Chapter 11 Gases 11.1 Gases & Pressure. Defining Gas Pressure How are number of particles and...

Chapter 11Chapter 11GasesGases

11.1 Gases & Pressure11.1 Gases & Pressure

Defining Gas PressureDefining Gas Pressure

• How are number of particles and How are number of particles and pressure related?pressure related?

• Pressure Pressure –force per unit area –force per unit area that particles exert on walls of that particles exert on walls of their containertheir container

• Gas particles collide with walls = Gas particles collide with walls = greater pressuregreater pressure

• Pressure is directly proportional Pressure is directly proportional to number of particles.to number of particles.

Nu

mb

er of P

articles Increases

Pressu

re Increases

Temperature & PressureTemperature & Pressure• Higher temperature results in more Higher temperature results in more

kinetic energy!kinetic energy!• IF the volume of container remains IF the volume of container remains

constant and IF the amount of gas constant and IF the amount of gas remains constant: remains constant:

the pressure of a gas increases in the pressure of a gas increases in direct proportion to the Kelvin direct proportion to the Kelvin temperaturetemperature..

(Kelvin Temp = Celsius Temp + 273)(Kelvin Temp = Celsius Temp + 273)• Volume of a gas at constant pressure is Volume of a gas at constant pressure is

directly proportional to Kelvin temp. directly proportional to Kelvin temp.

Pressu

re of G

as Increases

Kelvin

Tem

peratu

re Increases

Temperature ConversionsTemperature Conversions

• Kelvin & CelsiusKelvin & Celsius

TTkk = (T = (Tcc + 273) + 273)

TTcc = (T = (Tkk - 273) - 273)

• Fahrenheit & CelsiusFahrenheit & Celsius

TTff = (9/5 T = (9/5 Tcc) + 32) + 32

TTcc = (T = (Tff - 32) 5/9 - 32) 5/9

Devices to Measure PressureDevices to Measure Pressure

• BarometerBarometer: an instrument that : an instrument that measures pressure exerted by measures pressure exerted by the atmosphere.the atmosphere.

Invented in 1600’s by an Italian Invented in 1600’s by an Italian scientist, scientist,

Evangelista Torricelli Evangelista Torricelli • Height of column of Height of column of

mercury shows the mercury shows the atmospheric pressure. atmospheric pressure.

(atm)(atm)

Atmospheric PressureAtmospheric Pressure

• We live at the bottom of an We live at the bottom of an ocean of air; highest ocean of air; highest pressure occurs at the pressure occurs at the lowest altitudes!lowest altitudes!

• Standard Atmosphere is Standard Atmosphere is pressure that supports a pressure that supports a 760 mm column of mercury.760 mm column of mercury.

• 1.00 atm = 760 mm Hg1.00 atm = 760 mm Hg

Devices to Measure PressureDevices to Measure Pressure• Pressure GaugePressure Gauge::

instrument used to instrument used to measure pressure measure pressure inside a tire or oxygen inside a tire or oxygen tank.tank.

Tire Pressure

Blood Pressure

Absolute PressureAbsolute Pressure

• When measuring tire pressure; you measure When measuring tire pressure; you measure pressure pressure ABOVEABOVE atmosphere pressure. atmosphere pressure. Recommended pressures for tires are gauge Recommended pressures for tires are gauge pressures.pressures.

• Absolute pressureAbsolute pressure – the TOTAL pressure of – the TOTAL pressure of all gases including the atmosphere. all gases including the atmosphere.

Q:Q: How would you figure it for an inflated tire? How would you figure it for an inflated tire?A:A: Add barometric pressure to the gauge Add barometric pressure to the gauge

pressure.pressure.

Pressure UnitsPressure Units• SI unit for measuring pressure is the SI unit for measuring pressure is the

pascalpascal (pa) after the French physicist (pa) after the French physicist Blaise Pascal (1600s)Blaise Pascal (1600s)

• A A kilopascalkilopascal (kPa) is 1000 pascals and is (kPa) is 1000 pascals and is more commonly used.more commonly used.

Equivalent PressuresEquivalent Pressures

1.00 atm1.00 atm 101300 Pa101300 Pa 101.3 kPa101.3 kPa760 mm Hg760 mm Hg

760 Torr760 Torr14.7 psi14.7 psi

Sample CalculationsSample Calculations

1.1. Express 1.56 atm in kPa.Express 1.56 atm in kPa.

2.2. Convert 801 mm Hg to Pa.Convert 801 mm Hg to Pa.

3.3. How many psi are equivalent to 95.6 kPa?How many psi are equivalent to 95.6 kPa?

AnswersAnswers1)1) 1.56 atm1.56 atm

2)2) 801 mm Hg801 mm Hg

3)3) 95.6 kPa95.6 kPa

XX 101.3 kPa101.3 kPa

1.00 atm1.00 atm== 158 kPa158 kPa

XX 101300 Pa101300 Pa

760 mm Hg760 mm Hg== 107000 Pa107000 Pa

XX 14.7 psi14.7 psi

101.3 kPa101.3 kPa13.9 psi13.9 psi==

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

• Partial Pressure Partial Pressure

The pressure exerted by each gas in a mixture The pressure exerted by each gas in a mixture

The total pressure of a

The total pressure of a

mixture of gases is equal to

mixture of gases is equal to

the sum of the partial

the sum of the partial

pressures of the component

pressures of the component gasesgases

Dalton’s Law:Dalton’s Law:

PPTT = P = P11 + P + P22 + P + P33 + + ……

PracticePractice• Calculate the partial pressure in mm Hg Calculate the partial pressure in mm Hg

exerted by the four main gases in air at 760 exerted by the four main gases in air at 760 mm Hg: nitrogen, oxygen, argon and carbon mm Hg: nitrogen, oxygen, argon and carbon dioxide. Their abundance by volume is dioxide. Their abundance by volume is 78.08%, 20.95%, 0.934% and 0.035%, 78.08%, 20.95%, 0.934% and 0.035%, respectively. respectively.

NN22= 593.4 mm Hg= 593.4 mm Hg

OO22= 159.2 mm Hg= 159.2 mm Hg

Ar = 7.10 mm HgAr = 7.10 mm Hg

COCO22= 0.27 mm Hg= 0.27 mm Hg

Gases Collected by Water Gases Collected by Water DisplacementDisplacement

• Gases produced in the lab are often collected by Gases produced in the lab are often collected by the displacement of water in a collection bottle the displacement of water in a collection bottle

• Water vapor will be present in the collected gas, Water vapor will be present in the collected gas, and it exerts a pressure and it exerts a pressure

• Water vapor pressure = PWater vapor pressure = PH20H20

• Water vapor pressure increases with temperature Water vapor pressure increases with temperature (Appendix A, Table-8) (Appendix A, Table-8)

• Pressure of the dry gas Pressure of the dry gas

P P atmatm = P = P gasgas + P + P H20H20

so… P so… P gasgas = P = P atmatm – P – P H2OH2O

PracticePractice• A student has stored 100.0 mL of neon gas A student has stored 100.0 mL of neon gas

over water on a day when the temperature over water on a day when the temperature if 27.0 °C. If the barometer in the room if 27.0 °C. If the barometer in the room reads 743.3 mm Hg, what is the pressure of reads 743.3 mm Hg, what is the pressure of the neon gas in its container? the neon gas in its container?

P P atmatm = P = P NeNe + P + P H2OH2O

P P NeNe = P = P atmatm – P – P H2OH2O

P P NeNe = 743.3 mm Hg – 26.7 mm Hg = 743.3 mm Hg – 26.7 mm Hg

=716.6 mm Hg=716.6 mm Hg


11.2 The Gas Laws11.2 The Gas Laws

Pressure & VolumePressure & Volume• In the 1600s, Robert Boyle did many experiments In the 1600s, Robert Boyle did many experiments

involving gases.involving gases.• He did these experiments at constant temperature.He did these experiments at constant temperature.

if pressure increases, volume if pressure increases, volume decreasesdecreasesif pressure decreases, volume if pressure decreases, volume increasesincreases

Pressure & Pressure & Volume are Volume are Inversely Inversely

Proportional!Proportional!

Boyle’s Law GraphBoyle’s Law Graph

Boyle’s LawBoyle’s Law

VV11PP11=V=V22PP22

Where:Where: VV1 1 = initial volume= initial volume

PP11 = initial pressure = initial pressure

VV22 = final volume = final volume

PP22 = final pressure = final pressure

Kinetic Explanation of Boyle’s LawKinetic Explanation of Boyle’s Law

• As volume is reduced, As volume is reduced, number of particles number of particles and temperature and temperature remains constant but remains constant but number of collisions number of collisions with the walls of the with the walls of the container increases. container increases.

• There is a smaller area of space for the same There is a smaller area of space for the same number of particles to move around, so pressure number of particles to move around, so pressure increases.increases.

PracticePractice• If you have 5.5 L of gas at a pressure of If you have 5.5 L of gas at a pressure of

1.6 atm, and the pressure changes to 1.2 1.6 atm, and the pressure changes to 1.2 atm, what is your new volume?atm, what is your new volume?

VV11PP11 = V = V22PP22

(5.5 L) x (1.6 atm) = ((5.5 L) x (1.6 atm) = (xx L) x (1.2 atm) L) x (1.2 atm)

xx = 7.3 L = 7.3 L

Temperature & VolumeTemperature & Volume• Jacques Charles did experiments concerning gases Jacques Charles did experiments concerning gases

held at constant pressure, while varying held at constant pressure, while varying temperature.temperature.

Temperature & Temperature & Volume are Directly Volume are Directly

Proportional!Proportional!

As Kelvin temperature As Kelvin temperature increases, volume increases.

increases, volume increases.As Kelvin temperature As Kelvin temperature decreases, volume

decreases, volume decreases.decreases.

Charles’s Law GraphCharles’s Law Graph

Charles’s LawCharles’s Law

VV11 V V22

TT11 T T22

Where:Where: VV11 = initial volume = initial volume

TT11 = initial temperature = initial temperature

VV22 = final volume = final volume

TT22 = final temperature = final temperature

==

Kinetic Explanation of Charles’s LawKinetic Explanation of Charles’s Law

• When a gas is When a gas is heated, its heated, its temperature temperature increases, which increases, which means the kinetic means the kinetic energy of the energy of the particles has particles has increased. increased.

• Then the particles begin to move faster, which causes Then the particles begin to move faster, which causes its volume to increase. its volume to increase.

• The reverse occurs as the temperature begins to fall. The reverse occurs as the temperature begins to fall.

PracticePractice• 3.0 L of Helium gas is in a balloon at 22 3.0 L of Helium gas is in a balloon at 22 C and a C and a

pressure of 760 mm Hg. If the temperature rises to pressure of 760 mm Hg. If the temperature rises to 31 31 C and the pressure remains constant, what will C and the pressure remains constant, what will the new volume be?the new volume be?

(remember to convert any temperatures to KELVIN!!!)(remember to convert any temperatures to KELVIN!!!)

VV11 V V22

TT11 T T22

3.0 L3.0 L VV22

(273 + 22(273 + 22C) (273 + 31C) (273 + 31C )C )

VV22 = (3.0 L x 304 K) / 295 K = (3.0 L x 304 K) / 295 K

VV22 = 3.1 L = 3.1 L

==

==

Pressure & TemperaturePressure & Temperature• From the prior relationships of volume & pressure, From the prior relationships of volume & pressure,

and temperature & volume, it could be concluded and temperature & volume, it could be concluded that a relationship exists between pressure & that a relationship exists between pressure & temperature. temperature.

Pressure & Pressure & Temperature are Temperature are

Directly Directly Proportional!Proportional!

For a given mass of a dry

For a given mass of a dry gas, if the volume is gas, if the volume is constant, the pressure is

constant, the pressure is directly proportional to the

directly proportional to the Kelvin temperatureKelvin temperature

Gay-Lussac’s Law GraphGay-Lussac’s Law Graph

Gay-Lussac’s LawGay-Lussac’s Law

PP11 P P22

TT11 T T22

Where:Where: PP11 = initial pressure = initial pressure

TT11 = initial temperature = initial temperature

PP22 = final pressure = final pressure

TT22 = final temperature = final temperature

==

PracticePractice• At 27 At 27 C, Helium gas is in a balloon at C, Helium gas is in a balloon at

pressure of 760 mm Hg. If the temperature pressure of 760 mm Hg. If the temperature rises to 31 rises to 31 C, what will the new pressure be?C, what will the new pressure be?

(remember to convert any temperatures to KELVIN!!!)(remember to convert any temperatures to KELVIN!!!)

PP11 P P22

TT11 T T22

760 mm Hg P760 mm Hg P22

(273 + 27(273 + 27C) (273 + 31C) (273 + 31C )C )

PP22 = (760 mm Hg x 304 K) / 300 K = (760 mm Hg x 304 K) / 300 K

PP22 = 770 mm Hg = 770 mm Hg

==

==

Combined Gas LawCombined Gas Law

• All 3 Gas Laws require one variable to be held All 3 Gas Laws require one variable to be held constant. constant.

• How can we solve a problem when all 3 How can we solve a problem when all 3 variables; volume, pressure & temperature variables; volume, pressure & temperature change?change?

• Since 2 out of the 3 laws always have a Since 2 out of the 3 laws always have a variable in common, there should be a way to variable in common, there should be a way to relate these laws into one formula. relate these laws into one formula.

• This new formula is called the This new formula is called the Combined Gas Combined Gas LawLaw. .

Combined Gas LawCombined Gas Law

PP11 V V11 = P = P22 V V22

TT11 T T22

Where: Where: PP11, V, V11 & T & T11 are initial values are initial values

PP22, V, V22 & T & T22 are final values are final values

*0*0C & 1 atm = Standard Temperature & C & 1 atm = Standard Temperature & Pressure, or Pressure, or STPSTP

Boyle’s Boyle’s LawLaw

Charles’s Charles’s LawLaw

Gay-Gay-Lussac’s Lussac’s LawLaw

PracticePractice• 154 mL of Carbon Dioxide gas is at a pressure of 121 154 mL of Carbon Dioxide gas is at a pressure of 121

kPa and a temperature of 117kPa and a temperature of 117C. What volume would C. What volume would this gas occupy at STP? (Remember to convert your this gas occupy at STP? (Remember to convert your temps to Kelvin!!!)temps to Kelvin!!!)

1 atm = 101.3 kPa1 atm = 101.3 kPa

PP11VV11/ T/ T11 = P = P22VV22/ T/ T22

(154 mL)(121 kPa) = (101.3 kPa)(V(154 mL)(121 kPa) = (101.3 kPa)(V22) )

(117(117C + 273) (0C + 273) (0C + 273)C + 273)

VV22 = = (154 mL)(121 kPa)(273 K)(154 mL)(121 kPa)(273 K)

(390 K)(101.3 kPa)(390 K)(101.3 kPa)

VV22 = 129 mL = 129 mL


11.3 Gas Volumes & the Ideal 11.3 Gas Volumes & the Ideal Gas LawGas Law

The Law of Combining Gas VolumesThe Law of Combining Gas Volumes• If one volume of water, HIf one volume of water, H22O, is O, is

decomposed, one volume of decomposed, one volume of oxygen will be formed and 2 oxygen will be formed and 2 volumes of hydrogen will be volumes of hydrogen will be formed. formed.

• How can 3 volumes be formed How can 3 volumes be formed from only 1 initial volume?from only 1 initial volume?

1 L H2O

1 L O21 L H21 L H2

++

The Law of The Law of Combining Combining

Gas VolumesGas Volumes

• All of the gases are at the same temperature & pressure, All of the gases are at the same temperature & pressure, each of the identical flasks contains the same number of each of the identical flasks contains the same number of molecules. Notice how the combining ratio: molecules. Notice how the combining ratio: 2 volumes H2 volumes H22 : 1 volume O : 1 volume O22 : 2 volumes H : 2 volumes H22O leads to a O leads to a

result in which all the atoms present initially are accounted result in which all the atoms present initially are accounted for in the product. for in the product.

The law of combining volumes states that in

chemical reactions involving gases, the ratio of the gas volumes is a small whole number.

• Avogadro was the first to study this and concluded a Avogadro was the first to study this and concluded a water molecule is composed of particles. water molecule is composed of particles.

The Law of Combining Gas VolumesThe Law of Combining Gas Volumes

• We now know that a water We now know that a water molecule is composed of 2 molecule is composed of 2 hydrogen atoms & 1 oxygen atom. hydrogen atoms & 1 oxygen atom. When a molecule of water breaks When a molecule of water breaks down, it breaks down according to down, it breaks down according to the ratio of particles that compose the ratio of particles that compose it; 2 volumes of Hit; 2 volumes of H22 & 1 volume of & 1 volume of OO22 from 1 volume of H from 1 volume of H22O.O.

My principle states My principle states that equal volumes of

that equal volumes of gases at the same gases at the same temp & pressure temp & pressure contain equal contain equal numbers of particles.

numbers of particles.

• He reasoned that the volume of a He reasoned that the volume of a gas depends on the number of gas gas depends on the number of gas particles, provided the temperature particles, provided the temperature & pressure are constant.& pressure are constant.


• Under the same conditions of temperature Under the same conditions of temperature and pressure, the volumes of reacting gases and pressure, the volumes of reacting gases and their gaseous products are expressed and their gaseous products are expressed in ratios of small whole numbersin ratios of small whole numbers

2 L H2 L H22 + 1 L O + 1 L O22 → 2 L H → 2 L H22O (g) O (g)

2 volumes H2 volumes H22 + 1volume O + 1volume O22 → 2 volumes H → 2 volumes H22O (g) O (g)

1 volume H1 volume H22 + 1 volume Cl + 1 volume Cl22 → 2 volumes HCl → 2 volumes HCl

1 volume HCl + 1 volume NH1 volume HCl + 1 volume NH33 → NH → NH44Cl (s)Cl (s)


Avogadro’s LawAvogadro’s Law

For a gas at constant temperature and For a gas at constant temperature and pressure, the volume is directly pressure, the volume is directly proportional to the number of moles of gas proportional to the number of moles of gas (at low pressures).(at low pressures).

VV = = anan

aa = proportionality constant = proportionality constant

VV = volume of the gas = volume of the gas

nn = number of moles of gas = number of moles of gas

Standard Molar Standard Molar VolumeVolume

Equal volumes of all Equal volumes of all gases at the same gases at the same temperature and pressure temperature and pressure contain the same number contain the same number of moleculesof molecules. .

- Amedeo Avogadro- Amedeo Avogadro

Standard Molar VolumeStandard Molar Volume

PracticePractice• You are planning an experiment that requires You are planning an experiment that requires

0.0580 mol of nitrogen monoxide gas at STP. 0.0580 mol of nitrogen monoxide gas at STP. What volume would you need?What volume would you need?

0.0580 mol x 0.0580 mol x 22.4 L 22.4 L = 1.30 L= 1.30 L 1 mol1 mol

Gas StoichiometryGas StoichiometryVolume-Volume CalculationsVolume-Volume Calculations

• Assume: All products and reactants are Assume: All products and reactants are at the same temp and pressure at the same temp and pressure

• Unless otherwise stated, assume STP Unless otherwise stated, assume STP

• Solve by normal stoichiometric Solve by normal stoichiometric processes processes

• Volume ratios are the same as mole Volume ratios are the same as mole ratios ratios

Volume-Mass and Mass-Volume Volume-Mass and Mass-Volume CalculationsCalculations

• Order of Calculations Order of Calculations

• You are given a gas volume and asked You are given a gas volume and asked to find a mass:to find a mass:

gas volume A →moles A →moles B → mass B gas volume A →moles A →moles B → mass B

• You are given a mass and asked to find You are given a mass and asked to find a gas volume:a gas volume:

mass A → moles A →moles B →gas volume Bmass A → moles A →moles B →gas volume B

Ideal Gas LawIdeal Gas Law PVPV = = nRTnRTPP = pressure in atm = pressure in atmVV = volume in liters = volume in litersnn = moles = molesRR = proportionality constant = proportionality constant

= 0.08206 L∙ atm/ mol·= 0.08206 L∙ atm/ mol·For units of kPa, L & K:For units of kPa, L & K:

R R = 8.31 = 8.31 kPa ∙ LkPa ∙ L

Mol ∙ KMol ∙ KTT = temperature in Kelvin = temperature in Kelvin

Calculate the Value of RCalculate the Value of R• Use all standard values!Use all standard values!

P = 1 atmP = 1 atm

V = 22.4 LV = 22.4 L

n = 1 molen = 1 mole

T = 273 KT = 273 K

• Try substituting different standard Try substituting different standard pressures to obtain different values of Rpressures to obtain different values of R

PracticePractice• A 2.07 L cylinder contains 2.88 mol of helium A 2.07 L cylinder contains 2.88 mol of helium

gas at 22.0 °C. What is the pressure in gas at 22.0 °C. What is the pressure in atmospheres of the gas in the cylinder?atmospheres of the gas in the cylinder?

PV = nRTPV = nRTP = P = nRTnRT

VV= = 2.88 mol x 0.0821 (atm∙L/mol∙K) x 295 K2.88 mol x 0.0821 (atm∙L/mol∙K) x 295 K

2.07 L2.07 L= 33.7 atm= 33.7 atm

Gas DensityGas Density

molar mass

molar volume

massDensity

volume

so at STP…so at STP…

molar mass

22.4 LDensity

Variations on the Ideal Gas LawVariations on the Ideal Gas Lawn = mass (m)

molar mass (M)

If PV = nRT thenPV = mRT M

So replace n with m/M

M = mRT PV

So rearrange

for M

Variations on the Ideal Gas LawVariations on the Ideal Gas LawD = mass (m)

volume (V)

If M = mRT then VP

M = DRT P

So replace m / V with D

Density and the Ideal Gas LawDensity and the Ideal Gas Law

Combining the formula for density with the Ideal Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically:Gas law, substituting and rearranging algebraically:

MPD

RT

M = Molar MassM = Molar Mass

P = PressureP = Pressure

R = Gas ConstantR = Gas Constant

T = Temperature in KelvinT = Temperature in Kelvin

PracticePractice1)1) At 28°C and 0.974 atm, 1.00 L of a gas has a mass of At 28°C and 0.974 atm, 1.00 L of a gas has a mass of

5.16 g. What is the molar mass of this gas?5.16 g. What is the molar mass of this gas?

2)2) What is the molar mass of a gas if 0.427g of the gas What is the molar mass of a gas if 0.427g of the gas

occupies a volume of 125 mL at 20.0°C and 0.980 atm?occupies a volume of 125 mL at 20.0°C and 0.980 atm?

3)3) What is the density of a sample of ammonia gas if the What is the density of a sample of ammonia gas if the

pressure is 0.928 atm and the temp is 63.0°C?pressure is 0.928 atm and the temp is 63.0°C?

4)4) The density of a gas was found to be 2.0 g/L at 1.50 atm The density of a gas was found to be 2.0 g/L at 1.50 atm

and 27°C. What is the molar mass of the gas?and 27°C. What is the molar mass of the gas?

5)5) What is the density of argon gas at a pressure of 551 torr What is the density of argon gas at a pressure of 551 torr

and a temp of 25°C?and a temp of 25°C?

131 g/mol

83.8 g/mol

0.572 g/L

33 g/mol

1.18 g/L


11.4 Diffusion & Effusion11.4 Diffusion & Effusion

EffusionEffusionEffusion: describes the passage of gas into Effusion: describes the passage of gas into an evacuated chamber.an evacuated chamber.

Rate of effusion for gas 1Rate of effusion for gas 2

2

1

MM

Distance traveled by gas 1Distance traveled by gas 2

2

1

MM

Effusion:Effusion:

Diffusion:Diffusion:

Graham’s LawGraham’s LawRates of Effusion and DiffusionRates of Effusion and Diffusion

Graham’s LawGraham’s Law

• Density can replace molar mass in Density can replace molar mass in Graham’s formula, since density is Graham’s formula, since density is directly proportional to molar mass.directly proportional to molar mass.

• Isotopes of elements can be separated Isotopes of elements can be separated by vaporizing the element, and allowing by vaporizing the element, and allowing it to effuse. it to effuse.

• The heavier isotope effuses more The heavier isotope effuses more slowly than the lighter isotope slowly than the lighter isotope

PracticePractice• At 25 °C, the average velocity of oxygen molecules At 25 °C, the average velocity of oxygen molecules

is 420 m/s. What is the average velocity of helium is 420 m/s. What is the average velocity of helium atoms at the same temperature?atoms at the same temperature?

Rate of ORate of O22 is 420 m/s is 420 m/s = = √M√MHeHe

Rate of HeRate of He √M √MO2O2

420 m/s 420 m/s = = √4 g/mol√4 g/mol x √32 g/molx √32 g/mol

420 m/s 420 m/s = = 0.35360.3536 x 1x 1

x = 1188 m/s ≈ 1200 m/sx = 1188 m/s ≈ 1200 m/s

Chapter 11 Gases 11.1 Gases & Pressure. Defining Gas Pressure How are number of particles and...

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Transcript of Chapter 11 Gases 11.1 Gases & Pressure. Defining Gas Pressure How are number of particles and...