KMT and Gas Laws.notebook

KMT and Gas Laws.notebook
March 17, 2014
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). Mar 3­2:43 PM
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Standard air pressure = 760 mm Hg = 760 torr = 1 atm = 101.3 kPa
We measure the pressure exerted by a gas or gases in a container by using a manometer:
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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)?
Mar 3­2:44 PM
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.
• 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.
• 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: Mar 3­2:46 PM
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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 significant; 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).
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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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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
thus,
P = k(1/V)
k = constant
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 V = kT
k = constant
and
thus,
This is a statement of Charles’ law
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Charles’ law can be represented graphically as:
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At constant volume,
P α T temperature)
(pressure is directly proport
and
This is a statement of the law of Guy
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The Combined Gas Law (CGL):
Peas Vegies
Table
P1V1 = P2V2
T1 T2
recall, 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
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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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Avogadro’s Idea
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).
Consider H2 and O2
H2
O2
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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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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:
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To review:
V α n
(Avogadro)
V α 1/P
(Boyle)
V α T
(Charles)
­­­­­­­­­­­­­­­­­­
V α and
V = or
PV = nRT
Ideal Gas Law
R = universal gas law constant whose n
the units of pressure:
R = 0.0821 atm L/mol K = 62.4 mm L/mol K
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Problem 2. Calculate the volume occupied by 2.68 g of nitrogen dioxide gas at 22.0oC and 745 torr.
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In a gas mixture at constant volume and constant temperature,
Ptotal = PA + PB + Pc + etc
partial pressures
This is a statement of Dalton’s Law of Partial Pressures
recall PV = nRT
in this case V, T, and R are constant
and therefore P α n (pressure is directly proportional to moles of gas)
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Consider a 25.0 L container with the following gases and has a total pressure of 1500 torr.
What volume does the Xe gas occupy?
What volume does the He gas occupy?
What volume does the Ar gas occupy?
How does the partial pressure of Xe compare with the partial pressure of He?
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The partial pressure of a gas in a mixture may be represented as:
Partial pressure = mole fraction X total pressure
In the above example, what is the partial pressure of the argon gas?
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o C contains the following gases:
Problem 3. A 15.0 liter cylinder at 25.0
262.2 g of xenon
16.0 g of helium
120.0 g of argon
a)
Calculate the total pressure: b)
Calculate the partial pressure of each gas :
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Recall that in the laboratory we typically collect a gas by water displacement:
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.
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Ptotal = PX + PH2O
where
Px = pressure of gas x
PH2O = vapor pressure of water
Problem 4. When 81.4 mL of hydrogen gas is collected by water displacement, the water levels inside and outside the gas­
collecting 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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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, KEA = KEB
Or
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.
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Rearranging the above equation yields:
This is a statement of Graham’s Law of Diffusion
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Problem 6 refers to the illustration below:
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Problem 7. Compare the relative rates of diffusion of hydrogen gas and oxygen gas if both gases are at the same temperature and pressure.
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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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:
• 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)
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Problem 9. A 5.00 g sample of sodium carbonate reacts with excess acid to produce carbon dioxide:
Na2 CO 3 (s) + HCl(aq) à NaCl(aq) + H2 O(l) + CO2 (g)
o
Balance the above equation and calculate the volume of CO
2 produced at STP and at 222.0 C and 745 mm Hg.
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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?
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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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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.
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