Uniformly fills any container.
Mixes completely with any other gas Exerts pressure on its surroundings.

05_47 h = 760mm Hg for standard atmosphere Vacuum
Simple barometer invented by Evangelista Torricelli

Vacuum 1 atm Pressure
760 mm Hg
The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg.
1 atm = 760 mm Hg

Gas
h
Column of mercury to measure pressure.
h is how much lower the pressure is than outside.

Gas h
h is how much higher the gas pressure is than the atmosphere.

05_48 Atmospheric pressure (P atm ) Atmospheric pressure (P atm ) Gas pressure (P gas ) less than atmospheric pressure (P gas ) = (P atm ) - h (a)
h
Gas pressure (P gas ) greater than atmospheric pressure (P gas ) = (P atm ) + h (b)
Simple Manometer
h

1 atmosphere = 760 mm Hg 1 mm Hg = 1 torr 1 atm = 101,235 Pascals = 101.325 kPa 14.69 psi=1Atm Can make conversion factors from these.
What is 724 mm Hg in atm in torr?
in? kPa?

is equal to force/unit area SI units = Newton/meter 2 = 1 Pascal (Pa) 1 standard atmosphere = 101,325 Pa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

The pressure of a tire is measured to be 28 psi. What would the pressure in atmospheres, torr, and pascals.
(28 psi)(1.000 atm/14.69 psi) = 1.9 atm (28 psi)(1.000 atm/14.69 psi)(760.0 torr/1.000atm) = 1.4 x 10 3 torr (28 psi)(1.000 atm/14.69 psi)(101,325 Pa/1.000 atm) = 1.9 x 10 5 Pa

05_1541
P
ext Volume is decreased
P
ext
Volume of a gas decreases as pressure increases at constant temperature

*
P
1
V
1 = P 2
V
2 (T = constant) ( * Holds
precisely
only at very low pressures.) Boyle’s Law Pressure and volume are inversely related at constant temperature.

V P (at constant T)

A 1.5-L sample of gaseous CCl 2 F 2 has a pressure of 56 torr. If the pressure is changed to 150 torr, will the volume of the gas increase or decrease? What will the new volume be?
Decrease P 1 = 56 torr P 2 = 150 torr V 1 = 1.5 L V 2 = ?
V 1 P 1 V 2 = V = V 1 P 2 1 P /P 2 2 V V 2 2 = (1.5 L)(56 torr)/(150 torr) = 0.56 L

In an automobile engine the initial cylinder volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The mixture is 1.00 atm, what is the final pressure?
P 1 = 1.00 atm P V 2 1 = ?
= 0.725 L V 1 P 1 P 2 = V = V 1 P 1 2 P /V 2 2 P P 2 2 = (0.725 L)(1.00 atm)/(0.075 L) = 9.7 atm V 2 = 0.075 L Is this answer reasonable?

05_1543
P
ext Energy (heat) added
P
ext
Volume of a gas increases as heat is added when pressure is held constant.

The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin.

V
1
T
1 
V
2
T
2
P


What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant?

If the moles of gas remains constant, use this formula and cancel out the other things that don’t change.
.
P 1 V 1 T 1 = P 2 V 2 T 2

A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the can was heated to 100.ºC?

For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).

05_1544
P
ext Gas cylinder Moles of gas increases
P
ext Increase volume to return to original pressure
P
ext
At constant temperature and pressure, increasing the moles of a gas increases its volume.

V 1 /n 1 = V 2 /n 2

A 5.20L sample at 18.0 C and 2.00 atm pressure contains 0.436 moles of gas . If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be ?

A 12.2 L sample containing 0.50 mol of oxygen gas, O 2 , at a pressure of 1.00 atm and a temperature of 25 o C is converted to ozone, O 3 , at the same temperature and pressure, what will be the volume of the ozone? 3 O 2(g) ---> 2 O 3(g) (0.50 mol O 2 )(2 mol O 3 /3 mol O 2 ) = 0.33 mol O 3 V 1 = 12.2 L V 2 = ?
n 1 = 0.50 mol n 2 = 0.33 mol V V 2 2 = (12.2 L)(0.33 mol)/(0.50 mol) = 8.1 L

An equation of state for a gas.
“state” is the condition of the gas at a given time.
PV = nRT

1. Molecules are infinitely far apart.
2. Zero attractive forces exist between the molecules.
3. Molecules are infinitely small--zero molecular volume.

1. Molecules are relatively far apart compared to their size.
2. Very small attractive forces exist between molecules.
3. The volume of the molecule is small compared to the distance between molecules.

P V
= n
R T
R = proportionality constant = 0.0821 L atm   mol  P = pressure in atm V = volume in liters n = moles T = temperature in Kelvins Holds closely at P < 1 atm

A 1.5 mol sample of radon gas has a volume of 21.0 L at 33 o C. What is the pressure of the gas?
p = ?
V = 21.0 L n = 1.5 mol T = 33 o C + 273
pV = nRT p = nRT/V p = (1.5mol)(0.0821Latm/molK)(306K) (21.0L) p = 1.8 atm
T = 306 K R = 0.0821 Latm/molK

A sample of hydrogen gas, H 2 , has a volume of 8.56 L at a temperature of O o C and a pressure of 1.5 atm. Calculate the number of moles of hydrogen present.
p = 1.5 atm V = 8.56 L R = 0.0821 Latm/molK n = ?
T = O o C + 273 T = 273K
pV = nRT n = pV/RT n = (1.5 atm)(8.56L) (0.08206 Latm/molK)(273K) n = 0.57 mol

“STP” P = 1 atmosphere T =  C or 273K The molar volume of an ideal gas is 22.42 liters at STP

pV = nRT V = nRT/p V = (1.00 mol)(0.08206 Latm/molK)(273K) (1.00 atm) V = 22.4 L

A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N 2 are present?
(1.75L N 2 )(1.000 mol/22.4 L) = 7.81 x 10 -2 mol N 2

n = m/M n = number of moles m = mass M = molar mass

P = mRT/VM or P = DRT/M therefore: M = DRT/P

A gas at 34.0C and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass.

What is the density of 2L of O2 gas at STP?

1. Mass-Volume 2. Volume-Volume

Quicklime, CaO, is produced by heating calcium carbonate, CaCO 3 . Calculate the volume of CO 2 produced at STP from the decomposition of 152 g of CaCO 3 . CaCO 3(s) ---> CaO (s) + CO 2(g) (152g CaCO 3 )(1 mol/100.1g)(1mol CO 2 /1mol CaCO 3 ) (22.4L/1mol) = 34.1L CO 2
Note: This method only works when the gas is at STP!!!!!

If 25.0 L of hydrogen reacts with an excess of nitrogen gas, how much ammonia gas will be produced? All gases are measured at the
standard temperature and pressure.
2N 2(g) + 3H 2(g) ----> 2NH 3(g)

16.7 L NH3

p = 1.00 atm V = ?
n = 1.28 x 10 -1 mol R = 0.08206 Latm/molK T = 25 o C + 273 = 298 K
pV = nRT V = nRT/p V = (1.28 x 10 -1 mol)(0.08206Latm/molK)(298K) (1.00 atm) V = 3.13 L O 2

For a mixture of gases in a container,
P
Total = P 1 + P 2 + P 3 + . . .

A mixture of nitrogen gas at a pressure of 1.25 atm, oxygen at 2.55 atm, and carbon dioxide at .33 atm would have what total pressure?
P
Total = P 1 + P 2 + P 3
P
Total = 1.25 atm + 2.55 atm + .33 atm P total = 4.13 atm

2KClO 3(s) ----> 2KCl (s) + 3O 2(g) When a sample of potassium chlorate is decomposed and the oxygen produced is collected by water displacement, the oxygen has a volume of 0.650 L at a temperature of 22 o C. The combined pressure of the oxygen and water vapor is 754 torr (water vapor pressure at 22 o C is 21 torr). How many moles of oxygen are produced?
P ox = P total - P HOH P ox = 754 torr - 21 torr p ox = 733 torr

Dalton’s Law
Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46L He at 25ºC and 1.0 atm and 12 L O2 at 25ºC and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25ºC.

A Volume of 2.0 L of He at 46° C and 1.2 ATM pressure was added to a vessel that contained 4.5L of N 2 at STP. What is the total pressure of each gas at STP after the He is added .
RB P 128

-- the ratio of the number of moles of a given component in a mixture to the total number of moles of the mixture.

1 = n 1 / n total

1 = V 1 / V total

1 = P 1 / P total constant) (volume & temperature

The partial pressure of Oxygen gas was observed to be 156 torr in the air with a total atmosphere pressure of 743 torr . Calculate the mole fraction of O2 present.

The mole Fraction of nitrogen in the air is .7808. Calculate the partial pressure of N2 in the air when the atmospheric pressure is 760.


rms
 3
RT M
Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

Calculate the root mean square velocity of carbon dioxide at 25ºC.
Calculate the root mean square velocity of chlorine at 25ºC.

Diffusion : describes the mixing of gases. The rate of diffusion is the rate of gas mixing.
Effusion : describes the passage of gas into an evacuated chamber.

05_60 Pinhole Gas Vacuum
Effusion of a gas into an evacuated chamber

Effusion: Rate of effusion for gas 1 Rate of effusion for gas 2 
M
2
M
1 Diffusion: Distance traveled by gas 1 Distance traveled by gas 2 
M
2
M
1

Calculate the ratio of effusion rates of hydrogen gas and uranium hexafluoride .

A compound effuses through a porous cylinder 3.20 time faster than helium. What is it’s molar mass?

1. Volume of individual particles is  zero.
2. Collisions of particles with container walls cause pressure exerted by gas.
3. Particles exert no forces on each other.
4. Average kinetic energy  Kelvin temperature of a gas.

Real molecules do take up space and they do interact with each other (especially polar molecules).
Need to add correction factors to the ideal gas law to account for these.

The actual volume free to move in is less because of particle size.
More molecules will have more effect.
Corrected volume V’ = V - nb b is a constant that differs for each gas.
P’ = nRT (V-nb)

Because the molecules are attracted to each other, the pressure on the container will be less than ideal depends on the number of molecules per liter.
since two molecules interact, the effect must be squared.

Pressure correction 
Because the molecules are attracted to each other, the pressure on the container will be less than ideal

depends on the number of molecules per liter.

since two molecules interact, the effect must be squared.
P observed = P’ - a n ( ) 2

P obs = nRT - a n 2 V-nb V
( ) Called the Van der Wall’s equation if     rearranged P obs + a  n V   2       nRT Corrected Pressure Corrected Volume

a and b are determined by experiment.
Different for each gas.
Bigger molecules have larger b.
a depends on both size and polarity.
once given, plug and chug.

Calculate the pressure exerted by 0.5000 mol Cl 2 in a 1.000 L container at 25.0ºC Using the ideal gas law.
Van der Waal’s equation – a = 6.49 atm L 2 /mol 2 – b = 0.0562 L/mol

Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).

05_62 2.0
PV nRT
1.0
CH 4 N 2 H 2 CO 2 Ideal gas 0 0 200 400 600 P(atm) 800 1000
Plots of PV/nRT vs. P for several gases at 200 K.
Note the significant deviation from ideal behavior.

[
P
obs 
a n V
2 ]  
V

nb
 
nRT
 corrected pressure  corrected volume
P
ideal
V
ideal

05_68 0.5
0.4
0.3
0.2
0.1
0 Molecules of unburned fuel (petroleum) NO NO 2 O 3 Other pollutants Time of day
Concentration for some smog components vs. time of day

NO 2(g)
 NO
(g) + O (g) O (g) + O 2(g)
 O
3(g) NO (g) + 1/2 O 2(g)
 NO
2(g) 3/2 O 2(g)
 O
3(g) What substances represent intermediates?
Which substance represents the catalyst?