Transcript Chapter 9
Chapter 9
Gases
Gases and Gas Pressure
Gases
– constituent atoms and molecules that have little attraction for one another Free to move in available volume
Some properties of gases
Mixtures are always homogenous Very weak attraction between gas molecules Identity of neighbor is irrelevant Compressible – volume contracts when pressure is applied 0.10% of volume of gas is occupied by molecules Exert a measurable pressure on the walls of their container
Gases and Gas Pressure
Pressure – force exerted per unit area Force
Pressure
: Unit area SI unit equals Pascal (Pa) 1 Pa = 1 N/m 2 (1 N = 1 (kg•m)/s 2 ) Alternative units Millimeters of mercury (mmHg) Atmosphere (atm) 1.0 atm = 760 mmHg = 101, 325 Pa 1.0 atm = 760 torr
Atmospheric Pressure
- pressure created from the mass of the atmosphere pressing down on the earth’s surface Standard atmospheric pressure at sea level – 760 mmHg
Gases and Gas Pressure
Barometer: long thin mercury filled tube sealed at once end and inverted into a dish of mercury Downward pressure of Hg in column equals outside atmospheric pressure
Measuring Pressure
Manometer:
U-tube filled with mercury, with one end connected to the gas – filled container and the other end open to the atmosphere.
Pgas < Patm; liquid level in the arm connected to the gas-filled cylinder will be higher P gas + P Hg = P atm
Measuring Pressure
Pgas > Patm; liquid level in the arm connected to the gas – filled cylinder will be lower Pgas = Patm + PHg (P Hg = the difference in the heights of the two mercury columns)
Example
What is the pressure of the gas inside the following apparatus (in mm Hg) if the outside pressure is 750 mm Hg? (1 cm Hg = 10 mm Hg)
The Gas Laws
Ideal Gas: A gas whose behavior follows the gas laws exactly.
The physical properties of a gas can be defined by four variables:
P T
pressure temperature
n V
volume number of moles
Pressure and Volume: Boyle’s Law
Showing the relationship between pressure and volume P x V = k (constant value @ specific temp and constant moles of gas) k = 1.40 x 10 3 V = 1/P (inverse relationship)
Pressure and Volume: Boyle’s Law
Can predict a new volume of pressure is changed P 1 V 1 = k = P 2 V 2 P 1 V 1 = P 2 V 2
PV
=
k
The Gas Laws
V
a 1
Boyle’s Law
P
(constant
n
and
T)
Example
A sample of helium gas has a pressure of 3.54 atm in a container with a volume of 23.1 L. This sample is transferred to a new container and the pressure is measured to be 1.87 atm. Assume constant temperature. Will the volume of the gas increase of decrease?
What will be the new volume of the gas?
Volume and Temperature: Charles’s Law
Relationship between Volume and Temperature V = bT (b is a constant) V / T = b Can predict the new volume or temperature (V 1 /T 1 )= (V 2 /T 2 )
V
initial =
V
final
T
initial
T
final
V T
=
k
The Gas Laws
Charles’ Law
V
a
T
(constant
n
and
P)
Example
A 2.0 L sample of air is collected at 298K and then cooled to 278 K. The pressure is held constant at 1.0 atm.
Does the volume increase of decrease?
Calculate the volume of the air at 278 K?
Volume and Moles: Avogadro’s Law
Relationship between volume of gas and number moles of gas V is directly proportional to n V = an or V / n = a (a = constant) Can predict the new volume or new moles of gas at constant pressure and temperature (V 1 /n 1 ) = (V 2 /n 2 )
V
initial =
n
initial
V
final
n
final
V n
=
k
The Gas Laws
Avogadro’s Law
V
a
n
(constant
T
and
P)
Example
Consider two samples of nitrogen gas (composed N 2 molecules). Sample 1 contains 1.5 mol of N 2 25oC and 1 atm. Sample 2 has a volume of 16.0 L at 25 o C and 1 atm. Calculate the number moles of N 2 and has a volume of 36.7 L at in sample 2
Boyle’s Law
Summary
Charles’ Law Avogadro’s Law
constant T & n constant P & n constant P & T V = 1/P
P
initial
V
initial =
P
final
V
final V = T
V
initial =
T
initial
V
final
T
final V = n
V
initial =
n
initial
V
final
n
final
9.3 The Ideal Gas law
Different gasses show similar physical behavior (unlike solid or liquid) Relationship of variable – gas laws Ideal gas – behavior follows the gas laws exactly
9.3 The Ideal Gas law
Describes how the volume of a gas is affected by changes in pressure, temperature and moles.
PV = nRT;
R
L atm = 0.082058
K mol
9.3 The Ideal Gas Law
Ideal Gas Law:
PV
=
n R T R
is the
gas constant
and is the same for all gases.
R
L atm = 0.082058
K mol
Standard Temperature and Pressure (STP) for Gases
T
= 0 °C (
273.15 K
)
P
= 1 atm
The Ideal Gas Law
What is the volume of 1 mol of gas at STP?
V
=
nRT
=
P
L atm (1 mol) 0.082058
K mol (1 atm) (273.15 K) = 22.414 L
Example
A helium gas cylinder of the sort used to fill balloons have a volume of 43.8 L and pressure of 1.51 x 10 4 kPa at 25.0oC. How many moles of helium are in the tank?
What volume is occupied by 0.250 mol of carbon dioxide gas at 25.0
o C and 371 torr?
A 0.250 mol sample of argon gas has a volume of 9.00L at a pressure of 875 mmHg. What is the temperature (in o C) of the gas?
Stoichiometric Relationships with Gases
The reaction used in the deployment of automobile airbags is the high temperature decomposition of sodium azide, NaN 3 , to produce N 2 gas. How many liters of N atm and 30.0
o C are produced by the decompostion of 45.0g NaN 3 ?
2 at 1.15 2NaN 3 (
s
) 2Na(
s
) + 3N 2 (
g
)
Examples
Consider the reaction represented by the equation P 4 (s) + 6 H 2 (g) 4H 3 (g) What is the amount of P 4 is required to react with 5.39 L of hydrogen gas at 27.0
o C and 1.25 atm?
Example
Ammonia is commonly used as a fertilizer to provide a source of nitrogen for plants. A sample of NH 3 (g) occupies a volume of 5.00 L at STP. What moles will this sample occupy?
Combine Gas Law
is an expression obtained by mathematically combining Boyle’s and Charles’ law
P 1 V 1 T 1 = P 2 V T 2 2
@ constant n
can predict P, V or T when condition is changed
Examples
A sample of diborane gas B2H6, a substance that bursts into flames when exposed to air, has a pressure of 0.454 atm at a temperature of 15 o C and a volume of 3.48L. If condition are changed so that the temperature is 36 the sample?
o C and the pressure is 0.616 atm, what is the new volume of
Examples
Consider a sample of hydrogen gas of 63 is cooled to -35 o o C with a volume of 3.65L at a pressure of 4.55 atm. The pressure is changed to 2.75 atm and the gas C. Calculate the new volume of the gas
13.6 Dalton’s Law of Partial Pressure
A. Gas laws apply to mixtures of gases B. Dalton's law of partial pressure –
P
total = P 1 + P 2 + P 3 + ….. at
constant V, T
where P 1 , P 2 , ….refer to the pressure of the individual gases in the mixture Moles of component
Mole Fraction ( X ) =
Total moles in mixture
X
i
n
i =
n
total or
X
i
P
i =
P
total
Partial Presssure
C. Partial pressures refer to the pressure each individual gas would exert if it were alone in the container (P 1 , P 2 , …) 1. Total pressure depends on the total molar amount of gas present
P total = n total (RT/V)
Examples
A 2.0 L flask contains a mixture of nitrogen gas and oxgyen gas at 25.0
oxygen present o C. The total pressure of the gas mixture is 0.91 atm, and the mixture is known to contain 0.050 mol of N 2 . Calculate the partial pressure of oxygen and the moles of
Examples
Mixture of helium and oxygen are use in the “air” tanks of underwater divers for deep dives. For a particular dive, 12.0L of O 2 at 25.0
o C and 1.0 atm and 46.0 L of He at 25 o C and 1.0 atm were both pumped into a 5.0 L tank. Calculate the partial pressure, moles fraction of each gas and the total pressure in the tank at 25.0
o C
The Kinetic Molecular Theory of Gas
A.Model that can explain the behavior of gases .Assumptions 1.
A gas consists of particles in constant random motion 2.
Most of the volume of a gas is empty spaces 3.
The attractive and repulsive forces between molecules of gases are negligible 4.
The total kinetic energy of the gas particles is constant at constant T 5.
Average Ek α T
The Kinetic-Molecular Theory of Gases
molar mass average speed
The Kinetic-Molecular Theory of Gases
Graham’s Law: Diffusion and Effusion of Gases
Diffusion: The mixing of different gases by molecular motion with frequent molecular collisions
Graham’s Law: Diffusion and Effusion of Gases
Effusion: The escape of a gas through a pinhole into a vacuum without molecular collisions.
Graham’s Law
: Rate a 1
m
The Behavior of Real Gases
The volume of a real gas is larger than predicted by the ideal gas law.
The Behavior of Real Gases
Attractive forces between particles become more important at higher pressures.
The Behavior of Real Gases
van der Waals equation
Correction for intermolecular attractions.
P
+
a n
2
V
2
V
-
n b
=
nRT
Correction for molecular volume.
examples
Assume that you have 0.500 mol of N N2, a = 1.35 (L 2 ·atm)mol 2 2 in a volume of 0.600L at 300K. Calculate the pressure in the atmosphere using both the ideal gas law and the van der Waals equation. For , and b = 0.0387 L/mol