KMT and Gas Laws 15.notebook

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KMT and Gas Laws 15.notebook
March 17, 2015
Kinetic Molecular Theory
and Gas Laws
Pressure = force/unit area = lbs/inch2(English) = Newtons/meter2 = pascal (metric)
We will be interested in studying the pressure exerted by gas molecules. We use an instrument called a barometer to measure the pressure exerted by the gases in out atmosphere. This air pressure is due to the weight of air or the force of gravity pulling the air molecules towards the earth. A gas behaves like a liquid in a sense that it is considered a fluid and exerts a pressure in all directions. Note that when measuring air pressure with a barometer the units we use are in length (not pressure). Standard air pressure = 760 mm Hg = 760 torr = 1 atm = 101.3 kPa
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We measure the pressure exerted by a gas or gases in a container by using an open­ended manometer:
What is the pressure exerted by the gas in the manomerer?
What is the pressure exerted by the gas in the manometer?
What is the pressure exerted by the gas in the container?
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Problems
Kinetic Molecular Theory of Gases
The KMT was developed to explain the behavior of gases under most conditions. Like all theories it is a model which is useful in interpreting and explaining the observed behavior of a gas. The main postulates of the theory are:
Gases are composed of separate, tiny invisible particles called molecules. These molecules are so far apart, on average, that the total volume •
of the molecules is extremely small compared to the total volume of space they” fly around” in. Therefore, under ordinary conditions, the gas consists mostly of empty space. This assumption explains why gases are so easily compressed and why they can mix so readily.
Gas molecules are in constant, rapid straight­line motion and thus possess kinetic energy. Recall gas molecules exhibit translational, rotational •
and vibrational kinetic energies. This motion is constantly interrupted by collisions with other molecules or with the walls of the container. The pressure of a gas is the effect of these molecular impacts.
1.
What is the atmospheric pressure, in kPa, indicated by the barometer in figure (a)?
2.
What is the pressure, in kPa, of the confined gas as indicated by the open­ended manometer in figure (b)?
3.
What is the pressure, in kPa, of the confined gas indicated by the open­ended manometer in figure (c)?
4.
What is the pressure, in kPa, of the confined gas indicated by the open­ended manometer in figure (d)?
The collisions between molecules are completely elastic – no kinetic energy is changed to heat or other forms of energy as a result of collisions. •
The total kinetic energy of the molecules remains the same as long as the temperature and volume do not change. Therefore, the pressure of an enclosed gas remains the same if its temperature and volume do not change.
•
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The molecules of a gas display no attraction or repulsion for one another.
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KMT and Gas Laws 15.notebook
At any one moment, the molecules in a gas have different velocities and hence different kinetic energies. It is assumed that the •
average kinetic energy of the molecules is directly proportional to the Kelvin temperature of the gas. The statistical distribution of kinetic energy among the molecules of a gas at different temperatures was shown independently by James Maxwell and Ludwig Boltzmann. These are known as Maxwell­Boltzmann distributions and are represented graphically below: March 17, 2015
Chemists refer to gases that conform or behave according to the postulates of the KMT as ideal gases. Although there is no such thing as an ideal gas, most gases at room temperature and pressure behave as ideal gases. Under what conditions will gases deviate from ideal behavior?
high pressure – this decreases the volume the gas molecules “fly around in” and hence the volumes of the individual gas molecules •
themselves becomes significantÍž the gas molecules get more closely packed and the density of the gas increases
low temperature – this lowers the kinetic energy of the molecules – they slow down! At lower temperatures the IMFA’s can become •
effective (and might turn the gas into a liquid).
Which gases behave most ideally?
Those with small sizes (molecular weights) and weak IMFA. In general, the smaller the molecular weight the weaker the IMFA. Thus, gases such as hydrogen and helium will behave most ideally.
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behave most ideally.
The Gas Laws
What variables are needed to describe the physical behavior of gases?
P
T
V
n
(pressure) (temp) (volume) (moles)
At constant temperature,
P α 1/V (pressure is inversely proportional to V
or
P = k(1/V)
k = constant
thus, PV = k
and
P1V1 = k = P2V2
or
P1V1 = P2V2
This is a statement of Boyle’s Law
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Boyle’s law can be represented graphically as:
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At constant pressure,
V α T temperature)
or
(volume is directly proportional to the Kelvin
V = kT
k = constant
and
thus,
This is a statement of Charles’ law
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KMT and Gas Laws 15.notebook
Charles’ law can be represented graphically as:
March 17, 2015
At constant volume,
P α T temperature)
(pressure is directly proportional to the Kelvin
and
This is a statement of the law of Guy­Lusaac
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The Combined Gas Law (CGL):
Peas Vegies
Table
recall, P1V1 = P2V2
T1 T2
K = oC + 273
I L = 1000 mL = 1000 cm3 = 1 dm3
Also recall STP (standard temperature and pressure):
T = 0oC = 273 K
P = 1 atm = 760 mm Hg = 760 torr
Problem 1. A certain gas occupies a volume of 86.0 L at 20oC and 760 mm Hg. Calculate the volume the gas would occupy at STP
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Two gases that have the same volume at the same temperature and the same
pressure must have the same number of gas molecules (or moles of gas
molecules).
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Avogadro’s idea also means that at constant temperature and pressure, the
volume of a gas is directly proportional to the moles of gas present:
Consider H2 and O2
H2
O2
If a hydrogen molecule can be represented by , illustrate Avogadro’s idea in
the above boxes, representing equal volumes. On a relative scale, an equal number of oxygen molecules how many times more
than hydrogen, if hydrogen is assigned a relative value of 1.0? The first relative weights of the elements were derived using Avogadro’s idea. This revolutionized chemistry!
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KMT and Gas Laws 15.notebook
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To review concepts of P, V, T and n,
V α n
(Avogadro)
V α 1/P
(Boyle)
(Charles)
V α T
­­­­­­­­­­­­­­­­­­­­­­­
Gas Density
V α
You can determine the density of any gas at STP (standard temperature ­ 0oC or
273K and pressure ­ 1 atm or 101.3 kPa):
Dstp =
or molar mass =
Dstp X molar volume
and
V =
or
PV = nRT
Ideal Gas Law
R = universal gas law constant whose numeric value depends on
the units of pressure:
R = 0.0821 atm L/mol K = 62.4 mm Hg/L/mol K = 8.31 kPa
L/mol K
Recall molar volume = 22.4 L/mol at STP.
Determine the density of carbon dioxide gas as STP:
Problem 2. Calculate the volume occupied by 2.68 g of nitrogen dioxide gas at
22.0oC and 745 torr.
A gas has a density of 0.77 g/L at STP. What is its molar mass?
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Consider a container of fixed volume and temperature containing a mixture of gases The total pressure exerted by the gases is given by:
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What volume does the Xe gas occupy?
What volume does the He gas occupy?
Ptotal = PA + PB + Pc + etc
What volume does the Ar gas occupy?
partial pressures
This is a statement of Dalton’s Law of Partial Pressures
How does the partial pressure of Xe compare with the partial pressure of He?
recall PV = nRT
The partial pressure of a gas in a mixture may be represented as:
in this case V, T, and R are constant
and therefore P α n (pressure is directly proportional to moles of gas)
Consider a 25.0 L container with the following gases and has a total pressure of 1500 torr.
Xe
Xe
Partial pressure = mole fraction X total pressure
In the above example, what is the partial pressure of the argon gas?
Ar
He
He
Xe
Xe
Ar
Xe
He
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Problem 3. A 15.0 liter cylinder at 25.0oC contains the following gases:
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Recall that in the laboratory we typically collect a gas by water displacement:
262.2 g of xenon
16.0 g of helium
120.0 g of argon
a)
Calculate the total pressure: Whenever a gas is collected by water displacement some of the water will
evaporate and we will have a mixture of the gas collected and water vapor. The
pressure exerted by the water vapor is directly proportional to the temperature
and you will be given the vapor pressure at a given temperature.
Ptotal = PX + PH2O
where
Px = pressure of gas x
PH2O = vapor pressure of water
b)
Calculate the partial pressure of each gas (this can be done two ways):
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Problem 4. When 81.4 mL of hydrogen gas is collected by water displacement,
the water levels inside and outside the gas­collecting bottle are equal. This means
that the pressure of the mixture of gas in the bottle (hydrogen and water vapor) is
equal to barometric (air) pressure. The barometric pressure is 740.0 torr and the
temperature is 23.0oC. Determine the moles of hydrogen gas in this sample. Note: the vapor pressure of water at 23.0oC is 21.1 torr.
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KMT and Gas Laws 15.notebook
Diffusion of gases ­ movement through air (effusion is movement through a
vacuum)
March 17, 2015
Problem 6 refers to the illustration below:
Consider two gases, A and B, at the same temperature. If they both have the
same temperature then they both have the same average kinetic energy.
Recall kinetic energy (KE):
KE = 1/2mv2 m = mass
v = velocity
Thus, Or
KEA = KEB
1/2mAv2A = 1/2mBv2B
Problem 5. If A has a molar mass of 40.0 g/mol and gas B has a molar mass of
120 g/mol, which gas will move faster? Explain.
Rearranging the above equation yields:
This is a statement of Graham’s Law of Diffusion
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Gas Stoichiometry
We can apply our knowledge of stoichiometric relationships in equations to include volumes of reactants or products occurring in the gas phase. Note the following:
Problem 7. Compare the relative rates of diffusion of hydrogen gas and oxygen gas if both gases are at the same temperature and pressure.
If the conditions are standard temperature (0oC or 273K) and pressure (1 atm or 760 mm Hg), also known as STP, one mole of any gas occupies a volume of 22.4 L. This is known as molar •
volume.
For gases at the same temperature and pressure, the volumes are directly proportional to moles (coefficients)
•
Problem 9. A 5.00 g sample of sodium carbonate reacts with excess acid to produce carbon dioxide:
Na2CO3(s) + HCl(aq) à NaCl(aq) + H2O(l) + CO2(g)
Balance the above equation and calculate the volume of CO2 produced at STP and at 222.0oC and 745 mm Hg.
Problem 8. A sample of hydrogen gas diffuses 8.70 times as fast as an unknown gas at the same temperature and pressure. Determine the molar mass of the unknown gas.
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Problem 12. Hydrogen gas reacts with chlorine gas to produce hydrogen chloride gas. Calculate the volume of chlorine needed to react with excess hydrogen to produce 50.0 g of hydrogen chloride at STP.
Problem 10. Given the reaction of hydrogen gas and nitrogen gas to produce ammonia gas:
H2(g) + N2(g) à NH3(g)
Balance the above equation. If 10.0 liters of hydrogen gas reacts at constant temperature and pressure, what volume of nitrogen gas is required to completely react with the hydrogen? What volume of ammonia will be produced?
Problem 11. What mass of potassium chlorate must be decomposed to produce 250.0 mL of oxygen gas measured at STP?
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KMT and Gas Laws 15.notebook
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