1. GAS LAWS
Chapter 10

2. 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 law

3. 1 atmosphere = 760 mmHg = 760 Torr = 101325 Pascals =
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 equivalents
1 atmosphere = 760 mmHg = 760 Torr = Pascals =
kPa = inHg = 14.7 psi

4. VARIABLES ?atm = 30.o8 in Hg × 1 atm = 1.005 atm 1 29.92 in Hg
Example: Convert inHg to atmospheres and pascals
?atm = 30.o8 in Hg × atm = atm
in Hg
?Pa = in Hg × Pa = Pa
in Hg
sig figs = Pa

5. VARIABLES VOLUME (V) TEMPERATURE (T) NUMBER OF MOLES (n) Units: liters
Molar volume of a gas at STP = 22.4 L
STP = 1 atm; 0˚C
TEMPERATURE (T)
Units: Kelvin
Recall: Degrees C = Kelvin
NUMBER OF MOLES (n)
Units: moles
What if you’re given grams? Use MM

6. CONSTANT THE UNIVERSAL GAS CONSTANT R Depends on pressure units.
R = L atm/ mol K or
R = L kPa/ mol K

7. COMBINED GAS LAW P1V1 = P2V2 T1 T2
Combines Boyle’s, Charles, and Lussac’s laws
Assumes a constant number of moles
Example: 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?

8. COMBINED GAS LAW
P1=1.00 atm P2=755 mm Hg V1=10.0 mL V2=? T1=25.0˚C T2=295 K 25.0˚C = K 755 mm Hg× 1 atm = .993 atm 760 mm Hg (1.00 atm) (10.0 mL) = (.993 atm)(V2) K 295 K V2=9.96 mL

9. 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.

10. IDEAL GAS LAW
PV = nRT
Example: moles of a gas are held in a 20.0 L container at 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

11. IDEAL GAS LAW
A gas at 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)

12. 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 psi
V= (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 mmHg?
760.0 mm Hg × atm =1atm
mm Hg
T= PV = (1.000atm) (22.4 L) =273 K 273 K=0˚C
nR (1.00 mol) (.0821 L atm/mol K)

13. IDEAL GAS LAW DERIVATIONS
Including density and molar mass
MEMORIZE!
MM = g R T g=mass in grams PV
MM = d R T d=density (g/L) P

14. DENSITY and MOLAR MASS
Example: What is the molar mass of a gas if .250 grams of gas are held in a 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 inHg?
MM= dRT MM P=d= (44.01 g/mol) (.9960 atm) =1.76 g/L
P RT ( L atm/mol K) (303.2 K)

15. 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 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× atm ) + (450 torr × 1 atm )
mm Hg torr
1.O atm = PCO atm atm
PCO2 = .28 atm

16. 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 H20
Water vapor pressure increases with temperature, since more water evaporates at higher temperatures.
See table of water vapor pressure in the appendix.

17. 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 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)
in Hg
PO2= mm Hg – 21.1 mm Hg = mm Hg

18. PRACTICE PROBLEM
A gas was collected by water displacement over a water bath at 21.0 degrees C 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 = mm Hg× atm = .986 atm
mm Hg
NO there is no possibility that the gas is CO!

19. 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.

20. 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 =.33
moles

21. MOLE FRACTION
If the total pressure of the mixture above is 2.0 atm. What is the partial pressure of the O2 gas?
PO2 = PO2= .67 atm
2.0 atm

22. 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= = 260 mmHg
What is the mole fraction of O2?
260 mmHg= .34
760 mmHg
What is the % of O2 in the mixture?
34%

23. GRAHAM’S LAW Effusion – movement of a gas through a small orifice
Diffusion – spontaneous mixing of gas molecules
The 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 2
Rate gas √MM gas 1
Example: How much faster does helium travel compared to carbon dioxide?
Rate He = √MM CO rate = √(44.01 g/mol ÷ 4.00 g/mol)
Rate CO2 √MM He ratio
He travels 3.32 times faster than CO2

24. 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
( )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 CO √28.02 g/mol