Properties of Gases
Important properties of a Gas
Quantity
n = moles
Volume
V = container size (usually L or mL)
Temperature
T ≈ average kinetic energy of molecules
(must be in K for all “gas laws”)
Pressure
P = force/area
Units of pressure: SI unit is the pascal (Pa)
• 1 atm = 101,325 Pa (not commonly used) = 14.7 psi
More important:
1 mm Hg = 1 torr
1 atm = 760 torr = 760 mm Hg
Measuring Pressure
Barometer
Manometer
Pressure - Volume - Temperature Relationships
• Boyle’s Law
(at constant T and n)
V ∝ 1/P
or
PV = constant
• Charles’ Law
(at constant P and n)
V∝T
or
V/T = constant
• Gay-Lussac’s Law (at constant V and n)
P∝T
or
P/T = constant
• Combined Gas Law (for constant n)
PV/T = constant
or
P1V1
T1
=
P2V2
T2
(remember that T must be in units of K -- practice problems in
book!)
Ideal Gas Law
• Avogadro’s Principle
– At constant P and T, V ∝ n
– i.e. at constant T and P, equal volumes of gases contain equal
numbers of moles
• The Ideal Gas Equation
PV = nRT
where R = “universal gas constant”
= 0.0821 L•atm/mole•K memorize!
{useful in many different kinds of calculations involving gases!}
• Standard Molar Volume
– At Standard Temperature and Pressure (0 °C and 1 atm), 1
mole of any gas occupies 22.4 L (i.e. 22.4 L/mol)
Example Problems
1.
2.
At STP, the density of a certain gas is 4.29 g/L. What is the
molecular mass of the gas?
(4.29 g/L) x (22.4 L/mol) = 96.1 g/mol
Acetylene (welding gas), C2H2, is produced by hydrolysis of
calcium carbide.
CaC2(s) + 2 H2O --> Ca(OH)2(s) + C2H2(g)
Starting with 50.0 g of CaC2, what is the theoretical yield of
acetylene in liters, collected at 24 °C and a pressure of 745 torr?
1st find yield in moles:
1 mole CaC2
1 mole C2H2
50.0 g CaC2 x
x
= 0.780 mol C2H2
64.1 g CaC2
1 mole CaC2
Now use ideal gas law to find volume of C2H2:
PV = nRT
V =
nRT
P
(0.780 mol) x (0.0821 L atm /mole K) x (297 K)
V =
(745 torr) x (1 atm/760 torr)
=
19.4 L
Dalton’s Law of Partial Pressures
• For a mixture of gases: Ptotal = Pa + Pb + Pc + …
• Mole fraction: Xa = moles a/total moles = Pa/Ptotal
• Gases are often prepared and collected over water:
Ptotal = Pgas + Pwater
where Pwater = vapor pressure of water (depends on temperature)
e.g. at 25 °C, Pwater = 23.8 torr
at 50 °C, Pwater = 92.5 torr
Example Problem
A sample of N2 gas was prepared and collected over water at
15 °C. The total pressure of the gas was 745 torr in a
volume of 310 mL. Calculate the mass of N2 in grams.
(vapor pressure of water at 15 °C = 12.8 torr)
Answer:
Ptotal = Pgas + Pwater
745 torr = Pgas + 12.8 torr
Pgas = 732 torr
PV = nRT
n=
n =
PV
RT
(732 torr) x (1 atm/760 torr) x (0.310 L)
(0.0821 L atm/mol K) x (288 K)
= 0.012624 mol N2
mass N2 = (0.012624 mol N2) x (28.014 g N2/mol N2) = 0.354 g N2
Kinetic Theory of Gases -- READ BOOK
Basic Postulate -- A gas consists of a very large number of
very small particles, in constant random motion, that
undergo perfectly elastic collisions with each other and the
container walls.
There is a distribution of kinetic energies of the particles.
Temp ∝ average KE
The kinetic theory “explains” the gas laws, pressure, etc.
based on motion and kinetic energy of gas molecules.
e.g. Boyle’s Law (P ∝ 1/V) ~ at constant Temp (same average KE)
If volume of container is reduced, there are more gas particles
per unit volume, thus, more collisions with the container walls
per unit area.
 higher pressure
Temperature & Molecular Velocities
• Kinetic molecular theory states that all particles have the
same average kinetic energy at a given temperature.
KE = ½mv2
• If m is smaller, v is bigger! i.e. small particles move faster.
Quantitatively,
u
where
u
rms
rms =
3RT
M
= root mean square velocity (a kind of average),
M = formula mass (in kg/mol!),
and R = universal gas constant, but in J/mol∙K rather than
L∙atm/mol∙K!
R = 8.314 J/mol∙K = 0.0821 L∙atm/mol∙K
Graham’s Law of Effusion
• diffusion
“mixing” of gases throughout a given volume
• effusion
“leaking” of a gas through a small opening
• mean free path average distance between collisions
Graham’s Law: effusion rate ∝ 1/√ M
mass
where M = formula
So, effusion rates of two gases can be compared as a
proportion:
rateA
rateB
=
MB
MA
e.g. He (FM = 4.0 g/mol) effuses 2 times faster than CH4 (FM =
16.0)
Real Gases -- Deviations from Ideal Gas Law
For real gases, small corrections can be made to account for:
– Actual volume of the gas particles themselves, and
– intermolecular attractive forces
One common approach is to use the Van der Waals’ Equation:
n
P+a
V
2
(V – nb) = nRT
Don’t memorize!
where a and b are empirical parameters that are dependent on
the specific gas (e.g. Table 5.5)
a ≈ intermolecular attractive forces
b ≈ molecular size
Sample Problems
Hydrogen gas is produced when metals such as aluminum are
treated with acids. Calculate the volume (in mL) of 0.500 M
HCl solution that is required to produce a total gas pressure
of 725 torr in a 2.50-L vessel if the hydrogen gas (H2) is
collected over water at 25 °C. (The vapor pressure of water
at 25 °C is 24 torr.)
2 Al(s) + 6 HCl(aq) --> 2 AlCl3(s) + 3 H2(g)
A gas mixture contains 25.0 g of CH4, 15.0 g of CO and 10.0 g
of H2. If the total pressure of the mixture is 1.00 atm, what is
the partial pressure of CH4 in torr?
Sample Problems
Hydrogen gas is produced when metals such as aluminum are
treated with acids. Calculate the volume (in mL) of 0.500 M
HCl solution that is required to produce a total gas pressure
of 725 torr in a 2.50-L vessel if the hydrogen gas (H2) is
collected over water at 25 °C. (The vapor pressure of water
at 25 °C is 24 torr.)
2 Al(s) + 6 HCl(aq) --> 2 AlCl3 + 3 H2(g)
Answer: 377 mL
A gas mixture contains 25.0 g of CH4, 15.0 g of CO and 10.0 g
of H2. If the total pressure of the mixture is 1.00 atm, what is
the partial pressure of CH4 in torr?
Answer: 168 torr
Chemistry in the Atmosphere
• Air Pollutants
– SOx
• e.g 2 SO2(g) + O2(g) + 2 H2O(g)  2 H2SO4(aq) acid rain
– NOx
• e.g. 4 NO2(g) + O2(g) + 2 H2O(g)  4 HNO3(aq) acid rain
– O3 (ozone)
– CO
– solid particles
• Ozone Layer
– Stratospheric ozone
≠ ground-level ozone
– CFC’s produce Cl, and
• O3(g) + UV light  O2(g) + O(g)
• Cl(g) + O3(g)  ClO(g) + O2(g)
• ClO(g) + O(g)  Cl(g) + O2(g)
– Freons are being replaced
by other less harmful refrigerants.