Gases. What do we know? 1.Gases have mass. 2.Gases are easily compressed. 3.Gases uniformly and...
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Transcript of Gases. What do we know? 1.Gases have mass. 2.Gases are easily compressed. 3.Gases uniformly and...
Gases
What do we know?1. Gases have mass.2. Gases are easily compressed.3. Gases uniformly and completely fill their
containers.4. Different gases move through each other quite
rapidly.5. Gases exert pressure.6. The pressure of a gas depends on its
temperature.
Measuring Gases• Amount of gas (moles = n)
n = mass (g) / molar mass (g/mol)
• Volume (V)Vgas = Vcontainer
Measured in liters
Measuring Gases• Temperature (T)
Always measured in Kelvin (K)T(K) = T(oC) + 273
• Pressure (P)Gases exert pressure when particles collide with
walls and the force is spread over the area of the container
STP = 1 atm pressure, 0oC (273 K)
Measuring Pressure• Pressure of air is Pressure of air is
measured with a measured with a barometerbarometer (developed (developed by Torricelli in 1643)by Torricelli in 1643)
• Hg rises in tube until Hg rises in tube until force of Hg (down) force of Hg (down) balances the force of balances the force of atmosphere (pushing atmosphere (pushing up). up).
Measuring Pressure1.000 atm (standard atmosphere) =
760.0 mm Hg (millimeters of mercury) =760 torr (named after Torricelli) =14.69 psi (pounds per square inch) =101,325 Pa (pascals)
The Pascal is the SI unit for pressure, but it is used sparingly because it is so small.
Just use the equalities as conversion factors to get from unit to another.
The Gas Laws• Boyle’s Law
(pressure-volume relationship)• Charles’s Law
(volume-temperature relationship)• Gay-Lussac’s Law
(pressure-temperature relationship)• Combined Gas Law
(relates pressure, volume, and temperature)• Dalton’s Law of Partial Pressures• The Ideal Gas Law
Boyle’s LawThe pressure of a given sample of a gas is
inversely proportional to the volume of the gas at constant temperature.
If we know the volume of a gas at a given pressure, we can predict the new volume if the pressure is
changed, as long as the temperature and the amount of the gas remain the same.
P1V1 = P2V2
Boyle’s LawFreon-12 (the common name for the compound
CCl2F2) was once widely used in refrigeration systems, but has now been replaced by other
compounds that do not lead to the breakdown of the protective ozone in the upper atmosphere. Consider a 1.5-L sample of gaseous CCl2F2 at a pressure of 56 torr. If the pressure is changed to
150 torr at a constant temperature, will the volume of the gas increase or decrease? What
will be the new volume of the gas?
Boyle’s LawGiven:• P1 = 56 torr
• V1 = 1.5 L
• P2 = 150 torr• Temperature is constant
Pressure increases so volume decreases.
Boyle’s Law: P1V1 = P2V2
(56 torr)(1.5 L) = (150 torr) V2
V2 = (56 torr)(1.5 L) / (150 torr)
V2 = 0.56 L
Unknown:• V2 = ? L
Boyle’s LawIn an automobile engine the gaseous fuel-air
mixture enters the cylinder and is compressed by a moving piston before it is ignited. In a certain
engine the initial cylinder volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The
fuel-air mixture initially has a pressure of 1.00 atm. Calculate the pressure of the compressed
fuel-air mixture, assuming that both the temperature and the amount of gas remain
constant.
Boyle’s LawGiven:• P1 = 1.00 atm
• V1 = 0.725 L
• V2 = 0.075 L• Temperature and amount of gas are constant
Volume decreases so pressure increases.
Boyle’s Law: P1V1 = P2V2
(1.00 atm)(0.725 L) = P2(0.075 L)
P2 = (1.00 atm)(0.725 L) / (0.075 L)
P2 = 9.7 atm
Unknown:• P2 = ? atm
Charles’s LawThe volume of a given sample of a gas is directly proportional to the temperature of
the gas at constant pressure.
Initial values can also be used to find final values
V1/T1 = V2/T2
Charles’s Law A 2.0-L sample of air is collected at 298 K and
then cooled to 278 K. The pressure is held constant at 1.0 atm. Does the volume increase or decrease? Calculate the volume of the air at
278 K.
Charles’s LawGiven:• T1 = 298 K
• V1 = 2.0 L
• T2 = 278 K• Pressure is constant
Temperature decreases so volume decreases.
Unknown:• V2 = ? L
298 K
Charles’s Law
2.0 L
298 K 278 K
V2 =
V1/T1 = V2/T2
V2 = 2.0 L x 278 K
V2 = 1.9 L
(2.0 L)(278 K) = V2 (298K)
Charles’s LawA sample of gas at 15oC (at 1 atm) has a volume of
2.58 L. The temperature is then raised to 38oC (at 1 atm). Does the volume of the gas increase
or decrease? Calculate the new volume.
Charles’s LawGiven:• T1 = 15oC
• V1 = 2.58 L
• T2 = 38oC• Pressure is constant
Temperature increases so volume increases
Unknown:• V2 = ? L+ 273 = 288 K
+ 273 = 311 K
288 K
Charles’s Law
2.58 L
288 K 311 K
V2 =
V1/T1 = V2/T2
V2 = 2.58 L x 311 K
V2 = 2.79 L
(2.58 L)(311 K) = V2 (288K)
Equal volumes of gases at the same temperature and pressure contain an
equal number of particles.
Initial values can also be used to find final values
V1/n1 = V2/n2
Avogadro’s Law
The pressure of a given sample of a gas is directly proportional to the temperature of
the gas at constant volume.
Initial values can also be used to find final values
P1/T1 = P2/T2
Gay-Lussac’s Law
Gay-Lussac’s LawCalculate the final pressure inside a scuba tank after it cools from 1000 °C to 25.0 °C. The initial
pressure in the tank is 130.0 atm.
Gay-Lussac’s LawGiven:• P1 = 130 atm
• T1 = 1000 oC
• T2 = 25 oC • Volume is constant
Temperature decreases so pressure decreases.
Unknown:• P2 = ? atm
+ 273 = 1273 K+ 273 = 298 K
1273 K
Gay-Lussac’s Law
130.0 atm
1273 K 298 K
P2 =
P1/T1 = P2/T2
P2 = 130 atm x 298 K
P2 = 30.4 atm
(130 atm)(298 K) = P2 (1273 K)
Gay-Lussac’s LawIf 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?
Gay-Lussac’s LawGiven:• P1 = 15 atm
• P2 = 16 atm
• T1 = 25.0 oC • Volume is constant
Pressure increases so temperature increases.
Unknown:• T2 = ? K
+ 273 = 298 K
15 atm
Gay-Lussac’s Law
15.0 atm
298 K T2
16.0 atm =
P1/T1 = P2/T2
T2 = 16 atm x 298 K
T2 = 318 K
(15 atm)T2 = (16 atm)(298 K)
Summary of Laws
Law Statement Equation Constant
Boyle'sP inversely proportional to
V P1V1 =
P2V2 T, n
Charles's V directly proportional to T V1/T1 =
V2/T2 P, n
Avogadro's V directly proportional to nV1/n1 =
V2/n2 P, T
Gay-Lussac’s P directly proportional to TP1/T1 =
P2/T2 V, n
Combined Gas Law
Combined Gas Law• Combines Boyle’s, Charles’s, and Gay-Lussac’s
laws to relate the pressure, volume, and temperature of constant amounts of gases.
Initial values can also be used to find final values
P1V1/T1 = P2V2/T2
Combined Gas LawA toy balloon has an internal pressure of 1.05 atm
and a volume of 5.0 L. If the temperature above where the balloon is released is 20oC, what will
happen to the volume when the ballooon rises to an altitude where the pressure is 0.65 atm and
the temperature is -15oC?
Combined Gas LawGiven:• P1 = 1.05 atm
• V1 = 5.0 L
• T1 = 20 oC
• P2 = 0.65 atm
• T2 = -15 oC
Unknown:• V2 = ? L
+ 273 = 293 K
+ 273 = 258 K
(0.65 atm)(293 K)
Combined Gas Law
(1.05 atm)(5.0 L)
293 K 258 K
(0.65 atm)V2 =
P1V1/T1 = P2V2/T2
V2 = (1.05 atm)(5.0 L)(258 K)
V2 = 7.1 L
(1.05 atm)(5.0 L)(258 K) = (0.65 atm)V2(293 K)
Combined Gas LawIf I initially have 4.0 L of gas at a pressure of 1.1 atm and a temperature of 25oC, what will happen to the pressure if I decrease the volume to 3.0 L
and decrease the temperature to 20oC?
Combined Gas LawGiven:• V1 = 4.0 L
• P1 = 1.1 atm
• T1 = 25 oC
• V2 = 3.0 L
• T2 = 20 oC
Unknown:• P2 = ? L
+ 273 = 298 K
+ 273 = 293 K
(3.0 L)(298 K)
Combined Gas Law
(1.1 atm)(4.0 L)
298 K 293 K
P2(3.0 L)=
P1V1/T1 = P2V2/T2
P2 = (1.1 atm)(4.0 L)(293 K)
P2 = 1.4 atm
(1.1 atm)(4.0 L)(293 K) = P2(3.0 L)(298 K)
The Ideal Gas Law
The Ideal Gas LawCombines Boyle’s, Charles’s, and Avogadro’s Laws to
describe the behavior of gases as dependent upon volume, pressure, temperature, and the number of moles
present.
PV = nRTwhere R is the combined constant called the
universal gas constantR = 0.08206 L atm / mol K
The Ideal Gas LawA gas that obeys this equation is said to
behave ideally, thus the name “Ideal Gas Law”
There is no such thing as an ideal gas, but many real gases behave ideally at pressures
of approximately 1 atm and lower and temperatures of approximately 273 K or
higher.
The Ideal Gas LawA sample of hydrogen gas has a volume of 8.56 L
at a temperature of 0oC and a pressure of 1.5 atm. Calculate the number of moles of H2
present in this gas sample. Assume that the gas behaves ideally.
The Ideal Gas LawGiven:• P = 1.5 atm• T = 0oC = 273 K• V = 8.56 L
PV = nRT(1.5 atm)(8.56 L) = n(0.08206 L-atm/mol-K)(273 K)
Unknown:• n = ?
(1.5 atm)(8.56 L)
(0.08206 L-atm/mol-K)(273 K)n =
n = 0.57 mol
The Ideal Gas LawWhat volume is occupied by 0.250 mol of carbon
dioxide gas at 25oC and 371 torr?
Because the gas constant is measured in L-atm/mol-K, volume must be measured in liters, pressure in
atmospheres, amount of gas in moles, and temperature in Kelvin.
25oC + 273 = 298 K
371 torr x760 torr1.00 atm
= 0.488 atm
The Ideal Gas LawGiven:• P = 0.488 atm• T = 298 K• n = 0.250 mol CO2
PV = nRT(0.488 atm)V = (0.250 mol)(0.08206 L-atm/mol-K)(298K)
Unknown:• V = ? L
(0.250 mol)
(0.488 atm)
(0.08206 L-atm/mol-K)(298 K)V =
V = 12.5 L
Dalton’s Law of Partial Pressures
Dalton’s Law of Partial PressuresFor a mixture of gases in a container, the total pressure
exerted is the sum of the partial pressures of the gases present
The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container.
Ptotal = P1 + P2 + P3
Each gas is responsible for only part of the total pressure.
Dalton’s Law of Partial PressuresWhat is the atmospheric pressure if the partial pressure
of nitrogen, oxygen, and argon are 604.5 mm Hg, 162.8 mm Hg, and 0.5 mm Hg, respectively?
PT = PN2 + PO2
+ PAr
PT = 604.5 mm Hg + 162.8 mm Hg + 0.5 mm Hg
PT = 767.8 mm Hg
Dalton’s Law of Partial PressuresA gas mixture contains hydrogen, helium, neon, and argon. The total pressure of the mixture is 93.6 kPa. The partial pressures of helium, neon, and argon are 15.4 kPa, 25.7 kPa, and 35.6 kPa, respectively. What
is the pressure exerted by hydrogen?
PT = PH2 + PHe + PNe + PAr
93.6 kPa = PH2 + 15.4 kPa + 25.7 kPa + 35.6 kPa
PH2 = 93.6 – (15.4 + 25.7 +35.6) = 16.9 kPa
Dalton’s Law of Partial PressuresMixtures of helium and oxygen are used in the “air” tanks
of underwater divers for deep dives. For a particular dive, 12 L of O2 at 25oC and 1.0 atm and 46 L of He at
25oC and 1.0 atm were pumped into a 5.0-L tank. Calculate the partial pressure of each gas and the
total pressure in the tank at 25oC.
Dalton’s Law of Partial PressuresO2
• V1 = 12 L
• P1 = 1.0 atm
• T1 = 298 K
He• V1 = 46 L
• P1 = 1.0 atm
• T1 = 298 K
Mixture• V2 = 5.0 L
• T2 = 298 K
• P2 = ?
Since only V and P are changing, use Boyle’s Law to solve for the pressure of each
gas. (Do TWO Boyle’s Law problems to solve for the final pressure of each gas in the
mixture!)
O2
P1V1 = P2V2
(1.0 atm)(12 L) = P2 (5.0 L)
P2 = 2.4 atm(partial pressure of oxygen in the mixture)
HeP1V1 = P2V2
(1.0 atm)(46 L) = P2 (5.0 L)
P2 = 9.2 atm(partial pressure of oxygen in the mixture)
PT = 2.4 atm + 9.2 atm = 11.6 atm
Understanding the Laws
What do we want to know?What are the characteristics of the
individual gas particles that cause a gas to behave like it does?
Remember: Laws tell us what happens, not why. Scientists develop models or theories to explain
why nature behaves the way it does.
The Kinetic Molecular Theory of Gases (KMT)
• A simple model that attempts to explain the behavior of an ideal gas
• Based on the behavior of the individual particles (atoms or molecules) in a gas
Assumptions/Postulates of the KMT of Ideal Gases
1. Gases consist of tiny particles (atoms or molecules).2. These particles are so small, compared with the
distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).
3. The particles are in constant random motion, colliding with the walls of the container. These collisions cause the pressure exerted by the gas.
4. The particles are assumed not to attract or repel one another. In other words, the particles don’t interact.
5. The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas.
Postulates True for Real Gases?1. Yes, all gases are made up of atoms or molecules.2. Fair assumption, but not perfect. The particles are in fact
very small, but their volumes are not zero. 3. Yes, all gases have particles in constant, random motion
that produce pressure.4. Depends on the gas. Some gases have stronger
intermolecular forces than others.5. Yes, increased temperature does increase motion of the
particles in a real gas.
The KMT of Gases• Temperature reflects how rapidly the individual
particles of a gas are moving.• High temps = fast movement = lots of collisions with walls
of container• Low temps = slow movement = fewer collisions with walls
of container• Consider a container with an expandable volume.
What will happen if the external pressure remains the same and the temperature of the gas is increased?• Higher temp = faster movement = more collisions with the
walls of the container• Increased pressure on walls increase in volume• Proves Charles’s and Gay-Lussac’s Laws!