Chapter 11 Gas Laws. The Gas Phase Gases have no distinct volume or shape. Gases expand to fill the...

Chapter 11 Gas Laws

The Gas Phase• Gases have no distinct volume or shape.• Gases expand to fill the volume of their

container.

• Gas particles are miscible with each other.

• Evidence for gas particles being far apart : We can see through gases We can walk through gases Gases are compressible Gases have low densities

Composition of Earth’s AtmosphereCompound %(Volume) Mole Fractiona

Nitrogen 78.08 0.7808

Oxygen 20.95 0.2095

Argon 0.934 0.00934

Carbon dioxide 0.033 0.00033

Methane 2 x 10-4 2 x 10-6

Hydrogen 5 x 10-5 5 x 10-7

a. mole fraction = mol component/total mol in mixture.

The Air We Breathe

Kinetic Theory of GasesKinetic Theory Postulates:

• Gas particles are sizeless relative to the volume of the gas

• Gas particles are in constant rapid motion

• Gas particles have elastic collisions; means no kinetic energy is lost on impact.

• The absolute temperature is directly proportional to the kinetic energy of a gas.

• Gas particles have no attraction to each other; i.e. no inter particle froces.

Parameters Affecting Gases

• Pressure (P); atm, mmHg, torr, lbs/in2

• Volume (V); L, mL

• Temperature (T); K (only)

• Number of Moles (n)

PressurePressure is equal to force/unit area (P =F/A) lbs/in2

Force is a push which comes from gas particles striking a container wall

Pressure Units• SI units = Newton/meter2 = 1 Pascal (Pa)• 1 standard atmosphere (atm) = 101,325 Pa

1 atm =760 mm Hg 1 atm = 760 torr (torr is abbreviation of mmHg) 1 atm = 14.7 lbs/in2

1 atm = 1.013 barr Barr = 100 kPa

Measurement of Pressure

What is above mercury?

Measurement of Pressure

Elevation and Atmospheric Pressure

Units for Expressing Pressure

Unit Value

Atmosphere 1 atm

Pascal (Pa) 1 atm = 1.01325 x 105 Pa

Kilopascal (kPa) 1 atm = 101.325 kPa

mmHg 1 atm = 760 mmHg

Torr 1 atm = 760 torr

Bar 1 atm = 1.01325 bar

mbar 1 atm = 1013.25 mbar

psi 1 atm = 14.7 psi

Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas

in the canister pushing harder?

= 15 mm

Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas

in the canister pushing harder? Gas in the canister

If the atmospheric pressure is 766 mm, then what is the pressure of the canister?

= 15 mm

Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas in

the canister pushing harder? Gas in the canister

If the atmospheric pressure is 766 mm, then what is the pressure of the canister?P = 766 + 15 = 781 mm (torr)

= 15 mm

gas

Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas

in the canister pushing harder?

gas

Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas

in the canister pushing harder? The atmosphere

What is the pressure of the gas if the atmosphere is 766 mm?

= 13 mm

gas

Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas

in the canister pushing harder? The atmosphere

What is the pressure of the gas if the atmosphere is 766 mm? 753 mm

= 13 mm

gas

Pressure MeasurementOpen Tube Manometer

gas

Now what is pushing harder, the gas or the atomosphere?

Pressure MeasurementOpen Tube Manometer

gas

Now what is pushing harder, the gas or the atmosphere?Neither, both the same.

Pressure MeasurementOpen Tube Manometer

gas

Now what is pushing harder, the gas or the atmosphere?Neither, both the same.

Is the gas canister empty?

Pressure MeasurementOpen Tube Manometer

gas

Now what is pushing harder, the gas or the atmosphere?Neither, both the same.

Is the gas canister empty?No, completely full of gas!

Dalton’s Law of Partial Pressures

• For a mixture of gases in a container

• PTotal = P1 + P2 + P3 + . . .

Boyles Law

Atm

●

● ●

●

Consider a gas in a closed system containing a movable plunger. If the plunger is not moving up or down, what can be said about the pressure of the gas relative to the atmospheric pressure?

Boyles Law

Atm

●

● ●

●

Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same?

●●

●

Boyles Law

Atm

●

● ●

●

Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same?

More particles, more collisions, and more pressure.

What happens to the plunger?

●●

●

Boyles Law

Atm

●

● ●

●

Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same?

More particles, more collisions, and more pressure.

What happens to the plunger?

●●

●

Boyles Law

●●

●

●●●

●●

●●

The number of particles remain the same, but the surface area they have to strike increases, thus the number of collisions per square inch decrease as the plunger goes up exposing more surface area causing a decrease in pressure.

Boyle’s Law

• P 1/V (T and n fixed)

• P V = Constant

• P1V1 = P2V2

Pressure and volume are inversely proportional.

Charles’s Law• The volume of a gas is

directly proportional to Kelvin temperature, and extrapolates to zero at zero Kelvin.

V T (P & n are constant)

V1 = V2

T1 T2

Combined Gas Law• Combining the gas laws the relationship

P T(n/V) can be obtained.

• If n (number of moles) is held constant, then PV/T = constant.

P1V1

T1

= P2V2

T2

Volume: L, mL, cm3, …Pressure: Atm, mmHg, Torr, PSI, KPaTemperature, K (only)

Example A balloon is filled with hydrogen to a A balloon is filled with hydrogen to a

pressure of 1.35 atm and has a volume of pressure of 1.35 atm and has a volume of 2.54 L. If the temperature remains constant, 2.54 L. If the temperature remains constant, what will the volume be when the pressure is what will the volume be when the pressure is increased to 2.50 atm?increased to 2.50 atm?

P1V1

T1

= P2V2

T2

(1.35 atm)(2.54 L)T1

= (2.50atm)V2T1

Constant Temp. means T1=T2

(1.35 atm)(2.54 L)(2.50atm)V2 =

V2 = 1.37 L

Example

A sample of oxygen gas is at 0.500 atm and occupies a volume of 11.2 L at 00C, what volume will the gas occupy at 6.00 atm at room temperature (250C)?

Ideal Gas Law

PV = nRT

R = universal gas constant = 0.08206 L atm K-1 mol-1

P = pressure in atm V = volume in liters n = moles T = temperature in Kelvin

STP• “STP” means standard temperature and standard

pressure P = 1 atmosphere T = 0C The molar volume of an ideal gas is 22.42 liters

at STP (put 1 mole, 1 atm, R, and 273 K in the ideal gas law and calculate V)

Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of

methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.

Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of

methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.

0.0821 L-atmMole-K

Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of

methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.

0.0821 L-atmMole-K 3.3 L

Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of

methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.

0.0821 L-atmMole-K 3.3 L

298 K

Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of

methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.

0.0821 L-atmMole-K 3.3 L

298 K 1.2 mole

Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of

methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.

0.0821 L-atmMole-K 3.3 L

298 K 1.2 mole = 8.9 atm

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

mole-K0.0821 L-atm

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

mole-K0.0821 L-atm 0.495 g

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

mole-K0.0821 L-atm 0.495 g

10-3 LmL

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

mole-K0.0821 L-atm 0.495 g

10-3 LmL

127 mL

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

mole-K0.0821 L-atm 0.495 g

10-3 LmL

127 mL760 torratm

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

mole-K0.0821 L-atm 0.495 g

10-3 LmL

127 mL760 torratm 754 torr

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

mole-K0.0821 L-atm 0.495 g

10-3 LmL

127 mL760 torratm 754 torr

371 K

Example

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

mole-K0.0821 L-atm 0.495 g

10-3 LmL

127 mL760 torratm 754 torr

371 K

= 120 g/mole

Collecting a Gas Over Water

A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?

Hint: The gas collected is a mixture so use Hint: The gas collected is a mixture so use Dalton’s Law to calculate the pressure of oxygen Dalton’s Law to calculate the pressure of oxygen then the ideal gas law to find the number of then the ideal gas law to find the number of moles oxygen.moles oxygen.

PT = P + P

Practice

O2 H2O

Vapor Pressure of Water

A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?

Hint: The gas collected is a mixture so use Hint: The gas collected is a mixture so use Dalton’s Law to calculate the pressure of oxygen Dalton’s Law to calculate the pressure of oxygen then the ideal gas law to find the number of then the ideal gas law to find the number of moles oxygen.moles oxygen.

PT = P + P

Practice

O2 H2O

755 torr = PO2+ 23.8 torr

P = 755 – 23.8 = 731 torr O2

Practice• A sample of KClO3 is heated and decomposes to

produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?

mole-K0.0821 L-atm

Practice• A sample of KClO3 is heated and decomposes to

produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?

mole-K0.0821 L-atm

atm760 torr

Practice• A sample of KClO3 is heated and decomposes to

produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?

mole-K0.0821 L-atm

atm760 torr

731 torr

Practice• A sample of KClO3 is heated and decomposes to

produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?

mole-K0.0821 L-atm

atm760 torr

731 torr298 K

Practice• A sample of KClO3 is heated and decomposes to

produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?

mole-K0.0821 L-atm

atm760 torr

731 torr298 K mL

10-3 L

Practice• A sample of KClO3 is heated and decomposes to

produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?

mole-K0.0821 L-atm

atm760 torr

731 torr298 K mL

10-3 L 229 mL

= 9.00 X 10-3 mole

Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

13.4 g Zn

Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

13.4 g Zn

65.39 g Zn

mole Zn

Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

13.4 g Zn

65.39 g Zn

mole Znmole Znmole H2

Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

13.4 g Zn

65.39 g Zn

mole Znmole Znmole H2 0.08206 L-atm

mole H2-K

Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

13.4 g Zn

65.39 g Zn

mole Znmole Znmole H2 0.08206 L-atm

mole H2-K298.15 K

Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

13.4 g Zn

65.39 g Zn

mole Znmole Znmole H2 0.08206 L-atm

mole H2-K298.15 K 760 torr

Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

13.4 g Zn

65.39 g Zn

mole Znmole Znmole H2 0.08206 L-atm

mole H2-K298.15 K 760 torr

atm 766 torr

Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

13.4 g Zn

65.39 g Zn

mole Znmole Znmole H2 0.08206 L-atm

mole H2-K298.15 K 760 torr

atm 766 torr= 4.97 L

The End