Gas Laws. The 4 Gas Law Variables 1. Volume (V) 2. Temperature (T) 3. Pressure (P) 4. Amount of Gas (n)

Gas LawsGas Laws

The 4 Gas Law VariablesThe 4 Gas Law Variables

1. Volume (V)1. Volume (V)

2. Temperature (T)2. Temperature (T)

3. Pressure (P)3. Pressure (P)

4. Amount of Gas (n)4. Amount of Gas (n)

VOLUMEVOLUME

Whatever units of volume are used, use them all Whatever units of volume are used, use them all the way through the problemthe way through the problem

• Symbol = VSymbol = V

• The 3-dimensional space enclosed by The 3-dimensional space enclosed by the walls of a container is the the walls of a container is the volumevolume

• Usually measured in liters (L) or Usually measured in liters (L) or milliliters (mL)milliliters (mL)

TemperatureTemperature

Kelvin = celcius + 273 (32 Kelvin = celcius + 273 (32 ooC = 305 K)C = 305 K)

Standard Temperature is defined as 0 Standard Temperature is defined as 0 ooC or 273 KC or 273 K

• Symbol = TSymbol = T

• All gases have a temperature, usually All gases have a temperature, usually

measured in degrees Celsius= measured in degrees Celsius= ooCC

• Also measured in Kelvins = K note Also measured in Kelvins = K note there is no degree symbolthere is no degree symbol

• ALL GAS PROBLEMS WILL BE DONE IN KELVINSALL GAS PROBLEMS WILL BE DONE IN KELVINS

Conversion PracticeConversion Practice

Convert 25 Convert 25 oo

C to KelvinC to Kelvin

25 + 273 = 298 K

Convert 375 K to Convert 375 K to oo

CC

Convert -50 Convert -50 oo

C to KelvinC to Kelvin

375 – 273 = 102 oo

CC

-50 + 273 = 223 K

PressurePressure

Unfortunately all 4 will be usedUnfortunately all 4 will be used

Gas PressureGas Pressure (P) is created by the (P) is created by the molecules of a gas colliding with the walls of molecules of a gas colliding with the walls of a container.a container.• 4 units of pressure4 units of pressure

1. atmosphere (atm)1. atmosphere (atm)2. millimeters of Mercury (mmHg)2. millimeters of Mercury (mmHg)

3. kiloPascals (kPa)3. kiloPascals (kPa)4. torr4. torr

Pressure cont….Pressure cont….

Standard Pressure is defined as Standard Pressure is defined as 1 atm1 atm or or 760 mm Hg760 mm Hg or or 101.325 kPa 101.325 kPa oror 760 torr 760 torr

Standard Temperature and Pressure is a Standard Temperature and Pressure is a common chemistry phrase. It will be common chemistry phrase. It will be abbreviated STPabbreviated STP

Pressure Units ConversionsPressure Units Conversions *Remember**Remember*

1 atm = 760 mm Hg = 101.325 kPa1 atm = 760 mm Hg = 101.325 kPa

Convert 0.875 atm to mm HgConvert 0.875 atm to mm Hg 0.875 atm x 0.875 atm x 760 mm Hg 760 mm Hg 1 atm 1 atm

Convert 745 mm Hg to atmConvert 745 mm Hg to atm 745 mm Hg x 745 mm Hg x 1 atm 1 atm 760 mm Hg760 mm Hg

Convert 0.955 atm to kPaConvert 0.955 atm to kPa 0.955 atm x 0.955 atm x 101.325 kPa 101.325 kPa 1 atm1 atm

..

= 665 mm Hg

= 0.980 atm

= 96.8 kPa

Amount of GasAmount of Gas

Symbol= n Symbol= n (notice it lower case, all other symbols are upper case)(notice it lower case, all other symbols are upper case)

Amount of gas could be measured in moles or Amount of gas could be measured in moles or grams. If it is in grams you will have to grams. If it is in grams you will have to convert to moles at some point.convert to moles at some point.

Boyles LawBoyles Law: : The Pressure-Volume The Pressure-Volume RelationshipRelationship• Boyles Law shows the inverse Boyles Law shows the inverse

relationship between the pressure and relationship between the pressure and the volume of a gas. the volume of a gas. (Keeping T and n constant)(Keeping T and n constant)

A.A. If the volume of a container is increased, the pressure If the volume of a container is increased, the pressure decreasesdecreases

B.B. If the volume of a container is decreased, the pressure If the volume of a container is decreased, the pressure increasesincreases

P V or P VP V or P V

Boyles LawBoyles Law: : The Pressure-Volume The Pressure-Volume RelationshipRelationship

P V or P VP V or P V

WHY ? ? ? ? ?WHY ? ? ? ? ?

• If the volume of the container is increased, If the volume of the container is increased, gas molecules have to go further to collide gas molecules have to go further to collide with the wall. This means less pressure with the wall. This means less pressure b/c of less collisions.b/c of less collisions.

• If the volume is decreased, gas molecules If the volume is decreased, gas molecules have to go a shorter distance to strike the have to go a shorter distance to strike the walls, therefore, more collisions, increasing walls, therefore, more collisions, increasing the pressurethe pressure


• The mathematical form of Boyle's Law is: The mathematical form of Boyle's Law is: PV = kPV = k P = pressure V= volume k= a constantP = pressure V= volume k= a constant

• This means that the pressure-volume product will This means that the pressure-volume product will always be the same value if the temperature and always be the same value if the temperature and amount of gas remain constantamount of gas remain constant

• This is an This is an inverse mathematical relationshipinverse mathematical relationship. As . As one quantity goes up in the value, the other goes one quantity goes up in the value, the other goes downdown


• Suppose Suppose P1P1 and and V1V1 are a pressure- are a pressure-volume pair of data at the start of an volume pair of data at the start of an experiment. When you multiply experiment. When you multiply PP and and VV together, you get a number that is called together, you get a number that is called kk. .

• Now, if the volume is changed to a new Now, if the volume is changed to a new value called value called V2V2, then the pressure will , then the pressure will spontaneously change to spontaneously change to P2P2. It will do so . It will do so because the because the PV product must always PV product must always equal kequal k. .

Boyles LawBoyles Law: : The Pressure-Volume The Pressure-Volume RelationshipRelationship• So we know this: PSo we know this: P11VV11 = k = k

• And we know that the second data pair And we know that the second data pair equals the same constant: Pequals the same constant: P22VV22 = k = k

• Since k = k, we can conclude that PSince k = k, we can conclude that P11VV11 = = PP22VV22..

• This equation of This equation of PP11VV11 = P = P22VV22 will be very will be very helpful in solving Boyle's Law problems.helpful in solving Boyle's Law problems.

Boyles LawBoyles Law: : The Pressure-Volume The Pressure-Volume RelationshipRelationshipExample #1:Example #1:

2.00 L of a gas is at 740.0 mmHg 2.00 L of a gas is at 740.0 mmHg pressure. What is its volume at standard pressure. What is its volume at standard pressure?pressure?

** remember standard pressure = 760.0 mmHg**** remember standard pressure = 760.0 mmHg**

Use the equation PUse the equation P11VV11 = P = P22VV22 2.00 L x 740.0 mmHg = 760.0 mmHg x V2

V2 = 1.95 L


Example #2:Example #2:

5.00 L of a gas is at 1.08 atm. What 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume pressure is obtained when the volume is 10.0 L?is 10.0 L?

5.00 L x 1.08 atm = P5.00 L x 1.08 atm = P2 2 x 10.0 L x 10.0 L

Boyles LawBoyles Law: : The Pressure-Volume The Pressure-Volume RelationshipRelationshipExample #3:Example #3: 2.50 L of a gas was at an unknown pressure. 2.50 L of a gas was at an unknown pressure.

However, at standard pressure (101.325 However, at standard pressure (101.325 kPa), its volume was measured to be 8.00 kPa), its volume was measured to be 8.00 L. What was the unknown pressure?L. What was the unknown pressure?

insert into insert into PP11VV11 = P = P22VV22 for the solution for the solution..

Charles Law:The Volume-Temperature Relationship

This law gives the relationship between volume and temperature if pressure and amount are held constant

A. If the temperature increases, the volume of a gas is increased

B. If the temperature is decreased, the volume of a gas is decreases.


V T or V T

WHY ? ? ? Temperature increase causes molecules to

move faster, which makes more collisions with the wall of the container, which increases the volume


Charles' Law is a direct mathematical relationship

This means there are two connected values and when one goes up, the other also increases


The mathematical form of Charles' Law is: V / T = k

Let V1 and T1 be a volume-temperature pair of data at the start of an experiment. If the volume is changed to a new value called V2, then the temperature must change to T2.

The new volume-temperature data pair will preserve the value of k. The two different volume-temperature data pairs equal the same value and that value is called k.


So we know this: V1/T1 = k

AND

And we know this: V2/T2 = k

Since k = k, we can conclude that


Example #1:

A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature?


Example #1: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its

volume at standard temperature?

Answer: Convert 25.0°C to Kelvin and you get 298 K. Standard

temperature is 273 K. We plug into our equation like this:


Example #2:

4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C?


Example #2: 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling

to 25.0°C?

Answer: Convert 50.0°C to 323 K and 25.0°C to 298 K. Then plug

into the equation and solve for x, like this:


Example #3: 5.00 L of a gas is collected at 100 K and then

allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure?


Example #3: 5.00 L of a gas is collected at 100 K and then allowed to expand to

20.0 L. What must the new temperature be in order to maintain the same pressure?

Answer:

Gay-Lussac’s LawGay-Lussac’s LawPressure-Temperature RelationshipPressure-Temperature Relationship

► The relationship between pressure The relationship between pressure and temperature when volume and and temperature when volume and amount are held constant amount are held constant

A.A. If the temperature of a container is increased, the If the temperature of a container is increased, the pressure increases.pressure increases.

B.B. If the temperature of a container is decreased, the If the temperature of a container is decreased, the pressure decreases.pressure decreases.


P T OR P TP T OR P T

WHY ? ? ? ?WHY ? ? ? ? If Temperature is increased, that will cause If Temperature is increased, that will cause

the molecules to move faster and create the molecules to move faster and create more collisions between molecules and the more collisions between molecules and the wall of the container.wall of the container.


►Gay-Lussac's Law is a Gay-Lussac's Law is a direct direct mathematical relationship. mathematical relationship.

►The mathematical form of Gay-The mathematical form of Gay-Lussac's Law is: Lussac's Law is: P ÷ T = kP ÷ T = k

► the pressure-temperature fraction will the pressure-temperature fraction will always be the same value if the volume and always be the same value if the volume and amount remain constantamount remain constant


► Let PLet P11 and T and T11 be a pressure-temperature pair be a pressure-temperature pair of data at the start of an experiment. If the of data at the start of an experiment. If the temperature is changed to a new value temperature is changed to a new value called Tcalled T22, then the pressure will change to , then the pressure will change to PP22. .

Since k = k, we can conclude that Since k = k, we can conclude that

PP11 ÷ T ÷ T11 = P = P22 ÷ T ÷ T22

Gay-Lussac’s LawGay-Lussac’s LawPressure-Temperature RelationshipPressure-Temperature Relationship Example #1: Example #1:

10.0 L of a gas is found to exert 97.0 kPa 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required at 25.0°C. What would be the required temperature (in Celsius) to change the temperature (in Celsius) to change the pressure to standard pressure?pressure to standard pressure?

Gay-Lussac’s LawGay-Lussac’s LawPressure-Temperature RelationshipPressure-Temperature Relationship Example #1: Example #1:

10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the be the required temperature (in Celsius) to change the pressure to standard pressure?pressure to standard pressure?

►change 25.0°C to 298.0 K and change 25.0°C to 298.0 K and remember that standard pressure in remember that standard pressure in kPa is 101.325. Insert values into the kPa is 101.325. Insert values into the equationequation


►Example #2: Example #2:

5.00 L of a gas is collected at 22.0°C 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the and 745.0 mmHg. When the temperature is changed to standard, temperature is changed to standard, what is the new pressure?what is the new pressure?


►Example #2: Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 5.00 L of a gas is collected at 22.0°C and 745.0

mmHg. When the temperature is changed to mmHg. When the temperature is changed to standard, what is the new pressure?standard, what is the new pressure?


►Example #2: Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 5.00 L of a gas is collected at 22.0°C and 745.0

mmHg. When the temperature is changed to mmHg. When the temperature is changed to standard, what is the new pressure?standard, what is the new pressure?

= 689.4 mmHg= 689.4 mmHg

Avogadro's Law Avogadro's Law Amount of Gas-Volume Amount of Gas-Volume RelationshipRelationship

the relationship between volume and the relationship between volume and amount when pressure and temperature are amount when pressure and temperature are held constant held constant

Remember amount is measured in moles Remember amount is measured in moles


If the amount of gas in a container is If the amount of gas in a container is increased, the volume increases.increased, the volume increases.

n Vn V

If the amount of gas in a container is If the amount of gas in a container is decreased, the volume decreasesdecreased, the volume decreases

n Vn V


WHY ? ? ?WHY ? ? ? If we increase the amount of gas (n) then If we increase the amount of gas (n) then

we have increased the # of molecules. we have increased the # of molecules. More molecules means more collisions More molecules means more collisions which in-turn means more volumewhich in-turn means more volume


The mathematical form of Avogadro's The mathematical form of Avogadro's Law is: Law is: V ÷ n = kV ÷ n = k

Let VLet V11 and n and n11 be a volume-amount pair of be a volume-amount pair of

data at the start of an experiment. If the data at the start of an experiment. If the amount is changed to a new value called amount is changed to a new value called nn22, then the volume will change to V, then the volume will change to V22. .


We know this: We know this: VV11 ÷ n ÷ n1 1 = k= k

And we know this: And we know this: VV22 ÷ n ÷ n22 = k = k

Since k = k, we can conclude that Since k = k, we can conclude that

VV11 ÷ n ÷ n11 = V = V22 ÷ n ÷ n22


Example #1: Example #1: 5.00 L of a gas is known to contain 0.965 5.00 L of a gas is known to contain 0.965

mol. If the amount of gas is increased to mol. If the amount of gas is increased to 1.80 mol, what new volume will result 1.80 mol, what new volume will result


Example #1: Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is

increased to 1.80 mol, what new volume will result increased to 1.80 mol, what new volume will result

5.00 L5.00 L = = X X

0.965 mol 1.80 mol0.965 mol 1.80 mol


Example #1: Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is

increased to 1.80 mol, what new volume will result increased to 1.80 mol, what new volume will result

5.00 L5.00 L = = X X 0.965 mol 1.80 mol0.965 mol 1.80 mol

= 9.33 L= 9.33 L

The Ideal Gas LawThe Ideal Gas Law

Describes the behavior of a gas in Describes the behavior of a gas in terms of P, V, n and Tterms of P, V, n and T

This is a law the encompasses all the This is a law the encompasses all the other gas laws we have studied.other gas laws we have studied.


PV = nRTPV = nRT

The R in this equation is called the The R in this equation is called the gas constant.gas constant.

R = .0821 atm-L/ mol-KR = .0821 atm-L/ mol-K

atm-L/ mol-K is read as atmosphere atm-L/ mol-K is read as atmosphere liter per mole Kelvinliter per mole Kelvin


Example #1Example #1

How many moles of a gas at 100 How many moles of a gas at 100 oo

C C does it take to fill a 1.00 L flask to a does it take to fill a 1.00 L flask to a pressure of 1.50 atm?pressure of 1.50 atm?





V = 1.00 L P = 1.50 atmV = 1.00 L P = 1.50 atm

T = 100 T = 100 oo

C convert to K = 373 KC convert to K = 373 K

n = ? Remember R = n = ? Remember R = 0.08210.0821





Solve for n: n = PV/RTSolve for n: n = PV/RT

(1.50 atm)(1.00 L)(1.50 atm)(1.00 L)

(0.0821 atm-L/mol-K)(373K)(0.0821 atm-L/mol-K)(373K)





Solve for n: n = PV/RTSolve for n: n = PV/RT

(1.50 atm)(1.00 L)(1.50 atm)(1.00 L)

(0.0821 atm-L/mol-K)(373K)(0.0821 atm-L/mol-K)(373K)

= 0.0490 mol= 0.0490 mol



What is the volume occupied by What is the volume occupied by 9.45g of C9.45g of C22HH22 at STP? at STP?




First change grams to moles:First change grams to moles:

9.45g x 1 mol = 0.363 moles9.45g x 1 mol = 0.363 moles

26 g26 g




Next solve for V: V = nRTNext solve for V: V = nRT

P

R = 0.0821 atm-L/mol-K n = 0.363 moles

T = 273 K P = 1 atm




V = (0.363V = (0.363molmol)(0.0821)(0.0821atm-L/mol-Katm-L/mol-K)(273 )(273 KK))

1 1 atmatm

V = 8.14 LV = 8.14 L

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Gas Laws. The 4 Gas Law Variables 1. Volume (V) 2. Temperature (T) 3. Pressure (P) 4. Amount of Gas (n)