 and D2 in a
vessel and a small opening is introduced. What will be the initial
rates of effusion?
vH2/vD2 = (mD2/mH2)0.5 = (4/2)0.5 = 1.42
Will the relative rate change with time?
What is the density of natural gas (CH4) at STP?
PV = nRT
density is g/mL or g/L
We know the molar volume of any gas is 22.4 L at STP
How many g of methane in a mole?
16g/22.4 L = 0.714 g/l or 7.14*10-4 g/mL or
PV =(wt/mw)*RT;
mw*P/RT = (wt/V)
The surface temperature of Venus is about 1050 K and the pressure
is about 75 Earth atmospheres. Assuming these conditions
represent a “Venusian STP, what is the standard molar volume of a
gas on Venus?
PV = nRT
75 atmV =1mol*0.0821(Latm/K mol)*1050 K; V = 1.15 L
Natural gas is a mixture of a number of substances including
methane (mol fraction, 0.94); ethane (mol fraction, 0.04); propane
(mol fraction, 0.015). If the total pressure of the gases is 1.5 atm,
calculate the actual pressure contributed by each of the gases
described.
mol fraction = mol A/(mol A + mol B + ....)
PT = 1.5 = P CH4+ PC2H6 + ...
Px V = nxRT
CH4 = 0.94*1.5
C2H6 = 0.04*1.5
C3H8 = 0.015*1.5
nCH4/nC2H6 = PCH4/PC2H6 = 0.94/.04