The KineticMolecular Theory
(KEY POINTS ARE IN RED)
1. Gases consist of large numbers of molecules (or atoms, in the case of the noble gases) that are in continuous, random motion.
Usually there is a great distance between each other, so the molecules travel in straight lines between abrupt collisions at the walls and between each other.
These collisions randomize the motion of the molecules.
Most of the collisions between molecules are binary, in that only two molecules are involved.
2. The volume of the molecules of the gas is negligible compared to the total volume in which the gas is contained.
A common bond length between atoms is about 10^{10} m or 1 Angstrom.
Small molecules are therefore on the order of 10 Angstroms in diameter, or less than 10^{24} Liters in Molecular Volume, quite tiny indeed!
Remember, however that there can be a great many molecules in the sample of gas, perhaps on the order of a mole, or 6 x 10^{23}.
So that when concentrations of molecules exceed about 1 mol/liter, then the approximation that the volume of ALL the molecules in the container is much less than the volume of the container itself, fails.
In the case of an ideal gas, we will assume that molecules are point masses, i.e., the volume of a mole of gas molecules (as if they were at rest) is zero, so molecular and container volumes never become comparable.
3. Attractive forces between gas molecules are negligible.
We know that if these forces were significant, the molecules would stick together.
This happens when it rains and gaseous water molecules stick together to form a liquid.
Water vapor is a condensible gas, and this shows us that gas molecules are sticky, but at a high enough temperature they form only a permanent gas, because their stickiness can be considered negligible.
We will assume that in an ideal gas, molecular attractive forces are not just small, but identically zero.
Consequences:
Pressure is force per unit area, calculated by dividing the force by the area on which the force acts.
The earth's gravity acts on air molecules to create a force, that of the air pushing on the earth. This is called atmospheric pressure.
The units of pressure that are used are pascal (Pa), standard atmosphere (atm), and torr.
1 atm is the average pressure at sea level. It is normally used as a standard unit of pressure.
The SI unit though, is the pascal. 101,325 pascals equals 1 atm.
For laboratory work the atmosphere is very large. A more convient unit is the torr. 760 torr equals 1 atm.
A torr is the same unit as the mmHg (millimeter of mercury).
It is the pressure that is needed to raise a tube of mercury 1 millimeter.
The Molecular picture of Pressure
Check out this simulation of molecules in motion in a gas.
Initial Definitions
Temperature is a number that is related to the average kinetic energy of the molecules of a substance.
If temperature is measured in Kelvin degrees, then this number is directly proportional to the average kinetic energy of the molecules.
Heat is a measurement of the total energy in a substance.
That total energy is made up of not only of the kinetic energies of the molecules of the substance, but total energy is also made up of the potential energies of the molecules.
Celsius
Anders Celsius (17011744) was a Swedish astronomer credited with the invention of the centigrade scale in 1742.
Celsius chose the melting point of ice and the boiling point of water as his two reference temperatures to provide for a simple and consistent method of thermometer calibration.
Celsius divided the difference in temperature between the freezing and boiling points of water into 100 degrees (thus the name centi, meaning one hundred, and grade, meaning degrees).
After Celsius's death, the centigrade scale was renamed the Celsius scale and the freezing point of water was set at 0°C and the boiling point of water at 100°C.
The Celsius scale takes precedence over the Fahrenheit scale in scientific research because it is more compatible with the base ten format of the International System (SI) of metric measurement (see our The Metric System module).
In addition, the Celsius temperature scale is commonly used in most countries of the world other than the United States.
Kelvin
William Kelvin (18241907) was a Scottish physicist who devised the Kelvin (K) scale in 1854.
The Kelvin scale is based on the idea of absolute zero, the theoretical temperature at which all molecular motion stops and no discernable energy can be detected (see our Matter: States of Matter module for more information).
In theory, the zero point on the Kelvin scale is the lowest possible temperature that exists in the universe: 273.15ºC.
The Kelvin scale uses the same unit of division as the Celsius scale; however, it resets the zero point to absolute zero: 273.15ºC.
The freezing point of water is therefore 273.15 Kelvins (graduations are called Kelvins on the scale and neither the term degree nor the symbol º are used) and 373.15 K is the boiling point of water.
The Kelvin scale, like the Celsius scale, is a standard SI unit of measurement used commonly in scientific measurements.
Since there are no negative numbers on the Kelvin scale (because theoretically nothing can be colder than absolute zero), it is very convenient to use Kelvins when measuring extremely low temperatures in scientific research.
Comparison of three different temperature scales 
Although it may seem confusing, each of the three temperature scales discussed allows us to measure heat energy in a slightly different way.
A temperature measurement in any of the three scales can be easily converted to another scale using the simple formulas below.
From  to Fahrenheit  to Celsius  to Kelvin 
ºF  (ºF  32)/1.8  (ºF32)*5/9+273.15  
ºC  C  ºC + 273.15  
K  K  273.15  K 
Microscopic equilibrium between gas and liquid at low temperature. Note the small number of particles in the gas.  Microscopic equilibrium between gas and liquid at high temperature. Note the large number of particles in the gas. 
Molecular Speed, Averaging, and The MaxwellBoltzmann Speed Distribution
The following graph shows the Distribution of Speeds for Gases, i.e.the fraction of the sample of gas molecules that have a given speed is shown by the height of the curve above the speed axis.
There are no molecules exactly at rest.
Volume is how much threedimensional space a substance (solid, liquid, gas, or plasma) or shape occupies or contains,^{ }often quantified numerically using the SI derived unit, the cubic meter.
The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.
Three dimensional mathematical shapes are also assigned volumes.
Volumes of some simple shapes, such as regular, straightedged, and circular shapes can be easily calculated using arithmetic formulas.
The volumes of more complicated shapes can be calculated by integral calculus if a formula exists for the shape's boundary.
Onedimensional figures (such as lines) and twodimensional shapes (such as squares) are assigned zero volume in the threedimensional space.
When a solid or a liquid evaporates to a gas in a closed container, the molecules cannot escape.
Some of the gas molecules will eventually strike the condensed phase and condense back into it.
When the rate of condensation of the gas becomes equal to the rate of evaporation of the liquid or solid, the amount of gas, liquid and/or solid no longer changes.
The gas in the container is in equilibrium with the liquid or solid.
Microscopic equilibrium between gas and liquid. Note that the rate of evaporation of the liquid is equal to the rate of condensation of the gas.  Microscopic equilibrium between gas and solid. Note that the rate of evaporation of the solid is equal to the rate of condensation of the gas. 
Boyle's law or the pressurevolume law states that the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant.
Another way to describing it is saying that their products are constant.
PV = C
When pressure goes up, volume goes down. When volume goes up, pressure goes down.
From the equation above, this can be derived:
P_{1}V_{1} = P_{2}V_{2} = P_{3}V_{3} etc.
This equation states that the product of the initial volume and pressure is equal to the product of the volume and pressure after a change in one of them under constant temperature.
For example, if the initial volume was 500 mL at a pressure of 760 torr, when the volume is compressed to 450 mL, what is the pressure?
Plug in the values:
P_{1}V_{1} = P_{2}V_{2}
(760 torr)(500 mL) = P_{2}(450 mL)
760 torr x 500 mL/450 mL = P_{2} 844 torr = P_{2}
The pressure is 844 torr after compression.
VIDEO: CHARLES LAW CALCULATION
This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature.
V €T
Same as before, a constant can be put in:
V / T = C
As the volume goes up, the temperature also goes up, and viceversa.
Also same as before, initial and final volumes and temperatures under constant pressure can be calculated.
V_{1} / T_{1} = V_{2} / T_{2} = V_{3} / T_{3} etc.
This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
P T
Same as before, a constant can be put in:
P / T = C
As the pressure goes up, the temperature also goes up, and viceversa.
Also same as before, initial and final volumes and temperatures under constant pressure can be calculated.
P_{1} / T_{1} = P_{2} / T_{2} = P_{3} / T_{3} etc.
Now we can combine everything we have into one proportion:
The volume of a given amount of gas is proportional to the ratio of its Kelvin temperature and its pressure.
Same as before, a constant can be put in:
PV / T = C
As the pressure goes up, the temperature also goes up, and viceversa.
Also same as before, initial and final volumes and temperatures under constant pressure can be calculated.
P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2} = P_{3}V_{3} / T_{3} etc.
1. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C.
2. A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard temperature?
3. A gas has a pressure of 699.0 mm Hg at 40.0 °C. What is the temperature at standard pressure? (HONOR’S ONLY)
4. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mm Hg? (HONOR’S ONLY)
5. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be?
The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the partial pressure of that gas.
Dalton himself stated this law in the simple and vivid way shown at the left.
The usual way of stating Dalton's Law of Partial Pressures is
Expressed algebraically,
or, equivalently,
Problem Example
Calculate the mass of each component present in a mixture of fluorine (MW 19.0) and xenon (MW 131.3) contained in a 2.0L flask. The partial pressure of Xe is 350 torr and the total pressure is 724 torr at 25°C.
Solution: From Dalton's law, the partial pressure of F_{2} is (724 – 350) = 374 torr:
The mole fractions are X_{Xe} = 350/724 = .48 and X_{F2} = 374/724 = .52 . The total number of moles of gas is
The mass of Xe is (131.3 g mol^{–1}) × (.48 × .078 mol) = 4.9 g
The Ideal gas law is the equation of state of a hypothetical ideal gas.
It is a good approximation to the behavior of many gases under many conditions, although it has several limitations.
It was first stated by Émile Clapeyron in 1834 as a combination of Boyle's law and Charles's law.
It can also be derived from kinetic theory, as was achieved (apparently independently) by August Krönig in 1856^{ }and Rudolf Clausius in 1857.
The state of an amount of gas is determined by its pressure, volume, and temperature.
The modern form of the equation is:
where p is the absolute pressure of the gas; V is the volume; n is the amount of substance; R is the gas constant; and T is the absolute temperature.
In SI units, p is measured in pascals; V in cubic meters; n in moles; and T in kelvin.
R has the value 8.314472 J·K^{−1}·mol^{−1} in SI units).
However, in our calculations we use the following units:
In SI units, p is measured in atmospheres (atm); V in Liters (L); n in moles; and T in kelvin.
R has the value 0.0821 atm · L·K^{−1}·mol^{−1}.
EXAMPLE:
Suppose you have 1.00 mol of a gas at 0^{o}C, occupying a container which is 500 mL in size. What is the pressure of this gas in atmospheres?
A. To solve this problem, consider that moles, temperature, volume and the ideal gas law constant, R, are known.
B. Pressure is the only unknown variable.
C. Recall that R will dictate the units.
D. The temperature is given in Celsius.
This must be converted to the Kelvin scale.
E. To convert Celsius to Kelvin, add 273 to the Celsius temperature:
F. Also, the volume must be in liters, not milliliters. Convert as follows:
Now we are ready to insert the known values into the ideal gas law:
Solve for pressure by dividing both sides of the equation by the volume, 0.500 L:
Notice how the units cancel to give atmospheres.
Next, let's find the volume of 2.50 mol of gas which is at 730 mm Hg of pressure, and at a temperature of 127 ^{o}C.
As before, we need to convert temperature from Celsius to Kelvins:
Additionally, we need to convert the pressure units of mm Hg to atmospheres. In one atmosphere of pressure, there are 760 mm Hg:
Remember, the reason for these conversions was to make the units consistent with the units of the ideal gas law constant, R.
Substitute the known values for pressure, moles, R, and temperature into the ideal gas law:
Again, notice how the units cancel to give volume units of liters.
Unit conversions of pressure and volume must be done, first.
We see here another pressure unit, torr. There are 760 torr in one atmosphere, and as one can see, the torr pressure unit is the same as the units of mm Hg.
Once again, since the pressure and volume units are now consistent with the units of R, the values may be substituted into the ideal gas law:
Notice how the units cancel to give temperature in K.
Convert the Kelvin temperature into Celsius:
Now let's find how many moles of gas are present when the gas is occupying a volume of 5.00 L at a pressure of 10.0 atmospheres and a temperature of 310 K. Substitute the pressure, volume, temperature and the gas constant, R, into the ideal gas law equation:
Divide both sides of the equation by the gas law constant, R, and the temperature to solve for n:
Now let's find how many moles of gas are present when the gas is occupying a volume of 5.00 L at a pressure of 10.0 atmospheres and a temperature of 310 K. Substitute the pressure, volume, temperature and the gas constant, R, into the ideal gas law equation:
Divide both sides of the equation by the gas law constant, R, and the temperature to solve for n:
From the ideal gas law, PV = nRT, one can derive other useful expressionsones that relate the molar mass and density of Gases to pressures and temperatures.
This is often done just by substituting a different known expression for one of the variables in the ideal gas law.
For instance, you know that the moles of a gas, n, can also be expressed as the mass of the gas in grams over the molar mass of that gas:
Substitute this into the ideal gas law, and one obtains the equation:
Multiplying both sides by the molar mass, MM, obtains:
This equation is useful for determining the molar mass of a gas from experimental data, where the mass, pressure, volume and temperature of the gas is measured.
Now, let's divide both sides of the above expression by the volume, V:
Since we know that g/V is density, D, let's substitute density in for g/V in the above equation:
The molar volumes of all gases are the same when measured at the same temperature and pressure. But the molar masses of different gases will vary.
This means that different gases will have different densities (different masses per unit volume).
If we know the molecular weight of a gas, we can calculate its density.
Problem Example
Uranium hexafluoride UF_{6} gas is used in the isotopic enrichment of natural uranium. Calculate its density at STP.
Solution: The molecular weight of UF_{6} is 352.
(353 g mol^{–1}) ÷ (22.4 mol L^{–1}) = 15.7 g L^{–1}
Note: there is no need to look up a "formula" for this calculation; simply combine the molar mass and molar volume in such a way as to make the units come out correctly.
More importantly, if we can measure the density of an unknown gas, we have a convenient means of estimating its molecular weight.
This is one of many important examples of how a macroscopic measurement (one made on bulk matter) can yield microscopic information (that is, about molecularscale objects.)
Determination of the molecular weight of a gas from its density is known as the Dumas method, after the French chemist Jean Dumas (18001840) who developed it.
One simply measures the weight of a known volume of gas and converts this volume to its STP equivalent, using Boyle's and Charles' laws.
The weight of the gas divided by its STP volume yields the density of the gas, and the density multiplied by 22.4 mol^{–1} gives the molecular weight.
Pay careful attention to the examples of gas density calculations shown in your textbook.
You will be expected to carry out calculations of this kind, converting between molecular weight and gas density.
Calculate the approximate molar mass of a gas whose measured density is 3.33 g/L at 30°C and 780 torr.
Solution. From the ideal gas equation, the number of moles contained in one litre of the gas is
The molecular weight is therefore (33 g L^{–1})/(.0413 mol L^{–1}) = 80.6 g mol^{–1}
Since gases are mostly empty space, the densities of gases are reported in g/L, not g/mL as found for solids and liquids.
As you’re probably aware, density is equal to mass per unit of volume.
To calculate the density of a gas at standard temperature and pressure, you take the molecular formula weight of the gas (grams per mole—from the periodic table) and divide that by the standard molar volume for a gas, which is 22.4 L per mole:
where the formula weight (FW) is in g/mol, and the standard molar volume is 22.4 L/mol.
Now try using this in a problem.
Example
What is the density of helium gas at STP?
Explanation
If the density of the gas is equal to , then d = 4.00 g/mol ∏ 22.4 L/mol, so the density = 0.179 g/L.
If conditions are not standard, we can use this expanded version of the ideal gas equation:
The van der Waals equation is an equation of state for a fluid composed of particles that have a nonzero size and a pairwise attractive interparticle force (such as the van der Waals force.)
It was derived by Johannes Diderik van der Waals in 1873, who received the Nobel prize in 1910 for "his work on the equation of state for gases and liquids".
The equation is based on a modification of the ideal gas law and approximates the behavior of real fluids, taking into account the nonzero size of molecules and the attraction between them.
Upon introduction of the Avogadro constant N_{A}, the number of moles n, and the total number of particles nN_{A}, the equation can be cast into the second (better known) form
where
A careful distinction must be drawn between the volume available to a particle and the volume of a particle.
In particular, in the first equation refers to the empty space available per particle. That is, is the volume V of the container divided by the total number nN_{A} of particles.
The parameter b', on the other hand, is proportional to the proper volume of a single particle—the volume bounded by the atomic radius.
This is the volume to be subtracted from because of the space taken up by one particle.
In van der Waals' original derivation, given below, is four times the proper volume of the particle.
Observe further that the pressure p goes to infinity when the container is completely filled with particles so that there is no void space left for the particles to move.
This occurs when V = n b...
Kineticmolecular theory states that the average kinetic energy of a mole of molecules molecules is proportional to absolute temperature, and the proportionality constant is R, the universal gas constant
where M is the molar mass in kg/mole, R is the gas constant in J/K^{.}mole, and T is the absolute temperature in K.
Numerical Example:
Calculate the rms speed, u, of an N_{2} molecule at room temperature (25°C) Be careful of your UNITS!
T = (25+273)K = 298K
M = 28 g/mol = 0.028 kg/mol
R = 8.314 J/mol K = 8.314 kg m^{2}/s^{2} mol K
u = 515 m/s
Note: this is equal to 1,150 miles/hour!
Where does this number lie on our Graph of the speed distribution? To find out, the root mean square 'average' of the distribution must be taken (defined) in a specific way.
It is defined exactly how it sounds.
First your square all the speeds, then average those numbers, and take the square root of that average.
The mean of a distribution is just the average of all the numbers in the distribtion.
The most probable value of a distribution is also exactly what it sounds like, the speed of that the largest fraction of molecules are travelling.
In general, the mean, the root mean square and the most probable value in a distribution are all different.
A Note on Distributions a simple numerical example:
Suppose we have four molecules in our gas sample.
Their speeds are 3.0, 4.5, 5.2 and 8.3 m/s.
The rms speed as well as the entire distribution of speeds of gas molecules are a function of temperature.
Below, the blue line is a cold gas and the red line is a hot gas.
Note that the rms speed, u as well as the entire speed distribution changes with temperature for a given gas.
The rms speed for a given speed distribution (which is determined by the temperature and molecular weight of the gas) is greater in magnitude than the most probable speed or the mean speed.
Effusion
The escape of a gas through a tiny pore or pinhole in its container is called EFFUSION.
The effusion rate, r, has been found to be inversely proportional to the square root of its molar mass: (Why?)
Thus, comparison of the effusion rates of two gases with different masses will follow the relation:
This effect was observed in the 19^{th} century by Graham and is sometimes called GRAHAM's LAW
A note on Rates and Times:
The effusion time (the time it takes for a certain amount of gas to escape a vessel) is inversely proportional to the effusion rate (the amount of gas effusing from the hole per unit time).
Be careful that you understand whether it is a rate or a time that you are calculating.
Gas may effuse, but for this to happen a molecule must pass through a pore or pinhole and escape to the outside. In effect, a molecule must 'collide' with an escape hole.
The number of such collisions will be linearly proportional to the average speed of the molecules in the gas and thus the effusion rate.
The effusion time should be inversely proportional to the average speed of the molecules or proportional to the square root of the ratio of the molecular masses.
The ratio of effusion rates, r_{i}, for two gases labelled by i, is proportional to the ratio of the RMS speeds of the gases, u_{i}
Diffusion
Similarly to effusion, the process of diffusion is the spontaneous intermingling (mixing) of dissimilar gases (fluids) that are initially spatially separated.
If you put a drop of ink in a glass of water and you see the ink gradually spread out to fill the glass, this is diffusion
The average distance traveled by a molecule between collisions with another molecule is called the mean free path
Gas Laws and Kinetic Theory
The Ideal Gas Equation of State follows directly from the Kinetic Theory of Gases. Here is a PseudoDerivation
Avogadro's law (sometimes referred to as Avogadro's hypothesis or Avogadro's principle) is a gas law named after Amedeo Avogadro who, in 1811,^{ }hypothesized that
"Equal volumes of ideal or perfect gases, at the same temperature and pressure, contain the same number of molecules."
Thus, the number of molecules in a specific volume of gas is independent of the size or mass of the gas molecules.
As an example, equal volumes of molecular hydrogen and nitrogen would contain the same number of molecules, as long as they are at the same temperature and pressure and observe ideal or perfect gas behavior.
In practice, for real gases, the law only holds approximately, but the agreement is close enough for the approximation to be useful.
The law can be stated mathematically as:
The most significant consequence of Avogadro's law is that the ideal gas constant has the same value for all gases.
This means that the constant
Where:
has the same value for all gases, independent of the size or mass of the gas molecules.
One mole of an ideal gas occupies 22.414 litres (dm³) at STP, and occupies 24.45 litres at SATP (Standard Ambient Temperature and Pressure = 273K and 1 atm).
This volume is often referred to as the molar volume of an ideal gas.
Real gases may deviate from this value.
TO SUMMARIZE:
Avogadro's Law states that for a gas at constant temperature and pressure the volume is directly proportional to the number of moles of gas.
According to Avogadro's Law:
V1 / n1 = constant
After the change involume and mole number,
V2 / n2 = constant
Combine the two:
V1 / n1 = V2 / n2
When any three of the four quantities in the equation are known, the fourth can be calculated. For example, if n1, V1 and V2 are known, the n2 can be solved by the following equation:
n2 = V2 x (n1/V1)
Ex 1. Method for preparing and collecting a gas less dense (lighter) than air by heating solid reactants. The less dense gas rises into, and displaces, the more dense air downwards. This is called upward delivery. e.g. Heating a mixture of ammonium chloride and calcium hydroxide solids gives ammonia which has a very pungent odour! and turns red litmus blue.
See also Ex 7. method. 2NH_{4}Cl_{(s)} + Ca(OH)_{2(s)} ==> CaCl_{2(s)} + 2H_{2}O_{(l)} + 2NH_{3(g)} To make dry ammonia you need a U tube packed with granules of calcium oxide between the horizontal pyrex tube and the vertical inverted collection test tube.  
Ex 2. Method for preparing and collecting a gas less dense (lighter) than air by reacting a liquid and a solid. The less dense gas rises into, and displaces, the more dense air downwards. This is called upward delivery.
e.g. A mixture of zinc and hydrochloric acid makes hydrogen. Hydrogen gives a squeaky pop! with a lit splint. See also methods Ex 5., Ex 6. and Ex 7. Zn_{(s)} + 2HCl_{(aq)} ==> ZnCl_{2(aq)} + H_{2(g)}  
 Ex 3. Method for preparing and collecting a gas more dense (heavier) than air by heating the reactants. The more dense gas sinks down into, and displaces, the less dense air upwards. This is called downward delivery.
e.g. Making nasty brown nitrogen dioxide by heating lead(II) nitrate crystals (thermal decomposition). The solid 'deflacrates', it crackles as the gas formed splits the crystals apart. See also method Ex 7. 2Pb(NO_{3})_{2(s)} ==> 2PbO_{(s)} + 4NO_{2(g)} + O_{2(g)} 
Ex 4. Method for preparing and collecting a gas more dense (heavier) than air by reacting a solid and a liquid. The more dense gas sinks down into, and displaces, the less dense air upwards. This is called downward delivery. Examples:
 (i) Calcium carbonate (limestone/marble chips) with hydrochloric acid makes carbon dioxide. Can also be done via Ex 6. but carbon dioxide is moderately soluble and does make 'carbonated water. See also Ex 8. for carbonate test. CaCO_{3(s)} + 2HCl_{(aq)} ==> CaCl_{2(aq)} + H_{2}O_{(l)} + CO_{2(g)} (ii) Sulphur dioxide from solid sodium metabisulphite or sodium sulphite and excess dilute hydrochloric acid. The nasty choking gas turns potassium dichromate(VI) paper from orange to green. Should be done in fume cupboard. Na_{2}S_{2}O_{5(s)} + 2HCl_{(aq)} ==> 2NaCl_{(aq)} + H_{2}O_{(l)} + 2SO_{2(g)} or Na_{2}SO_{3(s)} + 2HCl_{(aq)} ==> 2NaCl_{(aq)} + H_{2}O_{(l)} + SO_{2(g)} (iii) Nasty acrid Hydrogen chloride is formed when conc. sulphuric acid is mixed with solid sodium chloride. Should be done in fume cupboard. NaCl_{(s)} + H_{2}SO_{4(l)} ==> NaHSO_{4(s)} + HCl_{(g)} (iv) Chlorine from conc. sodium chlorate(I) and conc. hydrochloric acid. Very dangerous and should be done in a fume cupboard. NaClO_{(aq)} + 2HCl_{(aq)} ==> NaCl_{(aq)} + H_{2}O_{(l)} + Cl_{2(g)} or from the chemical reaction conc. hydrochloric acid + damp potassium manganate(VII) crystals ==> chlorine gas + other products. The potassium manganate(VII), old name potassium permanganate, acts as an oxidising agent  it oxidises the chloride ion (Cl^{}) to chlorine molecules (Cl_{2}). All of these can be done via Ex. 5, Ex. 6 (its not too soluble and a way of making small amounts of 'chlorine water') or Ex 7. described below. 
Ex 5. Method for preparing and collecting a gas of any density by reacting a solid and a liquid at room temperature. e.g. making carbon dioxide Ex 4., chlorine Ex 4., hydrogen Ex 2., hydrogen chloride Ex 4., oxygen Ex 6., sulfur dioxide Ex 4.  
This is called collecting over water, or displacement of water or pneumatic trough collection.  Ex 6. Method for preparing and collecting a gas of any density by reacting a solidliquid, as long as the gas is not too soluble in water! (dissolving or reacting). All gases are less dense than liquid water and will displace the water downwards. No good for soluble gases like ammonia, hydrogen chloride, nitrogen dioxide or sulphur dioxide. You can collect in inverted gas jar if bigger sample required. You have to watch for 'sucking back' effects. Examples: (i) Making oxygen from hydrogen peroxide solution using manganese dioxide catalyst. Oxygen has similar density to air so must be collected by methods Ex 5., Ex 6. or Ex 7. 2H_{2}O_{2(aq)} ==> 2H_{2}O_{(l)} + O_{2(g)} (ii) Hydrogen, (iii) carbon dioxide and (iv) chlorine (moderately soluble, makes 'chlorine water') 
Ex 7. Method for preparing and collecting a gas of any density by reacting a solidliquid or heating solids. The angled boiling tube minimises the risk of contaminating the gas syringe with solids or liquids eg making ammonia or nitrogen dioxide. Its a smaller scale alternative to Ex 5. and using a Pyrex tube suitable for small scale heated experiments.  
Ex 8. A simple way to test for a carbonate. Add acid to the suspected carbonate. Collect a sample of the gas in a teat pipette from just above the reaction mixture. Bubble the gas sample into calcium hydroxide solution (limewater) and a milky white confirms the gas is carbon dioxide formed from the original carbonate.  
U tube dreschel bottle  Ex 9. Use of a U tube. This is useful if the dry gas is needed. It is inserted in the apparatus setup between the reaction container and the gas collection system. It is packed with a solid water absorbing drying agent e.g. anhydrous calcium chloride (not for ammonia), calcium oxide (not acidic gases like sulphur dioxide, carbon dioxide and chlorine), anhydrous sodium sulphate. A dreschel bottle can also be used e.g. the gas is bubbled through concentrated sulphuric acid which will dry the gas. It cannot be used to dry alkaline gases like ammonia, with which it will react exothermically to form the solid salt ammonium sulphate. 
Ex 10. A cracking experiment. This diagram outlines a way of demonstrating high temperature thermal cracking of larger hydrocarbon alkanes like paraffin oil/wax into smaller molecules i.e. smaller/lower alkanes and alkenes. This is a thermal decomposition reaction catalysed by aluminium oxide (or broken porous pot). The gases produced can be tested with (i) a match! and (ii) bromine water, if it is decolorized from orange to colourless, then unsaturated alkenes were formed. This is another example of over water, or displacement of water collection. The dreschel bottle is to collect any sucked back water if the hot gasses cool and contract. Cold water on the hot Pyrex tube has very nasty effect! plus the risk of fire! Gas preparation (general)The diagram below shows a typical set of apparatus which can be used to prepare a range of gases.

Boyles’ Law
Use Boyles’ Law to answer the following questions:
1) 1.00 L of a gas at standard temperature and pressure is compressed to 473 mL. What is the new pressure of the gas?
2) In a thermonuclear device, the pressure of 0.050 liters of gas within the bomb casing reaches 4.0 x 10^{6} atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. What is the volume of the gas after the explosion?
3) Synthetic diamonds can be manufactured at pressures of 6.00 x 10^{4} atm. If we took 2.00 liters of gas at 1.00 atm and compressed it to a pressure of 6.00 x 10^{4} atm, what would the volume of that gas be?
4) The highest pressure ever produced in a laboratory setting was about 2.0 x 10^{6} atm. If we have a 1.0 x 10^{5} liter sample of a gas at that pressure, then release the pressure until it is equal to 0.275 atm, what would the new volume of that gas be?
1) Three flasks are connected to each other, separated only by a threeway stopcock.
· Flask 1 has a volume of 3.000 liters and holds helium gas at a pressure of 3.500 atmospheres
· Flask 2 has a volume of 2.000 liters and holds nitrogen gas at a pressure of 2.000 atmospheres
· Flask 3 has a volume of 1.800 liters and holds oxygen gas at a pressure of 4.000 atmospheres
If the stopcock separating the flasks were to be opened, what would the partial pressure of each gas in the apparatus be?
2) What would the total pressure in the apparatus be?
3) What would the mole fraction of oxygen be inside the apparatus after the stopcock was opened?
4) If liquid water is added to the mixture, what will the mole fraction of each of the gases in the mixture be? The vapor pressure of water at 25^{0} C is 0.031 atm.
1) Three flasks are connected to each other, separated only by a threeway stopcock.
· Flask 1 has a volume of 3.000 liters and holds helium gas at a pressure of 3.500 atmospheres
· Flask 2 has a volume of 2.000 liters and holds nitrogen gas at a pressure of 2.000 atmospheres
· Flask 3 has a volume of 1.800 liters and holds oxygen gas at a pressure of 4.000 atmospheres
If the stopcock separating the flasks were to be opened, what would the partial pressure of each gas in the apparatus be?
Use the combined gas law to treat each gas as if it were expanding from its flask into a vacuum.
P_{He} = 1.54 atmospheres
P_{N}_{2} = 0.74 atmospheres
P_{O2} = 1.06 atmospheres
2) What would the total pressure in the apparatus be?
Dalton’s Law states that the total pressure will be equal to the sum of the partial pressures, so the final pressure will be 1.54 + 0.74 + 1.06 = 3.34 atm.
3) What would the mole fraction of oxygen be inside the apparatus after the stopcock was opened?
Mole fraction is equal to the partial pressure of a gas divided by the total pressure of a system. In this case, the mole fraction of oxygen is 0.32
4) If liquid water is added to the mixture, what will the mole fraction of each of the gases in the mixture be? The vapor pressure of water at 25^{0} C is 0.031 atm.
To solve, use
In the second step, divide the partial pressure of each gas by the total pressure to find that:
x_{He} = 0.46 x_{N2} = 0.22 x_{O2} = 0.31 x_{H2O} = 0.01
Practice Problems
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Name_________________________________________________  Section____________________________ 
This page is designed to help students practice written problems, and is meant to be printed out. Hit the print command and show all work in the spaces provided. Final answers must show units, and must be rounded to the correct number of significant digits.
1. What volume will a sample of hydrogen occupy at 28.0 ^{o}C if the gas occupies a volume of 2.23 dm^{3} at a temperature of 0.0 ^{o}C? Assume that the pressure remains constant. (remember to change to Kelvin).
2. If a gas occupies 733 cm^{3} at 10.0 ^{o}C, at what temperature will it occupy 950 cm^{3}? Assume that pressure remains constant.
3. A gas occupies 560 cm^{3} at 285 K. To what temperature must the gas be lowered to, if it is to occupy 25.0 cm^{3}? Assume a constant pressure.
next set
1. A sample of neon has a volume of 1.83 L at 23.5 ^{o}C. At what temperature would the gas occupy 5.00 L? Assume pressure is constant.
2. What volume would a sample of helium occupy at standard temperature, if it occupies 250 cm^{3} at 185 K? Assume constant pressure.
3. A sample of argon is collected in a 5.00 x 10^{2 }cm^{3} bottle at a temperature of 12.0 ^{o}C. Assuming the pressure remains the same, what volume would the gas occupy at 2.0 ^{o}C?
Return to Charles Law Explainations
Boyles’ Law
Use Boyles’ Law to answer the following questions:
1) 1.00 L of a gas at standard temperature and pressure is compressed to 473 mL. What is the new pressure of the gas?
2) In a thermonuclear device, the pressure of 0.050 liters of gas within the bomb casing reaches 4.0 x 10^{6} atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. What is the volume of the gas after the explosion?
3) Synthetic diamonds can be manufactured at pressures of 6.00 x 10^{4} atm. If we took 2.00 liters of gas at 1.00 atm and compressed it to a pressure of 6.00 x 10^{4} atm, what would the volume of that gas be?
4) The highest pressure ever produced in a laboratory setting was about 2.0 x 10^{6} atm. If we have a 1.0 x 10^{5} liter sample of a gas at that pressure, then release the pressure until it is equal to 0.275 atm, what would the new volume of that gas be?
5) Atmospheric pressure on the peak of Mt. Everest can be as low as 150 mm Hg, which is why climbers need to bring oxygen tanks for the last part of the climb. If the climbers carry 10.0 liter tanks with an internal gas pressure of 3.04 x 10^{4} mm Hg, what will be the volume of the gas when it is released from the tanks?
6) Part of the reason that conventional explosives cause so much damage is that their detonation produces a strong shock wave that can knock things down. While using explosives to knock down a building, the shock wave can be so strong that 12 liters of gas will reach a pressure of 3.8 x 10^{4} mm Hg. When the shock wave passes and the gas returns to a pressure of 760 mm Hg, what will the volume of that gas be?
7) Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be?
8) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
Boyles’ Law
Use Boyles’ Law to answer the following questions:
1) 1.00 L of a gas at standard temperature and pressure is compressed to 473 mL. What is the new pressure of the gas? 2.11 atm
2) In a thermonuclear device, the pressure of 0.050 liters of gas within the bomb casing reaches 4.0 x 10^{6} atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. What is the volume of the gas after the explosion?
2.0 x 10^{5} L
3) Synthetic diamonds can be manufactured at pressures of 6.00 x 10^{4} atm. If we took 2.00 liters of gas at 1.00 atm and compressed it to a pressure of 6.00 x 10^{4} atm, what would the volume of that gas be? 3.33 x 10^{5} L
4) The highest pressure ever produced in a laboratory setting was about 2.0 x 10^{6} atm. If we have a 1.0 x 10^{5} liter sample of a gas at that pressure, then release the pressure until it is equal to 0.275 atm, what would the new volume of that gas be? 72.7 L
5) Atmospheric pressure on the peak of Mt. Everest can be as low as 150 mm Hg, which is why climbers need to bring oxygen tanks for the last part of the climb. If the climbers carry 10.0 liter tanks with an internal gas pressure of 3.04 x 10^{4} mm Hg, what will be the volume of the gas when it is released from the tanks? 2.0 x 10^{3} L
6) Part of the reason that conventional explosives cause so much damage is that their detonation produces a strong shock wave that can knock things down. While using explosives to knock down a building, the shock wave can be so strong that 12 liters of gas will reach a pressure of 3.8 x 10^{4} mm Hg. When the shock wave passes and the gas returns to a pressure of 760 mm Hg, what will the volume of that gas be? 600 L
7) Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be? 72 L
8) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver? V = 0.25 L, yes