CHAPTER 10 GAS LAWS. THE LAWS: Find and define Boyle’s law Charles’ law Lussac’s law Avogadro’s law Combined gas law Ideal gas law Dalton’s law Graham’s.

CHAPTER 10

GAS LAWS

THE LAWS: Find and define

Boyle’s lawCharles’ lawLussac’s lawAvogadro’s lawCombined gas lawIdeal gas lawDalton’s lawGraham’s law

VARIABLES

PRESSURE (P) Defined as force per unit area Caused by collisions of gas molecules with the container

walls Measured using a barometer

Invented by Torricelli Measured atmospheric pressure at sea level using a

column of mercury which rose 760 mm Above sea level, atmospheric pressure decreases

Units and equivalents1 atmosphere = 760 mmHg = 760 Torr = 101325 Pascals

= 101.325 kPa = 29.92 inHg = 14.7 psi

VARIABLES

Example: Convert 30.08 inHg to atmospheres and pascals

?atm = 30.o8 in Hg × 1 atm = 1.005 atm

1 29.92 in Hg

?Pa = 30.08 in Hg × 101325 Pa = 101866 Pa

1 29.92 in Hgsig figs = 101900 Pa

VARIABLES

VOLUME (V) Units: liters Molar volume of a gas at STP = 22.4 L STP = 1 atm; 0˚C

TEMPERATURE (T) Units: Kelvin

Recall: Degrees C + 273.15 = Kelvin

NUMBER OF MOLES (n) Units: moles What if you’re given grams? Use MM

CONSTANT

THE UNIVERSAL GAS CONSTANTR Depends on pressure units.

R = 0.0821 L atm/ mol Kor

R = 8.314 L kPa/ mol K

COMBINED GAS LAW

P1V1 = P2V2T1 T2

Combines Boyle’s, Charles, and Lussac’s laws

Assumes a constant number of molesExample: 5.0 moles of gas are held in a flexible

container with an initial volume of 10.0 ml at 25.0 degrees C and 1.0 atm. If the pressure changes to 755 mmHg and the temperature is 295 K, what is the new volume of the container?

COMBINED GAS LAW

P1=1.00 atm P2=755 mm HgV1=10.0 mL V2=?T1=25.0˚C T2=295 K

25.0˚C + 273.15 = 298.2 K755 mm Hg× 1 atm = .993 atm

760 mm Hg

(1.00 atm) (10.0 mL) = (.993 atm)(V2) 298.2 K 295 K

V2=9.96 mL

IDEAL GAS LAW

Assume the gas behaves ideally. No attraction or repulsion between particles.

Also ignores the volume of the gas molecules.

Real gases behave most like ideal gases when the temperature is high and the pressure is low.

Less interaction between gas particles allows the gas molecules to be treated independently and the calculations to be simplified.

IDEAL GAS LAW

PV = nRT

Example: 3.00 moles of a gas are held in a 20.0 L container at 300.0 degrees C, what is the pressure inside the container?

P= (3.00 moles) (.0821 L atm/mol K) (573.2 K) = 7.06 atm 20.0 L

IDEAL GAS LAW

A gas at 105.0 kPa and 25.0 degrees C is held in a 250. ml container. How many moles are present?

PV =n (105.0 K Pa) (.250L) = 1.06×10-2

moles RT (8.314 L K Pa/mol K) (298.2 K)

IDEAL GAS LAW

A cylinder of gas holds 5.0 moles at 120. psi and 25.0

degrees C, what is the volume of the cylinder?120. Psi × 1 atm =8.16 atm

14.7 psiV= (5.0 moles) (.0821 L atm/mol K) (298.2 K) =15L

8.16 atm

What is the temperature of 1.00 mole of gas if 22.4 L has a pressure of 760.0 mmHg?

760.0 mm Hg × 1 atm =1atm 1 760.0 mm Hg

T= PV = (1.000atm) (22.4 L) =273 K 273 K=0˚C nR (1.00 mol) (.0821 L atm/mol K)

IDEAL GAS LAW DERIVATIONS

Including density and molar massMEMORIZE!

MM = g R T g=mass in grams PV

MM = d R T d=density (g/L) P

DENSITY and MOLAR MASS

Example: What is the molar mass of a gas if .250 grams of gas are held in a 250.0 ml container at 25.0 degrees C and 1.0 atmosphere of pressure?

MM=gRT= (.250 g) (.0821 L atm/mol K) (298.2 K) =2.448 g/mol PV (1 atm) (.2500 L)

sig figs=2.4 g/mol

Example: What is the density of carbon dioxide gas at 30.0 degrees celsius and 29.80 inHg?

MM= dRT MM P=d= (44.01 g/mol) (.9960 atm) =1.76 g/L P RT (1.0821 L atm/mol K) (303.2 K)

DALTON’S LAW

Partial pressure – the pressure that an individual gas exerts in a mixture of gases.

P total = P1 + P2 + P3 + P4 … Pn

Example: A container of gas at 1.0 atm contains O2 at 100.0 mmHg,

N2 at 450. torr and CO2. What is the pressure of the CO2

in atmospheres?Ptotal= PCO2 + PO2 + PN2

1.0 atm = PCO2 + (100.0 mm Hg× 1 atm ) + (450 torr × 1 atm )

1 760 mm Hg 1 760 torr1.O atm = PCO2 + .1316 atm + .592 atm

PCO2 = .28 atm

DALTON’S LAW

Necessary for gases that have been collected by water displacement.

H2O evaporates and causes water vapor to be mixed with the gas.

In order to find the pressure of the gas alone (dry gas), the water vapor pressure must be subtracted from the total pressure.

P atm = P total = P gas + P H20Water vapor pressure increases with temperature,

since more water evaporates at higher temperatures.

See table of water vapor pressure in the appendix.

DALTON’S LAW

Oxygen gas was collected by water displacement and the water level inside the collection flask was equal to the level in the water trough to ensure that the pressure inside the flask = atmospheric pressure. The water bath was at 23.0 degrees Celsius and the barometric pressure was 28.90 inHg. What is the pressure of dry oxygen gas?

PO2+PH2O= P atm (table value)

21.1 mm Hg = (28.90 in Hg×760 mm Hg) 1 29.92 in Hg

PO2= 734.2 mm Hg – 21.1 mm Hg = 713.1 mm Hg

PRACTICE PROBLEM

A gas was collected by water displacement over a water bath at 21.0 degrees C. 350.o ml of gas were collected and the barometric pressure was 755 mmHg. If the gas weighed 1.00 grams, what is the molar mass of the gas? Is there any possibility that the gas is CO?

MM= gRT (1.00 g) (.0821 L atm/mol K) (294.2 K) = 71.2 g/mol PV (.986 atm) (.350 L)

Pdry gas = P atm – PH2O

Pdry gas = 755 mm Hg – 18.6 mm Hg = 736.4 mm Hg× 1 atm = .986 atm

1 760 mm HgNO there is no possibility that the gas is CO!

MOLE FRACTION

Defined as: the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture

(chi) = mole fraction = n1

ntotal

= p1

ptotal Multiplying the mole fraction by __100_ results in a

__percent__ composition of the gas in the mixture. Multiplying the mole fraction by _the total pressure_

results in the __partial pressure__ of the gas.

MOLE FRACTION

Examples: A mixture of gases includes 3.0 moles of O2,

2.0 moles of N2, and 4.0 moles of CO2, what is the mole fraction of O2 in the mixture?

3.0 moles =.333.0+4.0+2.0 moles

MOLE FRACTION

If the total pressure of the mixture above is 2.0 atm. What is the partial pressure of the O2 gas?

PO2 = .33 PO2= .67 atm2.0 atm

MOLE FRACTION

A mixture of gases has a total pressure of 760 mmHg. If the mixture contains oxygen, hydrogen, and nitrogen, each with equal pressures, What is the partial pressure of O2?

X+X+X=760 mmHg PO2= 253.33= 260 mmHg What is the mole fraction of O2?

260 mmHg= .34 760 mmHg

What is the % of O2 in the mixture?

34%

GRAHAM’S LAW

Effusion – movement of a gas through a small orificeDiffusion – spontaneous mixing of gas moleculesThe rate at which a gas effuses or diffuses is inversely

proportional to the square root of the molar mass of the gas.

Rate gas 1 = √MM gas 2Rate gas 2 √MM gas 1

Example: How much faster does helium travel compared to carbon dioxide?

Rate He = √MM CO2 rate = √(44.01 g/mol ÷ 4.00 g/mol) Rate CO2 √MM He ratio

He travels 3.32 times faster than CO2

GRAHAM’S LAW

A gas travels 3.0 times faster than N2. What is the molar mass of the gas?

3.0 = rate gas = √28.02 g/mol 3.0 √MM gas = √28.02 rate N2 √MM gas

3.0 √MM gas = √28.02(√28.02)2 = 3.1 g/mol( 3.0 )2

If nitrogen gas travels at 30.0 m/sec at a certain temperature, how fast does carbon dioxide travel at the same temperature?

30.0 m/s = √44.01 g/mol rate CO2=23.9 m/s

rate CO2 √28.02 g/mol