Title: Chapter 5 Gases
1Chapter 5Gases
2Preview
- Properties of gases.
- Effects of temperature, pressure and volume.
- Boyles law.
- Charless law, and
- Avogadros law.
- The ideal gas law.
- Concepts of pressure
- and partial pressure.
- Real gases.
3States of Matter
Chapter 5 Section 1
Gas
Liquid
Solid
Unlike liquid and solid, gas fills up the
container and exerts force towards the inner
walls of the container.
4Pressure
Chapter 5 Section 1
- Gas is always exerts pressure on its
surroundings. - Gas always has the volume of the container it
occupies.
Water inside is boiling
After cooling water down
5The Atmospheric Pressure
Chapter 5 Section 1
- Pressure (P)
- At sea level average P is 760 mm Hg.
- At higher attitudes P goes
- down
- why??
- There is less air (and less gravitational force)
pushing the Hg inside the column.
Force
Area
Pressure exerted by the atmosphere
Barometer was invented by Torricelli
6Manometer
Chapter 5 Section 1
Patm
Patm
Pgas gt Patm
Pgas lt Patm
Pgas Patm h
Pgas Patm h
Try http//www.chem.iastate.edu/group/Greenbowe/s
ections/projectfolder/flashfiles/gaslaw/manometer4
-1.html
7Units of Pressure
Chapter 5 Section 1
Chapter 5 Section 1
- 1 mm Hg 1 torr (after Torricelli)
- 760 mm Hg 1 atm standard atmospheric pressure
- P pascal
(Pa) - 1 atm 101,325 Pa 1.013105 Pa
- 1 atm 14.7 lb/in2.
- Pressure conversion.
Force
Newton
Area
m2
8Important Gas Laws
Chapter 5 Section 2
- Boyles Law
- PV k at constant T and n
- Charless Law
- V bT at constant P and n
- Avogadros Law
- V an at constant T and P
- Where V is volume, P is pressure, T is
temperature, and n is number of moles.
9Boyles Law
Chapter 5 Section 2
- Boyle studied the relationship between pressure
(P) and volume (V) for gases. - Adding more Hg at constant T compresses the gas
(less V and higher P).
opened
closed
h
A j-tube
gas
V a 1/P
V P k
10Plotting Boyles Results
Chapter 5 Section 2
V
Slope k
P
k
When P drops by half, V is doubled. Inverse
relationship. V a 1/P
P
11Boyles Law
Chapter 5 Section 2
- Validity of Boyles law at different pressures.
- Not all gases obey Boyles law at higher
pressures. - A gas that strictly obeys Boyles law is called
an ideal gas. - You can get the new volume of a gas when pressure
is changed (or vise versa).
12Application of Boyles Law
Chapter 5 Section 2
- Sample Exercise 5.2
- 1.53 L of SO2 is initially at a pressure of
5.6104 Pa. If the pressure is changed to 1.5104
Pa, what will be the new volume? - PVk
- P1V1 k P2V2
- T is constant and the gas
- is assumed to be ideal.
- V2 0.57 L
P1V1
P2
13Charless Law
Chapter 5 Section 2
- He found that the V of a gas, at constant P,
increases linearly with the increase of T. - V bT
- The slope (b) for each gas is different because
each sample contains different number of
molecules (moles) - All the lines extrapolate to zero volume at T
-273.15 ºC.
14Charless Law
Chapter 5 Section 2
- T -273.15 ºC is defined as zero Kelvin or
absolute zero. - K ºC 273.15
- 110-6 K was reached in laboratories, but 0 K
could not be attained. - Another form of Charless law
-
- at constant pressure
- and T is in K
V1
V2
T1
T2
15Avogadros Law
Chapter 5 Section 2
- Avogadros law Equal volumes of gases at the
same T and P contain the same number of
particles.
V an
(linear relationship between V and n at
constant P and T)
V1
V2
n1
n2
16Avogadros Law
Chapter 5 Section 2
- Sample Exercise 5.5
- 12.2 L sample containing 0.50 moles of O2 at P
1 atm and T 25 ºC is converted into O3 at
the same T and P. What would be the
volume of O3? - 3O2 2O3
- mol of O3 produced 0.50 mol O2
0.33 mol O3 - Equal volumes of gases at the same T and P
contain the same number of particles. - V2 ( ) V1 8.1 L
0.50 mol O2
0.33 mol O3
12.2 L
8.1 L
17The Ideal Gas Law
Chapter 5 Section 3
- Boyles law V k 1/P
- Charless Law V b T
- Avogadros Law V a n
- V a gt V R
- R is the universal gas constant
- R 0.08206
- Ideal Gas Law PV nRT
T n
T n
P
P
L atm
K mol
18Applications of the Ideal Gas Law
Chapter 5 Section 3
- Sample Exercise 5.6
- A sample of H2 gas has V 8.56 L _at_ T 0 ºC and
P 1.5 atm. How many H2 molecules are present? - ? Dont forget to state T in K.
19Applications of the Ideal Gas Law
Chapter 5 Section 3
- Sample Exercise 5.8
- A sample methane with V 3.8 L _at_ 5ºC is heated
to 86ºC with constant P, what is the new volume? -
- (Charless
Law) - V2 V1T2 / T1 (359 K)(3.8 L) / 278 K
4.9 L - ? Again, dont forget to state T in K.
- ? Does the answer make physical sense to you?
V1
nR
V2
T2
T1
P
V1
V2
T1
T2
20The Molar Volume
Chapter 5 Section 4
- What is the volume of 1.000 mole of gas at
standard temperature and pressure (STP)? - STP T 0ºC P 1 atm.
- V
- 22.42 L (Molar volume)
- ? For an ideal gas
- 1 mol 22.42 L
nRT
(1.000 mol)(0.0821 L atm / K mol)(273.15 K)
T
(1.000 atm)
22.42L
21Ideal Gas Law
Chapter 5 Section 4
- Sample Exercise 5.12
- CaCO3 (s)
CaO (s) CO2 (g) - 152 g
V? -
_at_STP
22Molar Volume
Chapter 5 Section 4
-
- 2NaN3 (s)
2Na (s) 3N2 (g) - ? g
V 65L -
- NaN3 is an explosive material.
- It explodes very gast and
- completes the reaction in 40 ms.
_at_ 25oC and 1.0 atm
23Gas Stoichiometry
Chapter 5 Section 4
- Sample Exercise 5.13
- CH4 2O2
CO2 H2O - Pressure 1.65 atm
1.25 atm 2.50 atm - Volume 2.80 L
35.0 L ? - Temperature 298 K
304 K 398 K - n 0.189 mol 1.75 mol
- VCO2
PV RT
gt 0.189 mol
Limiting reactant because it requires 0.189 mol
2 0.378 mol of O2
nRT P
(0.189 mol)(0.0821 Latm/Kmol)(398K) (2.50 atm)
24Molar Mass of a Gas
Chapter 5 Section 4
- For a gas
- n
- P
- Since the density of a gas (d)
- Then, P d or MM d
mass MM
(mass)RT V (MM)
nRT V
(mass/MM)RT V
mass V
RT MM
RT P
g/L
25Daltons Law of Partial Pressures
Chapter 5 Section 5
- For a mixture of gases in a container at constant
V and T, the total pressure exerted (PTot) is the
sum of the pressures that each gas would exert if
it were alone.
Partial pressures
n1 RT V
n2 RT V
n3 RT V
PTot P1 P2 P3
PTot (n1 n2 n3 )
nTot ( )
RT V
RT V
26Important Observations from Daltons Theory
Chapter 5 Section 5
RT V
PTot nTot ( )
- When you calculate the total vapor pressure of a
mixture of gases - The identity or composition of each gas is not
important, only the total number of moles (nTot)
of particles are important. - The volume of each gas is not important, only the
total volume of gases is important. - The forces among the particles (attraction or
repulsion) are not important, as long as no
chemical reaction is taking place within them.
27Examples on Daltons Law
Chapter 5 Section 5
- Sample Exercise 5.15
- He 46 L _at_ 25ºC and 1.0 atm
- O2 12 L _at_ 25ºC and 1.0 atm
- They are pumped into a 5.0 L tank _at_ 25ºC.
Calculate PHe , PO2 and PTot . - nHe 1.9 mol.
- Similarly, nO2 0.49 mol.
P VHe RT
nHe RTHe Vtank
PHe 9.3 atm. Also PO2
2.4 atm. PTot PHe PO2 11.7 atm
28The Mole Fraction in a Mixture
Chapter 5 Section 5
n1 n1 n2 n3
- Mole fraction of substance 1 (?1)
-
- Also, simple derivation gives ?1
- or
P1 ?1 PTot - Look at the mathematical derivation on page 196.
- Also note that
n1 nTot
P1 PTot
29Examples on Daltons Law
Chapter 5 Section 5
- Calculate the mole fraction of oxygen in air
where the values above are pressures in kPa. - ?O2 0.206
- Also, P1 ?1 PTot (From the mole fraction of
one component you can calculate its partial
pressure)
PO2 Pair
20.9 kPa 101.3 kPa
30Collecting a Gas Over Water
Chapter 5 Section 5
and water vapor
2KClO3 (s) 2KCl (s) 3O2 (g)
31Collecting O2 Gas Over Water
Chapter 5 Section 5
- Sample Exercise 5.18
- 2KClO3 (s) 2KCl (s) 3O2 (g)
- O2 was collected _at_ 22ºC and PTot (PO2 PH2O) of
754 torr. - The volume of the gas collected is 0.650 L.
- Calculate PO2 and the mass of KClO3 that was
consumed. - PO2 PTot PH2O 754 torr 21 torr 733
torr. - nO2 2.5910-2 mol.
- Mass of KClO3 decomposed
- 2.5910-2 mol O2
2.12 g
PO2 V RT
2 mol KClO3 3 mol O2
122.6 g KClO3 1 mol KClO3
32The Kinetic Molecular Theory of Gases
Chapter 5 Section 6
- Theories (or models) attempt to describe physical
observations (laws). - The kinetic molecular theory (KMT) attempts to
explain the properties of ideal gas atoms or
molecules
1 mol N2(g)
1 mol N2(l )
Volume Density
35 mL 0.81 g/mL
22.4 L 0.0012 g/mL
33The Kinetic Molecular Theory of Gases
Chapter 5 Section 6
- The KMT for ideal (NOT real) gases can be stated
as follow - 1- The particles are so small comparing with the
distances between them. Thus, the volume of the
individual particle can be assumed to be
negligible (zero). - 2- The particles are constantly in random
motion. The collisions of the particles with the
walls of the container are the cause of the
pressure exerted by the gas. - 3- The particles are assumed neither to attract
nor to repel each other (No interaction). - 4- The average kinetic energy (KE) of a
collection of gas particles is assumed to be
directly proportional to the Kelvin temperature
of the gas.
34Pressure and Volume
Chapter 5 Section 6
- Boyles law (constant n and T).
- P (nRT)
- The decrease in V means the particles will hit
the wall of the container more often, resulting
in an increase of P.
1 V
constant
35Pressure and Temperature
Chapter 5 Section 6
- Constant n and V.
- P ( )T
- When T increases, the speed of particles
increases (increase in KE), causing the particles
to hit the wall of the container more often and
with greater force (higher P)
nR V
constant
36Volume and Temperature
Chapter 5 Section 6
- Charless law (constant n and P).
- V ( )T
- When T becomes higher, the only way to keep P
constant is to increase V to minimize collision
forces on the wall.
nR P
constant
37Volume and Number of Moles
Chapter 5 Section 6
- Avogadros law (constant P and T)
- V ( )n
- When n increases at constant V, P will become
larger. Thus, in order to keep P constant, V must
be increases
RT P
constant
38Mixture of Gases
Chapter 5 Section 6
- Daltons law.
- KMT account for Daltons law because it assumes
that all gas particles are independent of each
other (assumption 3 of KMT) and that the
volumes of individual particles are not important
(assumption 1 of KMT). Thus the identities of
the gas particles are not important. - The Total P the sum of the Ps exerted by each
type of particles, no matter what volumes they
occupy or what kinds of particles they are.
39Deriving the Ideal Gas Law from the KMT
Chapter 5 Section 6
- Deriving the ideal gas law from applying the
principles of physics to the KMT.
From physics background. Details are in appendix
2
40Deriving the Ideal Gas Law from the KMT
Chapter 5 Section 6
- From the assumption 4 of the KMT.
This result is identical to the ideal gas
equation if the proportionality constant is taken
to be R
41The Meaning of Temperature
Chapter 5 Section 6
- From the previous conclusion
The Kelvin temperature is an index of the random
motions of gas particles.
42Root Mean Square Velocity
Chapter 5 Section 6
Average of the square of the velocity of a
particle
Root mean square velocity
m is the mass of a particle in Kg. NA is 1 mole
of particles.
M is the molar mass in (Kg/mol). R 8.314 J /
K .mol
43Root Mean Square Velocity
Chapter 5 Section 6
- Sample Exercise 5.19
- Calculate urms for the atoms in a sample of He
gas _at_ 25oC.
44Molecular Velocity
Chapter 5 Section 6
- The path and speed of a particle in a gas are not
constant or systematic. - Collides with other particles.
- Collides with the walls.
- Velocity distributions of different gases at
constant T and P. - The peak value corresponds to the most probable
velocity not the root mean squared velocity.
Mean free path of order of 10-7 m
45Molecular Velocity
Chapter 5 Section 6
- The velocity distributions of N2 molecules at
three different temperatures. - The higher the temperature is, the greater the
most probable velocity is.
Interesting Experiment
(a)
(b)
(c)
46Effusion and Diffusion
Chapter 5 Section 7
Effusion is the passage of gas particles through
a pinhole to an evacuated chamber.
Diffusion is rate of mixing of gases with each
other.
47Effusion
Chapter 5 Section 7
Study the derivation of Grahams law form the KMT
as presented on page 207
Sample Exercise 5.20 Calculate the ratio of
effusion rates of H2 and UF6. MH2 2.016
g/mol. MUF6 352.02 g/mol.
13.2
48Diffusion
Chapter 5 Section 7
White solid forms after few minutes
NH3
CH4
49Diffusion
Chapter 5 Section 7
- Approximation to get the diffusion rate gives
- The velocities for HCl and NH3 are 450 and 660
m/s, respectively. - So, why does it take several minutes for them to
meet?
50Real Gases
Chapter 5 Section 8
- No gas exactly follows the ideal gas law.
- Many gases behave close to ideality at low P
and/or high T.
At constant T PV/nRT approaches 1 when P lt 1
atm. At higher T, real gases behave nearly
ideal.
51Real Gases
Chapter 5 Section 8
- PV nRT describes the behavior of a hypothetical
gas that - has no volume, and
- its particles dont interact with each other.
- Johannes van der Waals made some modification to
the ideal gas low. - He suggested that the volume of gas particles
are finite. - V (real gas) V (ideal gas) nb
- Number of moles of gas Empirical constant
52Real Gases
Chapter 5 Section 8
- Van der Waals also suggested that there are
attractions taking place between the real gas
particles, giving rise to Pobs. - Pobs lt Ptheor
53Real Gases
Chapter 5 Section 8
- Pobs P correction factor
- The correction factor is
proportional to (n/V)2. - The of possible pairs
- 10(9)/2 45 distinct pairs.
- For large number of particles
- n/V (n-1)/V, then
- n(n-1)/V2/2 n2/V2/2
54Real Gases
Chapter 5 Section 8
- At low P van der Waals model behaves nearly
ideal. - At high T the model also behaves nearly ideal.
- b constant a the size of the molecule particles.