Transcript How Do Gases Behave?

How Do Gases Behave?
What is a solid, liquid or gas?





Help Marvin the Martian
understand what a solid,
liquid and gas are!
Draw what solids, liquids,
gases look like
Describe
physical/chemical
properties
What would happen if we
changed pressure?
What would happen if we
changed temperature?
What is Pressure?
Pressure = Force/Area
 1 atmosphere (atm)
 760 Torr
 760 mmHg
 1.01 Bar
 101,327 Pascal
 101.3 Kpa
 14.7 lbs/in2
 Measured with
a barometer

MANOMETER


h
h


Column of mercury
to measure
pressure.
h is how much
lower or higher the
pressure is than
outside.
Pgas = Patm - h
Pgas = Patm + h
What is Temperature?






Average Kinetic Energy (1/2 mv2) of an atom or molecule
Measured in Fahrenheit, Celsius or Kelvin (SI)
F = (C x 1.8) + 32
K = C + 273
0 Kelvin: absolute zero (atom stops moving completely)
Is there a maximum temperature in the universe?
Kinetic Molecular Theory


Theory explains why ideal gases behave the way they
do.
Assumptions that simplify the theory, but don’t work in
real gases.
1. The particles are so small we can ignore their volume.
2. The particles are in constant motion and their collisions cause
3.
4.
5.
pressure.
The particles do not affect each other, neither attracting or
repelling.
The average kinetic energy is proportional to the Kelvin
temperature.
The molecules move in straight path and all collisions are
elastic
What is an Ideal Gas?

An ideal gas or perfect gas is a hypothetical gas
consisting of identical particles of:





Negligible volume
With no intermolecular forces
Atoms or molecules undergo perfectly elastic collisions with the
walls of the container
Ideal gas law calculations are favored at low pressures
and high temperatures.
Real gases existing in reality do not exhibit these exact
properties, although the approximation is often good
enough to describe real gases.
What is Boyle’s Law?





In the mid 1600's, Robert Boyle studied the
relationship between the pressure P and the
volume V of a confined gas held at a constant
temperature.
Boyle observed that the product of the pressure
and volume are observed to be nearly constant.
The product of pressure and volume is exactly a
constant for an ideal gas.
P * V = constant
This relationship between pressure and volume
is called Boyle's Law in his honor.
BOYLE’S LAW
V
P (at constant T)
V
Slope = k
1/P (at constant T)
22.41 L atm
PV
O2
CO2
P (at constant T)


20.5 L of nitrogen at 25ºC and 742 torr are compressed
to 9.8 atm at constant T. What is the new volume?
P1V1=P2V2
(0.98 atm)(20.5 L) = (9.8 atm)(V2)
20.09 atm*L/9.8 atm= V2
2.1 L = V2
30.6 mL of carbon dioxide at 740 torr is expanded at
constant temperature to 750 mL. What is the final
pressure in kPa?
(30.6mL)(740Torr)=(750 mL)(P2)
22,644 mL*Torr/750 mL= P2
30.2 Torr =P2
(30.2 Torr) (1 atm/760 Torr) (101.3 kPa/1 atm)=
3059.3 kPa/760= 4.03 kPa
What is Charles’ Law?



The relationship between temperature and
volume, at a constant number of moles and
pressure, is called Charles and Gay-Lussac's
Law in honor of the two French scientists who
first investigated this relationship.
Charles did the original work, which was verified
by Gay-Lussac. They observed that if the
pressure is held constant, the volume V is equal
to a constant times the temperature T, or:
V / T= constant
CHARLES LAW
He
CH4
V (L)
H2O
H2
-273.15ºC
T (ºC)


What would the final volume be if 247 mL of gas at 22ºC
is heated to 98ºC , if the pressure is held constant?
247 ml/295 K = X ml/371 K
91,637 mL * K = 295 X * K
91,637 mL * K/295 K= X
310 mL = X
If the volume of oxygen at 21 °C is 785 L, at what
temperature would oxygen occupy 804 L?
785 L/294 K = 804 L/X K
785 X = 236,376
X = 236,376/785
X= 301 K = 28 °C
Combined Gas Law






Combining Charles’s Law and Boyle’s Law in a
single statement:
P1V1/T1 = P2V2/T2
39.8 mg of caffeine gives 10.1 mL of nitrogen
gas at 23°C and 746 mmHg. What is the
volume of nitrogen at 0°C and 760 mmHg?
First change temperature to Kelvin
V1 = 10.1mL
P1 = 746 mmHg
K1 = 296 K
V2 = ?
P2 = 760 mmHg
K2 = 273 K
10.1 * 746/296 = V2 * 760/273
V2 = 9.14 mL
Other Gas Laws

Gay-Lussac Law



At constant volume, pressure and absolute
temperature are directly related.
P/T = k (constant)
Avogadro’s Law


At constant temperature and pressure, the
volume of gas is directly related to the
number of moles.
V /n= k (n is the number of moles)
Gas Law Summary
Law
Statement Equation
Constant
Boyle’s
P inversely
proportional
to V
PV = k1
T, n
Charle’s
V directly
proportional
to T
V/T = k2
P, n
Gay-Lussac
P directly
proportional
to T
Avogadro’s
V directly
proportional
to n
P/T = k3
V/n = k4
V, n
P, T
What equation would we get if we combined them all?
What is the Ideal Gas Law?

Combining Boyle’s Law, Charles’ law & Avogadro’s Law
we derive the Ideal Gas Law:
P








V=nRT
P = Pressure (atm)
V = Volume (L)
n = # moles (0.0821 L atm /mol K)
R = Gas Constant
T = Temperature (K)
Ideal gas law calculations are favored at low pressures
and high temperatures
Tells you about a gas NOW.
The other laws tell you about a gas when it changes.
Let’ Try It!







Example:
If we had 1.0 mol of gas at 1.0 atm of pressure
at 0°C (STP), what would be the volume?
PV = nRT
V = nRT/P
V = (1.0 mol)(0.0821 L atm/mol K)(273 K)/(1.0
atm)
V = 22.41 L
1 mole of ANY gas at STP will occupy 22.4 Liters
of volume
Gas Density and Molar Mass









D = m/V
Let M stand for molar mass
M = m/n
n = m/M
PV = nRT
PV = (m/M) RT
P = mRT/VM = (m/V)(RT/M)
P = d RT/M
PM/RT = d (density)
Examples
What is the density of ammonia at 23ºC
and 735 torr?
 Units must be: atm, K
 735 torr(1 atm/760 torr) = 0.967 atm
 23 + 273 = 296 K
 Molar mass of NH3 = 17.0 g
d=
0.967 * 17.0 g
(0.0821 L* atm/mol * K)(296 K)
d = 0.676 g / L

Gases and Stoichiometry



Reactions happen in moles
At Standard Temperature and Pressure
(STP, 0ºC and 1 atm) 1 mole of gas
occupies 22.42 L.
If not at STP, use the ideal gas law to
calculate moles of reactant or volume of
product.
Examples

Consider the following reaction:
2KClO3 (s)  2KCl(s) + 3O2 (g)
heat


Suppose you heat 0.0100 mol of
potassium chlorate, KClO3, in a test tube.
How many liters of oxygen can you
produce at 298 K and 1.02 atm?
Break it into 2 problems, one involving
stoichiometry and the other using the
ideal gas law
0.0100 mol KClO3 X 3 mol O2/2 mol KClO3
= 0.0150 mol O2
Now that you have the moles of oxygen use the
ideal gas law to calculate the volume:
V = nRT/P
0.0150 mol x 0.0821 L * atm (K * mol) x 298 K
1.02 atm
V = 0.360 L

Using the following reaction
NaHCO 3 (s) + HCl 
NaCl(aq) + CO 2 (g) +H 2 O(l)

Calculate the mass of sodium hydrogen carbonate
necessary to produce 2.87 L of carbon dioxide at 25ºC and
2.00 atm.
n = PV/RT = (2.00 atm)(2.87 L)
(0.0821 L*atm/K*mol)(298 K)
n= 0.235 mol CO2
0.235 mol CO2 (1 mol NaHCO3) ( 84.0 g)
(1 mol CO2 ) (1 mol NaHCO3)
19.7 g NaHCO3
Dalton’s Law




The total pressure in a container is the
sum of the pressure each gas would exert
if it were alone in the container.
The total pressure is the sum of the partial
pressures.
PTotal = P1 + P2 + P3 + P4 + P5 ...
For each P = nRT/V
Dalton's Law




PTotal = n1RT + n2RT + n3RT +...
V
V
V
In the same container R, T and V are the
same.
PTotal = (n1+ n2 + n3+...)RT
V
PTotal = (nTotal)RT
V
The Mole Fraction




Ratio of moles of the substance to the
total moles.
symbol is Greek letter chi
c
Because pressure of a gas is proportional
to moles, for fixed volume and
temperature then,
c1 =
n1
= P1
nTotal
PTotal
Calculating the Partial Pressure and Mole
Fraction of a Gas Mixture

A 1.00 L sample of dry air at 25°C and
786 mmHg contains 0.925 g N2, plus other
gases including oxygen, argon and carbon
dioxide.


What is the partial pressure (in mmHg) of N2
in the air sample?
What is the mole fraction and mole percent of
N2 in the mixture?
Convert grams into moles
0.925 g N2 x (1 mol N2/28.0g N2)
=0.0330 mol N2
 Substitute into ideal gas law
PN2 = nN2RT/V
=0.0330mol x 0.0821 L*atm/K*mol x 298
1.00 L
=0.807 atm = 613 mmHg




The mole fraction of N2 in the air is
= PN2/P = 613 mmHg/786 mmHg
=0.780
Mole percent equals mole fraction x 100
=0.780 x 100 = 78%
Air contains 78.0 mole percent of N2
Vapor Pressure




Water evaporates!
When that water evaporates, the vapor
has a pressure.
Gases are often collected over water so
the vapor pressure of water must be
subtracted from the total pressure.
Vapor pressure varies by temperature and
must be given in the problem or in a
table.

Hydrogen gas is produced by the reaction
of hydrochloric acid, HCl, on zinc metal:
2HCl (aq) + Zn (s) —> ZnCl2 (aq) + H2 (g)

The gas is collected over water. If 156 mL
of gas is collected at 19°C and 769 mmHg
total pressure, what is the mass of
hydrogen collected?


First find the Partial Pressure. The vapor
pressure of water at 19°C is 16.5 mmHg
P = PH2 + PH2O
PH2 = P - PH2O
PH2 =769 – 16.5 = 752 mmHg
Use the ideal gas law to find the moles of
hydrogen collected.
P: 752 mmHg x (1 atm/760 mmHg) = 0.989 atm
V: 156 mL x (1 L/1000 mL) = 0.156 L
T: 19 + 273 = 292 K
R: 0.0821 L*atm/K*mol
n: ?


Solve for moles:
n = PV/RT
= 0.989 x 0.156/0.0821 x 292
=0.00644 mol H2
Convert moles to grams:
0.00644 mol H2 x (2.02g/1 mol H2)
=0.0130 g H2
What’s Diffusion and Effusion?



Only a few physical properties of gases depends on the
identity of the gas.
Diffusion - The rate at which two gases mix.
Effusion - The rate at which a gas escapes through a
pinhole into a vacuum.
What is Graham’s Law?





We know that Kinetic energy = 1/2 mv2
If two bodies of unequal mass have the same
kinetic energy, which moves faster?
The lighter one!
Thus, for two gases at the same temperature,
the one with lower molecular mass will
diffuse/effuse faster.
The rate of effusion/diffusion of a gas is
inversely proportional to the square root of its
mass.
Rate of effusion for gas 1

Rate of effusion for gas 2


M2
M1
Calculate the ratio of effusion rates of molecules
of carbon dioxide and sulfur dioxide from the
same container and at the same temperature
and pressure:
Rate of effusion of CO2 = √Mm SO2
Rate of effusion of SO2 √Mm CO2
= √64.1/44.0 = 1.21
In other words, carbon dioxide effuses 1.21
times faster than sulfur dioxide.