unit 9 - states of matter and gas behavior chemistry chapters 12-13

Unit 9 - States of Matter and Gas Behavior

Chemistry Chapters 12-13

The Behavior of GasesThe Behavior of Gases

• Kinetic-molecular theory explains the different properties of solids, liquids, and gases.

• Atomic composition affects physical and chemical properties.

• The kinetic-molecular theory describes the behavior of matter in terms of particles in motion.

• Kinetic-molecular theory explains the different properties of solids, liquids, and gases.

• Atomic composition affects physical and chemical properties.

• The kinetic-molecular theory describes the behavior of matter in terms of particles in motion.

Gases expand, diffuse, exert pressure, and can be compressed because they are in a low density state consisting of tiny, constantly-moving particles.


• Gases consist of small particles separated by empty space.

• Gas particles are too far apart to experience significant attractive or repulsive forces.

• Gas particles are in constant random motion.• An elastic collision is one in which no kinetic

energy is lost.• Kinetic energy of particles is directly

proportional to temperature

• Gases consist of small particles separated by empty space.

• Gas particles are too far apart to experience significant attractive or repulsive forces.

• Gas particles are in constant random motion.• An elastic collision is one in which no kinetic

energy is lost.• Kinetic energy of particles is directly

proportional to temperature

• Kinetic energy of a particle depends on mass and velocity.

• Temperature is a measure of the average kinetic energy of the particles in a sample of matter.

• Kinetic energy of a particle depends on mass and velocity.

• Temperature is a measure of the average kinetic energy of the particles in a sample of matter.



• Great amounts of space exist between gas particles.

• Compression reduces the empty spaces between particles.

• Great amounts of space exist between gas particles.

• Compression reduces the empty spaces between particles.

Gas PressureGas Pressure

• Pressure is defined as force per unit area.• Gas particles exert pressure when they

collide with the walls of their container.• The particles in the earth’s atmosphere exert

pressure in all directions called air pressure.• There is less air pressure at high altitudes

because there are fewer particles present, since the force of gravity is less.

• Pressure is defined as force per unit area.• Gas particles exert pressure when they

collide with the walls of their container.• The particles in the earth’s atmosphere exert

pressure in all directions called air pressure.• There is less air pressure at high altitudes

because there are fewer particles present, since the force of gravity is less.


• Barometers are instruments used to measure atmospheric air pressure.

• Barometers are instruments used to measure atmospheric air pressure.


• The average height of a mercury column in a barometer at sea level is 760 mm (760 mmHg)

• There are other units of pressure as well with the following relationships:• 1.00 atm = 760 mmHg = 760 torr =

101.3 kPa

• The average height of a mercury column in a barometer at sea level is 760 mm (760 mmHg)

• There are other units of pressure as well with the following relationships:• 1.00 atm = 760 mmHg = 760 torr =

101.3 kPa


• Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases of the mixture.

• The partial pressure of a gas depends on the number of moles, size of the container, and temperature and is independent of the type of gas.

• Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases of the mixture.

• The partial pressure of a gas depends on the number of moles, size of the container, and temperature and is independent of the type of gas.


• Ptotal = P1 + P2 + P3 +…Pn

• Partial pressure can be used to calculate the amount of gas produced in a chemical reaction.

• Ptotal = P1 + P2 + P3 +…Pn

• Partial pressure can be used to calculate the amount of gas produced in a chemical reaction.


• Gases are often collected by water displacement, but water vapor is introduced to the gas sample (mixture of gases)

• You can use Dalton’s law to get the partial pressure of the gas without the water vapor

• Gases are often collected by water displacement, but water vapor is introduced to the gas sample (mixture of gases)

• You can use Dalton’s law to get the partial pressure of the gas without the water vapor

Forces of AttractionForces of Attraction

• Attractive forces between molecules cause some materials to be solids, some to be liquids, and some to be gases at the same temperature.

• Attractive forces between molecules cause some materials to be solids, some to be liquids, and some to be gases at the same temperature.

Intermolecular forces—including dispersion forces, dipole-dipole forces, and hydrogen bonds—determine a substance’s state at a given temperature.


Prefix INTRA means “within”, where INTER means “between”


• Dispersion forces are weak forces that result from temporary shifts in the density of electrons in electron clouds.

• Weakest of all intermolecular forces• Dispersion forces become increasingly

stronger as molecule gets more electrons, which explains why iodine is a solid but fluorine is a gas at room temperature

• Dispersion forces are weak forces that result from temporary shifts in the density of electrons in electron clouds.

• Weakest of all intermolecular forces• Dispersion forces become increasingly

stronger as molecule gets more electrons, which explains why iodine is a solid but fluorine is a gas at room temperature


• Dipole-dipole forces are attractions between oppositely charged regions of polar molecules.

• Dipole-dipole forces are attractions between oppositely charged regions of polar molecules.


• Hydrogen bonds are special dipole-dipole attractions that occur between molecules that contain a hydrogen atom bonded to a small, highly electronegative atom with at least one lone pair of electrons, typically fluorine, oxygen, or nitrogen.

• Hydrogen bonds are special dipole-dipole attractions that occur between molecules that contain a hydrogen atom bonded to a small, highly electronegative atom with at least one lone pair of electrons, typically fluorine, oxygen, or nitrogen.

Liquids and SolidsLiquids and Solids

• Liquids and solids are called “condensed phases” because their particles are very close together and do not have the energy necessary to “escape” the sample…their motion is limited

• Liquids and solids are called “condensed phases” because their particles are very close together and do not have the energy necessary to “escape” the sample…their motion is limited

The particles in solids and liquids have a limited range of motion and are not easily compressed.


• Solid particles are arranged in a crystal arrangement where they are fixed in position.

• This arrangement results in a substance that has a defined shape and volume.

• Solid particles are arranged in a crystal arrangement where they are fixed in position.

• This arrangement results in a substance that has a defined shape and volume.


• Liquid particles have enough energy to “slide” by other particles, but do not have the energy to escape the sample itself.

• This results in a substance that has a fixed volume but takes the shape of its container.

• Liquid particles have enough energy to “slide” by other particles, but do not have the energy to escape the sample itself.

• This results in a substance that has a fixed volume but takes the shape of its container.

Phase ChangesPhase Changes

• Heat is the transfer of energy from an object at a higher temperature to an object at a lower temperature.

• Heat is the transfer of energy from an object at a higher temperature to an object at a lower temperature.

Matter changes phase when energy is added or removed.


• When ice is heated, the ice eventually absorbs enough energy to break the hydrogen bonds that hold the water molecules together.

• When the hydrogen bonds break, the particles move apart and ice melts into water.

• The melting point of a crystalline solid is the temperature at which the forces holding the crystal lattice together are broken and it becomes a liquid.

• When ice is heated, the ice eventually absorbs enough energy to break the hydrogen bonds that hold the water molecules together.

• When the hydrogen bonds break, the particles move apart and ice melts into water.

• The melting point of a crystalline solid is the temperature at which the forces holding the crystal lattice together are broken and it becomes a liquid.


• Vaporization is the process by which a liquid changes to a gas or vapor.

• Evaporation is vaporization only at the surface of a liquid.

• Vaporization is the process by which a liquid changes to a gas or vapor.

• Evaporation is vaporization only at the surface of a liquid.


• In a closed container, the pressure exerted by a vapor over a liquid is called vapor pressure.

• In a closed container, the pressure exerted by a vapor over a liquid is called vapor pressure.


• The boiling point is the temperature at which the vapor pressure of a liquid equals the atmospheric pressure.

• The boiling point is the temperature at which the vapor pressure of a liquid equals the atmospheric pressure.


• Evaporation vs Boiling Summary:• Evaporation is the result of surface

level molecules obtaining enough energy to escape the sample as a gas

• Boiling occurs throughout the sample when the vapor pressure of the liquid is the same as the atmospheric pressure

• Evaporation vs Boiling Summary:• Evaporation is the result of surface

level molecules obtaining enough energy to escape the sample as a gas

• Boiling occurs throughout the sample when the vapor pressure of the liquid is the same as the atmospheric pressure

Phase ChangesPhase Changes• Sublimation is the process by which a solid changes into a gas

without becoming a liquid.• As heat flows from water to the surroundings, the particles lose

energy.• The freezing point is the temperature at which a liquid is

converted into a crystalline solid.• As energy flows from water vapor, the velocity decreases.

• The process by which a gas or vapor becomes a liquid is called condensation.

• Deposition is the process by which a gas or vapor changes directly to a solid, and is the reverse of sublimation.

• Sublimation is the process by which a solid changes into a gas without becoming a liquid.

• As heat flows from water to the surroundings, the particles lose energy.

• The freezing point is the temperature at which a liquid is converted into a crystalline solid.

• As energy flows from water vapor, the velocity decreases.

• The process by which a gas or vapor becomes a liquid is called condensation.

• Deposition is the process by which a gas or vapor changes directly to a solid, and is the reverse of sublimation.


• A phase diagram is a graph of pressure versus temperature that shows in which phase a substance will exist under different conditions of temperature and pressure.

• A phase diagram is a graph of pressure versus temperature that shows in which phase a substance will exist under different conditions of temperature and pressure.


• The triple point is the point on a phase diagram that represents the temperature and pressure at which all three phases of a substance can coexist.

• The critical point is the highest temperature the substance can exist as a liquid, regardless of pressure

• Normal melting point and Normal boiling point are the temperatures at which these phase changes occur under standard pressure (760 torr)

• The triple point is the point on a phase diagram that represents the temperature and pressure at which all three phases of a substance can coexist.

• The critical point is the highest temperature the substance can exist as a liquid, regardless of pressure

• Normal melting point and Normal boiling point are the temperatures at which these phase changes occur under standard pressure (760 torr)

Gas LawsGas LawsFor a fixed amount of gas, a change in one variable—pressure, temperature, or volume—affects the other two.

Gas LawsGas Laws• Boyle’s law states that the volume of a fixed amount

of gas held at a constant temperature varies inversely with the pressure.

• Boyle’s law states that the volume of a fixed amount of gas held at a constant temperature varies inversely with the pressure.

P1V1 = P2V2 where P = pressure and V = volume

Gas LawsGas Laws

• Example Problem. A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature is constant?

• Example Problem. A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature is constant?

Gas LawsGas Laws

• Givens and Unknown• P1=0.947 atm

• V1=150. mL

• P2=0.987 atm

• V2=?

• Givens and Unknown• P1=0.947 atm

• V1=150. mL

• P2=0.987 atm

• V2=?

Gas LawsGas Laws• Equation P1V1=P2V2

• Solve for unknown, V2

• Equation P1V1=P2V2

• Solve for unknown, V2

P1V1P2

=P2V2

P2

V2 =P1V1

P2

V2 =(0.947atm)(150.mL)

0.987atm=144mL

Gas LawsGas Laws

• As temperature increases, so does the volume of gas when the amount of gas and pressure do not change.

• Kinetic-molecular theory explains this property.

• As temperature increases, so does the volume of gas when the amount of gas and pressure do not change.

• Kinetic-molecular theory explains this property.

Gas LawsGas Laws

Gas LawsGas Laws

• Celsius and Fahrenheit use ARBITRARY zero points, where any temperature lower than these points are considered to be negative

• Absolute zero is zero on the Kelvin scale.• Remember - temperature is a measure of

kinetic energy of the particles. Particles CAN NOT HAVE less kinetic energy than zero. Therefore, at absolute zero, all motion ceases.

• Celsius and Fahrenheit use ARBITRARY zero points, where any temperature lower than these points are considered to be negative

• Absolute zero is zero on the Kelvin scale.• Remember - temperature is a measure of

kinetic energy of the particles. Particles CAN NOT HAVE less kinetic energy than zero. Therefore, at absolute zero, all motion ceases.

Gas Laws Gas Laws

• Conversions:• Kelvin T (K) = Celsius T (C) + 273• Celsius T (C) = Kelvin T (K) - 273

• Conversions:• Kelvin T (K) = Celsius T (C) + 273• Celsius T (C) = Kelvin T (K) - 273

Gas LawsGas Laws

• Charles’s law states that the volume of a given amount of gas is directly proportional to its kelvin temperature at constant pressure.

• Charles’s law states that the volume of a given amount of gas is directly proportional to its kelvin temperature at constant pressure.

Gas LawsGas Laws

• IMPORTANT! Your temperatures MUST BE IN KELVIN when using Charles’ Law (or any gas law).

• A rearrangement of the Charles’ Law equation that is often easier to use (no fraction) is V1T2=V2T1

• IMPORTANT! Your temperatures MUST BE IN KELVIN when using Charles’ Law (or any gas law).

• A rearrangement of the Charles’ Law equation that is often easier to use (no fraction) is V1T2=V2T1

Gas LawsGas Laws

• Example - A sample of oxygen gas has a volume of 150. mL when its temperature is 295 K. What will the volume of the gas be at a temperature of 309 K if the pressure is constant?

• Example - A sample of oxygen gas has a volume of 150. mL when its temperature is 295 K. What will the volume of the gas be at a temperature of 309 K if the pressure is constant?

Gas LawsGas Laws

• Givens and Unknown• V1=150. mL

• T1=295 K

• V2=?

• T2=309 K

• Givens and Unknown• V1=150. mL

• T1=295 K

• V2=?

• T2=309 K

Gas LawsGas Laws• Equation V1T2=V2T1

• Solve for V2

• Equation V1T2=V2T1

• Solve for V2

V1T2T1

=V2T1

T1

V2 =V1T2

T1

V2 =(150.mL)(309K )

295K=157mL

Gas LawsGas Laws

• Example # 2 - A sample of oxygen gas has a volume of 277 mL when its temperature is 52C. If the gas volume is increased to 344 mL, at what Celsius temperature will the gas be?

• Example # 2 - A sample of oxygen gas has a volume of 277 mL when its temperature is 52C. If the gas volume is increased to 344 mL, at what Celsius temperature will the gas be?

Gas LawsGas Laws

• Givens and Unknown• V1=277 mL

• T1=52C - convert to K - 52 + 273 = 325 K

• V2=344 mL

• T2=?

• Givens and Unknown• V1=277 mL

• T1=52C - convert to K - 52 + 273 = 325 K

• V2=344 mL

• T2=?

Gas LawsGas Laws• Equation - V1T2=V2T1

• Solve for T2

• Equation - V1T2=V2T1

• Solve for T2V1T2V1

=V2T1

V1

T2 =V2T1

V1

T2 =(344mL)(325K )

277mL=404K

404K - 273 = 131C

Gas LawsGas Laws

• Gay-Lussac’s law states that the pressure of a fixed amount of gas varies directly with the kelvin temperature when the volume remains constant.

• Gay-Lussac’s law states that the pressure of a fixed amount of gas varies directly with the kelvin temperature when the volume remains constant.

Gas LawsGas Laws

Gas LawsGas Laws

• The combined gas law states the relationship among pressure, temperature, and volume of a fixed amount of gas.

• We can rearrange the equation to remove the fraction and get P1V1T2=P2V2T1

• The combined gas law states the relationship among pressure, temperature, and volume of a fixed amount of gas.

• We can rearrange the equation to remove the fraction and get P1V1T2=P2V2T1

Gas LawsGas Laws• Example: A balloon is inflated to 20.0 L at 31

deg C and 89.0 kPa. Determine the new volume of the balloon if it is brought to conditions of 4 deg C and 104.4 kPa.

• P1 = 89.0 kPa• V1 = 20.0 L• T1 = 31 deg C + 273 = 304 K• P2 = 104.4 kPa• V2 = ?• T2 = 4 deg C + 273 = 277 K

• Example: A balloon is inflated to 20.0 L at 31 deg C and 89.0 kPa. Determine the new volume of the balloon if it is brought to conditions of 4 deg C and 104.4 kPa.

• P1 = 89.0 kPa• V1 = 20.0 L• T1 = 31 deg C + 273 = 304 K• P2 = 104.4 kPa• V2 = ?• T2 = 4 deg C + 273 = 277 K

Gas LawsGas Laws

• P1V1T2 = P2V2T1…solve for V2

• V2 = P1V1T2/P2T1

• P1V1T2 = P2V2T1…solve for V2

• V2 = P1V1T2/P2T1

V2 =(89.0kPa)(20.0L)(277K )

(104.4kPa)(304K )=15.5L

The Ideal Gas LawThe Ideal Gas Law

• Avogadro’s principle states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

• Avogadro’s principle states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

The ideal gas law relates the number of particles to pressure, temperature, and volume.


• The molar volume of a gas is the volume 1 mol occupies at 0.00°C and 1.00 atm of pressure.

• 0.00°C and 1.00 atm are called standard temperature and pressure (STP).

• At STP, 1 mol of gas occupies 22.4 L.

• The molar volume of a gas is the volume 1 mol occupies at 0.00°C and 1.00 atm of pressure.

• 0.00°C and 1.00 atm are called standard temperature and pressure (STP).

• At STP, 1 mol of gas occupies 22.4 L.


• Ideal gas particles occupy a negligible volume and are far enough apart to exert minimal attractive or repulsive forces on each other.

• Combined gas law to ideal gas law

• Ideal gas particles occupy a negligible volume and are far enough apart to exert minimal attractive or repulsive forces on each other.

• Combined gas law to ideal gas law


• The ideal gas constant is represented by R and is 0.08206 L•atm/mol•K when pressure is in atmospheres. It is expressed as 8.314 L•kPa/mol•K when pressure is in kilopascals

• The ideal gas law describes the physical behavior of an ideal gas in terms of pressure, volume, temperature, and amount.

• We call the constant “R”

• The ideal gas constant is represented by R and is 0.08206 L•atm/mol•K when pressure is in atmospheres. It is expressed as 8.314 L•kPa/mol•K when pressure is in kilopascals

• The ideal gas law describes the physical behavior of an ideal gas in terms of pressure, volume, temperature, and amount.

• We call the constant “R”


• To rearrange this equation to get rid of the fraction, we get the most common expression of the law

• PV = nRT• n = number of moles

• To rearrange this equation to get rid of the fraction, we get the most common expression of the law

• PV = nRT• n = number of moles


• Remember! The ideal gas law DOES NOT deal with changing conditions. Given three of the four variables that describe a gas, you can solve for the fourth!

• Remember! The ideal gas law DOES NOT deal with changing conditions. Given three of the four variables that describe a gas, you can solve for the fourth!


• Sample Problem - A sample of carbon dioxide with a mass of 0.250 g was placed in a 350 mL container at 127 deg C. What is the pressure exerted by the gas in kPa?

• P = ?• V = 350 mL x (1L/1000mL) = 0.35 L• N = 0.250 g x (1 mole/44.01g) = 0.0568

mol• R = 8.314 LkPa/(Kmol)• T = 127 C + 273 = 400. K

• Sample Problem - A sample of carbon dioxide with a mass of 0.250 g was placed in a 350 mL container at 127 deg C. What is the pressure exerted by the gas in kPa?

• P = ?• V = 350 mL x (1L/1000mL) = 0.35 L• N = 0.250 g x (1 mole/44.01g) = 0.0568

mol• R = 8.314 LkPa/(Kmol)• T = 127 C + 273 = 400. K


• PV=nRT, divide by V• P = nRT/V

• PV=nRT, divide by V• P = nRT/V

P =(0.0568mol)(8.314L • kPa)(400K )

(0.35L)(K • mol)=540kPa

Ideal Gas LawIdeal Gas Law

• You Try: An oxygen sample has a volume of 4.0 L at 37C and 1.086 atm. Calculate the number of moles of oxygen in the sample.

• You Try: An oxygen sample has a volume of 4.0 L at 37C and 1.086 atm. Calculate the number of moles of oxygen in the sample.

Ideal Gas LawIdeal Gas Law

• P = 1.086 atm• V = 4.0 L• N = ?• R = 0.08206 Latm/(Kmol)• T=310 K• PV=nRT, n=PV/RT

• P = 1.086 atm• V = 4.0 L• N = ?• R = 0.08206 Latm/(Kmol)• T=310 K• PV=nRT, n=PV/RT

n =(1.086atm)(4.0L)(K • mol)(0.08206L • atm)(310K )

=0.17mol

Real GasesReal Gases

• Ideal gases follow the assumptions of the kinetic-molecular theory.

• Ideal gases experience:– There are no intermolecular attractive or repulsive forces

between particles or with their containers.

– The particles are in constant random motion.

– Collisions are perfectly elastic.

– No gas is truly ideal, but most behave as ideal gases at a wide range of temperatures and pressures.

• Ideal gases follow the assumptions of the kinetic-molecular theory.

• Ideal gases experience:– There are no intermolecular attractive or repulsive forces

between particles or with their containers.

– The particles are in constant random motion.

– Collisions are perfectly elastic.

– No gas is truly ideal, but most behave as ideal gases at a wide range of temperatures and pressures.

Real GasesReal Gases

• Real gases deviate most from ideal gases at high pressures and low temperatures.

• Polar molecules have larger attractive forces between particles.

• Polar gases do not behave as ideal gases.

• Large nonpolar gas particles occupy more space and deviate more from ideal gases.

• Real gases deviate most from ideal gases at high pressures and low temperatures.

• Polar molecules have larger attractive forces between particles.

• Polar gases do not behave as ideal gases.

• Large nonpolar gas particles occupy more space and deviate more from ideal gases.

Gas StoichiometryGas Stoichiometry

• The gas laws can be applied to calculate the stoichiometry of reactions in which gases are reactants or products.

• The gas laws can be applied to calculate the stoichiometry of reactions in which gases are reactants or products.

When gases react, the coefficients in the balanced chemical equation represent both molar amounts and relative volumes.


2H2(g) + O2(g) → 2H2O(g)

• 2 mol H2 reacts with 1 mol O2 to produce 2 mol water vapor.

• Coefficients in a balanced equation represent volume ratios for gases.

2H2(g) + O2(g) → 2H2O(g)

• 2 mol H2 reacts with 1 mol O2 to produce 2 mol water vapor.

• Coefficients in a balanced equation represent volume ratios for gases.


• Solving Volume to Volume Problems• Volume of a gas is DIRECTLY

PROPORTIONAL to the number of moles.• Since volume is dependant on

temperature, pressure, and moles, and NOT TYPE OF GAS, then a VOLUME RATIO can be used INSTEAD of a mole ratio.

• Solving Volume to Volume Problems• Volume of a gas is DIRECTLY

PROPORTIONAL to the number of moles.• Since volume is dependant on

temperature, pressure, and moles, and NOT TYPE OF GAS, then a VOLUME RATIO can be used INSTEAD of a mole ratio.


3 H2 (g) + N2 (g) 2 NH3 (g)

If 2.0 L of hydrogen reacts with excess nitrogen, calculate the volume of ammonia produced in L at 20 deg C and 102.0 kPa.

3 H2 (g) + N2 (g) 2 NH3 (g)

If 2.0 L of hydrogen reacts with excess nitrogen, calculate the volume of ammonia produced in L at 20 deg C and 102.0 kPa.2.0Lhydrogen1

x2Lammonia

3Lhydrogen=1.3Lammonia


• Solving other types of Gas Stoichiometry Problems• Unless Volume - Volume, you must

always use a MOLE RATIO.• YOU GOTTA BE IN MOLES TO USE

THAT MOLE RATIO!• GOT MOLES?

• Solving other types of Gas Stoichiometry Problems• Unless Volume - Volume, you must

always use a MOLE RATIO.• YOU GOTTA BE IN MOLES TO USE

THAT MOLE RATIO!• GOT MOLES?

Gas StoichiometryGas Stoichiometry• When given the volume of a gas, use the ideal gas law to

find MOLES of the gas.

2 Cu2S + 3 O2 2 Cu2O + 2 SO2

What mass of copper (I) sulfide is required to react completely with 46.6 L of Oxygen at 25°C and 103.56 kPa?

• When given the volume of a gas, use the ideal gas law to find MOLES of the gas.

2 Cu2S + 3 O2 2 Cu2O + 2 SO2

What mass of copper (I) sulfide is required to react completely with 46.6 L of Oxygen at 25°C and 103.56 kPa?

n=PVRT

n=(103.56kPa)(46.6L)(Kxmol)

(8.314LxkPa)(298K)=1.95moloxygen

1.95moloxygen1

x2molcopper(I )sulfide

3moloxygenx159.17gcopper(I )sulfide

1molcopper(I )sulfide

=207g


• When given an amount other than volume of a gas and asked to SOLVE for volume of a gas, first solve for MOLES of the gas (using stoichiometry relationships), then use ideal gas law to solve for volume.

• When given an amount other than volume of a gas and asked to SOLVE for volume of a gas, first solve for MOLES of the gas (using stoichiometry relationships), then use ideal gas law to solve for volume.


CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

What volume of oxygen gas at 29°C at 1.06 atm is needed for the combustion of 6.25 g of methane?

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

What volume of oxygen gas at 29°C at 1.06 atm is needed for the combustion of 6.25 g of methane?

6.25gmethane1

x1molmethane16.042g

x2moloxygen1molmethane

=0.779moloxygen

V =nRTP

=(0.779mol)(0.08206Lgatm)(302K )

(1.06atm)(Kgmol)=18.2L


• When you are at STP, you can use the molar volume of a gas and avoid using the ideal gas law altogether!2 H2O (l) 2 H2 (g) + O2 (g)

If 25.0g of water is decomposed, what volume of oxygen gas is produced at STP?

• When you are at STP, you can use the molar volume of a gas and avoid using the ideal gas law altogether!2 H2O (l) 2 H2 (g) + O2 (g)

If 25.0g of water is decomposed, what volume of oxygen gas is produced at STP?25.0gwater

1x

1molwater18.016gwater

x1moloxygen2molwater

x22.4Loxygen1moloxygen

=15.5Loxygen

Remember! Molar volume of any gas at STP = 22.4 L/mol!

unit 9 - states of matter and gas behavior chemistry chapters 12-13

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