GasesGases

Chapter 13Chapter 13

Some basicsSome basicsGases have properties that are very Gases have properties that are very

different from solids and liquids.different from solids and liquids.They are very sensitive to changes in They are very sensitive to changes in

pressure and temperature and have low pressure and temperature and have low density.density.

Gases Gases dodo have mass( although small). have mass( although small).Think of properties of gases when your Think of properties of gases when your

ears pop while driving to Tahoe , deep-sea ears pop while driving to Tahoe , deep-sea divers or your kid sister’s birthday divers or your kid sister’s birthday balloons.balloons.

Pressure and TemperaturePressure and Temperature Pressure of gases is Pressure of gases is

usually denoted in usually denoted in atm or mmHg.atm or mmHg.

1atm =760 mmHg1atm =760 mmHg 1atm =760 torr1atm =760 torr Other units include Other units include

Pascal and psi.Pascal and psi. Measured by a Measured by a

barometer barometer

Temperature of gases Temperature of gases is measured in Kelvinis measured in Kelvin

Kelvin = Kelvin = °C +273°C +273 Other unit for Other unit for

temperature is temperature is Fahrenheit.Fahrenheit.

Measured by a Measured by a thermometer thermometer

Kinetic Molecular TheoryKinetic Molecular Theory

Volume of the molecules is negligible (zero) Volume of the molecules is negligible (zero) compared to the volume of the container.compared to the volume of the container.

The molecules are colliding with the walls of the container and this causes the pressure of the gas.

The molecules do not attract or repel each The molecules do not attract or repel each other.other.

The average kinetic energy of the molecules is directly proportional to the Kelvin temperature.

Boyle’s LawBoyle’s Law

When pressure of a gas is increased the volume of a gas When pressure of a gas is increased the volume of a gas decreases provided the temperature remains the same. decreases provided the temperature remains the same.

PV = kPV = k (constant temperature) (constant temperature)

Pressure and volume are inversely proportional Pressure and volume are inversely proportional

Or as P goes up, V must go down to keep the above Or as P goes up, V must go down to keep the above equation correct. equation correct.

P1V1 =P2V2 is a relationship that helps us solve problems is a relationship that helps us solve problems

when conditions are changed for a gas. when conditions are changed for a gas.

boyleslaw.mov

Sample problemSample problem What is the new pressure if 0.500L of oxygen at What is the new pressure if 0.500L of oxygen at

pressure 0.87atm is changed to 0.750L at constant pressure 0.87atm is changed to 0.750L at constant temperature?temperature?

Solution Solution Since this is at constant temperature, Boyle’s law Since this is at constant temperature, Boyle’s law

applies,applies,

P1V1 =P2V2

P1= 0.87atm, V1 = 0.500L, P2 = ?,V2 = 0.750L

P2 = P1V1 = 0.87atm X 0.500 L = 0.58atm

V2 0.750 L

Practice problemsPractice problems

The volume of a sample of hydrogen is 250.mL at 3.5 The volume of a sample of hydrogen is 250.mL at 3.5 atm pressure. What will be the volume when the atm pressure. What will be the volume when the pressure is reduced to 0.75atm, assuming that pressure is reduced to 0.75atm, assuming that temperature remains constant?temperature remains constant?

1200mL or 1.2 L

The pressure of a 2.34L sample of helium is 785 torr. The pressure of a 2.34L sample of helium is 785 torr. Calculate the pressure in atm if volume is decreased to Calculate the pressure in atm if volume is decreased to 2.04L and temperature kept constant.2.04L and temperature kept constant.

1.18 atm

Charles’s LawCharles’s Law

According to Charles’s Law, the volume of a gas According to Charles’s Law, the volume of a gas increases when temperature increases, provided the increases when temperature increases, provided the pressure is kept constant.pressure is kept constant.

V = k or when temperature increases, volume must or when temperature increases, volume must TT increase as wellincrease as well

VV11 = = VV22 all temperatures have to be in KELVIN scale!!!!

TT11 T T22

This law can now be used to solve problemsThis law can now be used to solve problems

Sample problemSample problem

The volume of a gas at 25The volume of a gas at 25ººC is 234mL. What will its C is 234mL. What will its volume be at 50volume be at 50ººC if pressure is kept constant?C if pressure is kept constant?

SolutionSolution

2525ººC + 273 = 298K = TC + 273 = 298K = T11 , V , V11 = 234mL = 234mL

5050ººC + 273 = 323K = TC + 273 = 323K = T22 , , VV22 = ? = ?

VV11 = = VV22 , V , V22 = = VV11 X T X T2 2 = 234mL X 323K = 254mL TT11 T T2 2 TT1 1 298K298K

Practice problemsPractice problemsA gas occupies 670mL at 45A gas occupies 670mL at 45ººC. At what C. At what

temperature (in temperature (in ººC) will it occupy 750mL if C) will it occupy 750mL if pressure is kept unchanged? pressure is kept unchanged?

8383ººCC

If methane gas occupies 58.0L at 17If methane gas occupies 58.0L at 17ººC, C, what volume will it occupy at 27what volume will it occupy at 27ººC if C if pressure is left the same? pressure is left the same?

60.L60.L

Avogadro’s LawAvogadro’s Law

When the number of moles of gas When the number of moles of gas increases, the volume increases too increases, the volume increases too (duh!!)(duh!!)

VV = k = k (at constant temperature and (at constant temperature and nn pressure) pressure)

VV11= = VV22

nn11 n n22

This above equation can be used to This above equation can be used to solve problems.solve problems.

ProblemsProblems If 6 moles of oxygen occupies 23L how If 6 moles of oxygen occupies 23L how

many L will 3.67moles occupy at constant many L will 3.67moles occupy at constant temperature and pressure?temperature and pressure?

VV11 = = VV22

nn11 n n22

VV22 = = VV11 X n X n22 = = 23L X 3.67moles23L X 3.67moles = = 14L14L

nn11 6 moles 6 moles

Ideal GasIdeal Gas

An ideal gas is one whose molecules are An ideal gas is one whose molecules are not attracted to each other and the volume not attracted to each other and the volume occupied by each molecule is too small to occupied by each molecule is too small to matter.matter.

Most gases behave ideally at Most gases behave ideally at high high temperatures and low pressures.temperatures and low pressures.

Gases are not ideal when compressed or Gases are not ideal when compressed or cooled down.cooled down.

Ideal Gas LawIdeal Gas Law If we combine all the three gas laws If we combine all the three gas laws

(Boyle’s, Charles’s and Avogadro's) we get (Boyle’s, Charles’s and Avogadro's) we get the ideal gas law:the ideal gas law:

PV =nRTPV =nRT where R =0.08206 where R =0.08206 L.atmL.atm K.molK.molIf any three of the properties are known the If any three of the properties are known the

fourth can be calculated.fourth can be calculated.This equation has some limits, can be used only in This equation has some limits, can be used only in

low pressures and high temperatures. If the low pressures and high temperatures. If the pressure goes up or temperature decreases, pressure goes up or temperature decreases, corrections need to be applied.corrections need to be applied.

ProblemsProblemsWhat volume is occupied by 0.250mol of What volume is occupied by 0.250mol of

COCO2 2 at 25at 25°°C and 371torr?C and 371torr?PV =nRTPV =nRT

P = 371/760 = 0.488atmP = 371/760 = 0.488atm

T = 298K R =0.08206 T = 298K R =0.08206 L.atmL.atm

K.molK.mol

V= V= nRTnRT = = .250mol X 298K X .08206 .250mol X 298K X .08206 L.atmL.atm

P 0.488 atmP 0.488 atm K.mol K.mol

= = 12.5L

PracticePracticeA 1.5mol of radon gas has a volume of 21.0L A 1.5mol of radon gas has a volume of 21.0L

at 33at 33°°C. What is the pressure of the gas?C. What is the pressure of the gas?1.8atmWhat is the mass of oxygen needed to fill a What is the mass of oxygen needed to fill a

tank of volume 22.7L at temperature 34tank of volume 22.7L at temperature 34°C °C and pressure 1.5 atm? and pressure 1.5 atm?

43.3g

Combined Gas LawCombined Gas LawWe can use the combined gas law if the We can use the combined gas law if the

conditions of pressure, volume or conditions of pressure, volume or temperature for the same amount of gas temperature for the same amount of gas are altered.are altered.

P1V1 = P2V2

T1 T2

Temperature must be in Kelvin !!!!

Sample problemSample problemA sample of neon gas has a volume of 27.5mL A sample of neon gas has a volume of 27.5mL

at 22.0at 22.0°°C and 740. torr pressure. What will its C and 740. torr pressure. What will its volume be at temperature 15.0volume be at temperature 15.0°°C and pressure C and pressure 755 torr?755 torr?

PP11VV11 = = PP22VV22

TT11 T T22

740.torr X 27.5mL740.torr X 27.5mL = = 755torr X 755torr X VV22

295K 288K295K 288K

VV22==26.3mL26.3mL

Dalton’s Law of partial pressuresDalton’s Law of partial pressuresAccording to Dalton’s law the pressure of According to Dalton’s law the pressure of

a mixture of gases is related to the number a mixture of gases is related to the number of moles of the gas in the mixture.of moles of the gas in the mixture.

PPtotal total = p= p11 +p +p22 + p + p33 +….. +…..

nntotal total = n= n11 + n + n22 + n + n33 + ….. + …..

PPtotal total V =nV =ntotaltotalRTRT

ProblemProblem A 6.00L tank contains 32.0g of oxygen gas and A 6.00L tank contains 32.0g of oxygen gas and

18.02g of water vapor. At 2018.02g of water vapor. At 20°°C what is the total C what is the total pressure of the tank?pressure of the tank?

32.0g = 1.00mol O32.0g = 1.00mol O22; 18.02g = 1.00mol H; 18.02g = 1.00mol H22OO

nntotaltotal = 1.00 + 1.00 = 2.00mol = 1.00 + 1.00 = 2.00mol

PPtotal total = = nntotaltotalRTRT = = 2.00* 0.0821* 293K2.00* 0.0821* 293K = = 8.01atm

V 6.00L

PracticePractice

If a 23.0L tank contains 3.00mol of HIf a 23.0L tank contains 3.00mol of H22, ,

2.00mol of He and 1.00mol of Ne gases, 2.00mol of He and 1.00mol of Ne gases, calculate the total pressure of the tank at calculate the total pressure of the tank at 4040°°C? Calculate the partial pressure of C? Calculate the partial pressure of each gas as well.each gas as well.

Ptotal = 6.7atm

pH2 = 3.35atm

pHe = 2.23atm

pNe = 1.12atm

Gas StoichiometryGas Stoichiometry First write a balanced chemical equation.First write a balanced chemical equation. Calculate moles of givenCalculate moles of given Calculate moles of askedCalculate moles of asked Calculate volume of gas producedCalculate volume of gas produced

MOLES OF GIVEN MOLES OF ASKEDn

Grams of Given

V=nRT P

Mole ratio

Sample problemSample problem Calculate the volume of NHCalculate the volume of NH3 produced when 43g produced when 43g

of Nof N2 reacts with excess H reacts with excess H2 at 29 at 29°°C and 0.997atm C and 0.997atm

pressure.pressure.

NN2 + 3HH2 2NHNH3

43g X 1molN2 X 2mol NH3 = 3.07mol NHNH3

28.02g 1molN2

V = nRT = 3.07mol* 0.0821* 302K = 76. L

P 0.997atm

Practice problemPractice problemCalculate the volume of oxygen produced Calculate the volume of oxygen produced

from the decomposition of 6.00g of KClOfrom the decomposition of 6.00g of KClO33

to KCl and Oto KCl and O22 at 30.0 at 30.0°°C and 100.kPa C and 100.kPa

pressure.pressure.1.85L

Graham’s Law of DiffusionGraham’s Law of Diffusion

According to Graham’s law, the rates of diffusion of two gases is inversely proportional to the square root of their molar masses. In other words, lighter gases travel faster under similar conditions of temperature.

RateA = √molar massB

RateB √ molar massA

PracticePractice Find the molar mass of a gas that diffuses Find the molar mass of a gas that diffuses

0.31 times as fast as oxygen gas.0.31 times as fast as oxygen gas. If oxygen travels at a rate of 1, then the If oxygen travels at a rate of 1, then the

other gas travels at 0.31,other gas travels at 0.31,

Rate Rate OO22 = = √molar mass√molar mass gas gas

Rate Rate gasgas √molar mass √molar mass OO22

1 1 = = √molar mass√molar mass gas gas

0.31 √32.0g/mol0.31 √32.0g/mol

molar mass molar mass gasgas = 330 g/mol = 330 g/mol

Suppose a gas diffuses 1.41 times faster Suppose a gas diffuses 1.41 times faster than sulfur dioxide. What is the molar than sulfur dioxide. What is the molar mass of this gas?mass of this gas?

RateRategasgas = =√64.0g/mol√64.0g/mol

RateRateSOSO2 2 √molar mass√molar massgasgas

1.41 = 1.41 = √64.0g/mol√64.0g/mol

√ √ molar massmolar massgasgas

molar mass molar mass gasgas = = 32.2g/mol32.2g/mol