Transcript THERMOCHEMISTRY or Thermodynamics

Energy & Chemistry
2H2(g) + O2(g) → 2H2O(g) + heat and light
This can be set up to provide
ELECTRIC ENERGY in a fuel cell.
Oxidation:
2 H2 → 4 H+ + 4 e-
Reduction:
4 e- + O2 + 2 H2O → 4 OHH2/O2 Fuel Cell Energy
Energy & Chemistry
ENERGY is the capacity to do work (w) or transfer heat (q).
HEAT is the thermal energy that can be transferred from an
object at one temperature to an object at another temperature
– Net transfer of thermal energy stops when the
two objects reach the same temperature.
Other forms of energy —
1. Radiant (light) — energy in light, microwaves, and radio
waves
2. Thermal (kinetic and potential) — results from atomic
and molecular motion – Temperature of an object is a
measure of the thermal energy content
3. Chemical — results from the particular arrangement of
atoms in a chemical compound; radiant and thermal
energy produced in this reaction due to energy released
during the breaking and reforming of chemical bonds
4. Nuclear — radiant and thermal energy released when
particles in the nucleus of the atoms are rearranged
5. Electrical — due to the flow of electrically charged
particles
Potential & Kinetic Energy
Kinetic energy
rotate
vibrate
translate
 Potential energy – Energy stored in an object because of

the relative positions
or orientations of its components – PE = Fd = mad = mgh = work
Kinetic energy – Energy due to the motion of an object – KE = ½ mv2
Potential & Kinetic Energy
Internal Energy (E)
• PE + KE = Internal energy (E or U)
• Internal Energy of a chemical system depends on
• number of particles
• type of particles
• temperature
• The higher the T the higher the internal energy
• So, use changes in T (∆T) to monitor changes in
E (∆E).
Internal Energy (E)
heat transfer in
(endothermic), +q
heat transfer out
(exothermic), -q
SYSTEM
∆E = q + w
w transfer in
(+w)
w transfer out
(-w)
Energy & Chemistry
All of thermodynamics depends on the law of
CONSERVATION OF ENERGY.
• The total energy is unchanged in a chemical reaction.
• If PE of products is less than reactants, the difference
must be released as KE.
Energy Change in Chemical Processes
PE
Reactants
Kinetic
Energy
Products
Potential Energy of system
dropped. Kinetic energy
increased. Therefore, you often
feel a Temperature increase.
Thermodynamics -Enthalpy
• Thermodynamics – the science of heat (energy) transfer.
Heat transfers until thermal equilibrium is established.
∆T measures energy transferred.
• SYSTEM – The object under study
1. Open system — can exchange both matter and energy
with its surroundings
2. Closed system — can exchange energy but not matter
with its surroundings
3. Isolated system — exchanges neither energy nor matter
with the surroundings; total energy of the system plus the
surroundings is constant
• SURROUNDINGS – Everything outside the system
FIRST LAW OF
THERMODYNAMICS
heat energy transferred
∆E = q + w
energy
change
work done
by the
system
Energy is conserved!
The First Law of
Thermodynamics
• Exothermic reactions
generate specific amounts
of heat.
• This is because the
potential energies of the
products are lower than
the potential energies of
the reactants.
The First Law of Thermodynamics
•
1.
2.
•
There are two basic ideas of importance for thermodynamic
systems.
Chemical systems tend toward a state of minimum potential
energy.
Chemical systems tend toward a state of maximum disorder.
The first law is also known as the Law of Conservation of
Energy.
– Energy is neither created nor destroyed in chemical
reactions and physical changes.
State of a System
•
•
The state of the system is a complete description of the system at a
given time, including its temperature and pressure, the amount of
matter it contains, its chemical composition, and the physical state of
the matter Some examples of state functions are:
– T (temperature), P (pressure), V (volume), E (change in energy),
H (change in enthalpy – the transfer of heat), and S (entropy)
Examples of non-state functions are:
– n (moles), q (heat), w (work)
∆H along one path = ∆H along another path
•
•
•
•
This equation is valid because ∆H is a STATE FUNCTION
These depend only on the state of the system and not how it got there.
V, T, P, energy — and your bank account!
Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.
Standard States and Standard Enthalpy Changes
•
•
Thermochemical standard state conditions
– The thermochemical standard T = 298.15 K.
– The thermochemical standard P = 1.0000 atm.
• Be careful not to confuse these values with STP.
Thermochemical standard states of matter
– For pure substances in their liquid or solid phase the standard state is
the pure liquid or solid.
– For gases the standard state is the gas at 1.00 atm of pressure.
• For gaseous mixtures the partial pressure must be 1.00 atm.
– For aqueous solutions the standard state is 1.00 M concentration.
∆Hfo = standard molar enthalpy of formation
• the enthalpy change when 1 mol of compound is formed from
elements under standard conditions.
State Function
• The properties of a system that depend only on the state of the system
are called state functions.
– State functions are always written using capital letters.
• The value of a state function is independent of pathway.
• An analog to a state function is the energy required to climb a
mountain taking two different paths.
– E1 = energy on the first floor of Heldenfels
– E1 = mgh1
– E2 = energy on the fourth floor of Heldenfels
– E2 = mgh2
– E = E2-E1 = mgh2 – mgh1 = mg(h)
ENTHALPY
Most chemical reactions occur at constant P, so
Heat transferred at constant P = qp
qp = ∆H
where H = enthalpy
and so ∆E = ∆H + w (and w is usually small)
∆H = heat transferred at constant P ≈ ∆E
∆H = change in heat content of the system
∆H = Hfinal - Hinitial
Directionality of Heat Transfer
• Heat always transfer from hotter object to cooler one.
• EXOthermic: heat transfers from SYSTEM to
SURROUNDINGS.
T(system) goes down
T(surr) goes up
Directionality of Heat Transfer
• Heat always transfers from hotter object to cooler one.
• ENDOthermic: heat transfers from SURROUNDINGS to
the SYSTEM.
T(system) goes up
T (surr) goes down
ENTHALPY
∆H = Hfinal - Hinitial
If Hfinal > Hinitial then ∆H is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then ∆H is negative
Process is EXOTHERMIC
USING ENTHALPY
Consider the formation of water
H2(g) + 1/2 O2(g) → H2O(g) + 241.8 kJ
Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ
USING ENTHALPY
Example of HESS’S LAW— Making liquid H2O from H2 + O2 involves
two exothermic steps.
H2 + O2 gas
H2O vapor
Liquid H2O
Making liquid H2O from H2 involves two steps.
H2(g) + 1/2 O2(g) → H2O(g) + 242 kJ
H2O(g) → H2O(l) + 44 kJ
H2(g) + 1/2 O2(g) → H2O(l) + 286 kJ
If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s
of the other rxns.
Enthalpy Values
Depend on how the reaction is written and on phases
of reactants and products
H2(g) + 1/2 O2(g) → H2O(g)
∆H˚ = -242 kJ
2H2(g) + O2(g) → 2H2O(g)
∆H˚ = -484 kJ
H2O(g) → H2(g) + 1/2 O2(g)
H2(g) + 1/2 O2(g) → H2O(l)
∆H˚ = +242 kJ
∆H˚ = -286 kJ
Standard Molar Enthalpies of Formation, Hfo
•
The standard molar enthalpy of formation is defined as the enthalpy for the reaction in
which one mole of a substance is formed from its constituent elements.
– The symbol for standard molar enthalpy of formation is Hfo.
•
The standard molar enthalpy of formation for MgCl2 is:
Standard Molar Enthalpies of Formation, Hfo
•
•
•
Standard molar enthalpies of formation have been determined for many
substances and are tabulated in Table 15-1 and Appendix K in the text.
Standard molar enthalpies of elements in their most stable forms at 298.15
K and 1.000 atm are zero.
Example 15-4: The standard molar enthalpy of formation for phosphoric acid
is -1281 kJ/mol. Write the equation for the reaction for whichHorxn = 1281 kJ.
P in standard state is P4
Phosphoric acid in standard state is H3PO4(s)
Standard Molar Enthalpies of Formation, Hfo
H2(g) + ½ O2(g) → H2O(g) ∆Hf˚ (H2O, g)= -241.8 kJ/mol
C(s) + ½ O2(g) → CO(g) ∆Hf˚ of CO = - 111 kJ/mol
By definition, ∆Hfo = 0 for elements in their standard states.
Use ∆H˚’s to calculate enthalpy change for
H2O(g) + C(graphite) → H2(g) + CO(g)
Standard Molar Enthalpies of Formation, Hfo
• Calculate the enthalpy change for the reaction of one mole of H2(g)
with one mole of F2(g) to form two moles of HF(g) at 25oC and one
atmosphere.
Standard Molar Enthalpies of Formation, Hfo
• Calculate the enthalpy change for the reaction in which 15.0 g of
aluminum reacts with oxygen to form Al2O3 at 25oC and one
atmosphere.
Hess’s Law – Enthalpies of
Formation and Reaction

Standard enthalpies of reaction (ΔH°xn)
– The enthalpy that occurs when a reaction is carried out with all
reactants and products in their standard states
– General reaction: aA + bB  cC + dD where A, B, C, and D
are chemical substances and a, b, c, and d are their stoichiometric
coefficients
– Magnitude of ΔH°rxn is the sum of the standard enthalpies of
formation of the products, each multiplied by its appropriate
coefficient , minus the sum of the standard enthalpies of
formation of the reactants, multiplied by their coefficients
ΔH°rxn = [cΔH°f(C) + dΔH°f (D)]  [aΔH°f(A) + bΔH°f(B)]
or
ΔH°rxn= mΔH°f(product)  nΔH°f(reactants)
where  means “sum of” and m and n are the stoichiometric
coefficients of each of the products and reactants, respectively
Hess’s Law & Energy
Level Diagrams
Forming H2O can occur in a single
step or in a two steps. ∆Htotal is the
same no matter which path is
followed.
Enthalpies of Formation

Standard enthalpies of formation
– The magnitude of ΔH for a reaction depends on the physical states
of the reactants and products, the pressure of any gases present, and
the temperature at which the reaction is carried out.
– A specific set of conditions under which enthalpy changes are
measured are used to ensure uniformity of reaction conditions and
data.
– Standard conditions: a pressure of 1 atmosphere (atm) for gases
and a concentration of 1 M for species in solution (1 mol/L), each
pure substance must be in its standard state, its most stable form at
a pressure of 1 atm at a specified temperature.
– Enthalpies of formation measured under standard conditions are
called standard enthalpies of formation (ΔH°f).
– Standard enthalpy of formation of any element in its standard state
is zero.
Enthalpies of Reaction
 Hess’s law allows the calculation of the enthalpy change for any
conceivable chemical reaction by using a relatively small set of
tabulated data
1. Enthalpy of combustion, ΔHcomb—enthalpy change that occurs
2.
3.
4.
5.
when a substance is burned in excess oxygen
Enthalpy of fusion, ΔH fus—enthalpy change that accompanies
the melting, or fusion, of 1 mol of a substance
Enthalpy of vaporization, ΔHvap—the enthalpy change that
accompanies the vaporization of 1 mol of a substance
Enthalpy of solution, ΔHsoln—enthalpy change when a specified
amount of solute dissolves in a given quantity of solvent
Enthalpy of formation, ΔHf—enthalpy change for the formation
of 1 mol of a compound from its component elements
Hess’s Law
• Hess’s Law of Heat Summation, Hrxn = H1 +H2 +H3 + ..., states that
the enthalpy change for a reaction is the same whether it occurs by one step
or by any (hypothetical) series of steps.
– Hess’s Law is true because H is a state function.
• If we know the following Ho’s
1 4 FeO(s)  O 2(g)  2 Fe2O3(s)
2 2 Fe(s)  O 2(g)  2 FeO(g)
3 4 Fe(s)  3 O 2(g)  2 Fe2O3(s)
H  560 kJ
o
H  544 kJ
o
H  1648 kJ
o
Hess’s Law
•
•
For example, we can calculate the Ho for reaction [1] by properly adding (or
subtracting) the Ho’s for reactions [2] and [3].
Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a product.
– Arrange reactions [2] and [3] so that they also have FeO and O2 as reactants and
Fe2O3 as a product.
• Each reaction can be doubled, tripled, or multiplied by half, etc.
• The Ho values are also doubled, tripled, etc.
• If a reaction is reversed the sign of the Ho is changed.
Hess’s Law
• Given the following equations and Ho values
H
o
 kJ 
[1] 2 N 2 g   O 2 g   2 N 2 O  g 
164.1
[2] N 2 g  + O 2 g   2 NO  g 
[3] N 2 g  + 2 O 2 g   2 NO 2  g 
180.5
66.4
calculate Ho for the reaction below.
N 2 O g  + NO2 g   3 NOg 
H o  ?
Hess’s Law
• Calculate the H o298 for the following reaction:
C3H 8(g)  5 O 2(g)  3 CO 2(g)  4 H 2 O (  )
Hess’s Law
•
•
Application of Hess’s Law and more algebra allows us to calculate the Hfo for a
substance participating in a reaction for which we know Hrxno , if we also know
Hfo for all other substances in the reaction.
Given the following information, calculate Hfo for H2S(g).
2 H2Sg + 3 O2g 2 SO2g +2 H2Ol
Hof ?
(kJ / mol)
0
-296.8 -285.8
Ho298 = -1124 kJ
HEAT CAPACITY
The heat required to raise an object’s T by 1 ˚C.
Which has the larger heat capacity?
Thermal energy cannot be measured easily.
Temperature change caused by the flow of thermal energy between objects or substances
can be measured.
Calorimetry is the set of techniques employed to measure enthalpy changes in chemical
processes using calorimeters.
Specific Heat Capacity
How much energy is transferred due to
Temperature difference?
The heat (q) “lost” or “gained” is related to
a) sample mass
b) change in T and
c) specific heat capacity
Specific heat capacity = heat lost or gained by substance (J)
(mass, g) (T change,K)
Table of specific heat capacities
cp J g-1 K-1
Cp J mol-1 K-1
Substance
Phase
Air (typ. room conditions)
gas
1.012
29.19
Aluminium
solid
0.897
24.2
Argon
gas
0.5203
20.7862
Copper
solid
0.385
24.47
Diamond
solid
0.5091
6.115
Ethanol
liquid
2.44
112
Gold
solid
0.1291
25.42
Graphite
solid
0.710
8.53
Helium
gas
5.1932
20.7862
 Specific heat (Cs) — amount of energy needed to
increase the temperature of 1 g of a substance by 1°C,
units are J/(g•°C)
Hydrogen
gas
14.30
28.82
Iron
solid
0.450
25.1
 The quantity of a substance, the amount of heat
transferred, its heat capacity, and the temperature
change are related in two ways:
q = nCpΔT where n = number of moles of
substance
q = mCsΔT where m = mass of substance in grams
Lithium
solid
3.58
24.8
Mercury
liquid
0.1395
27.98
Nitrogen
gas
1.040
29.12
Neon
gas
1.0301
20.7862
Oxygen
gas
0.918
29.38
Uranium
solid
0.116
27.7
gas (100 °C)
2.080
37.47
liquid (25 °C)
4.1813
75.327
2.114
38.09
 Heat capacity of an object depends on both its mass
and its composition.
 Molar heat capacity (Cp) — amount of energy
needed to increase the temperature of 1 mol of a
substance by 1°C; units of Cp are J/(mol•°C)
Water
solid (0 °C)
Aluminum
All measurements are at 25 °C unless noted. Notable minimums and
maximums are shown in maroon text.
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how
many joules of heat energy are lost by the Al?
Heat/Energy Transfer
No Change in State
q transferred = (sp. ht.)(mass)(∆T)
Heat Transfer
•
•
•
•
•
Use heat transfer as a way to find specific heat capacity, Cp
55.0 g Fe at 99.8 ˚C
Drop into 225 g water at 21.0 ˚C
Water and metal come to 23.1 ˚C
What is the specific heat capacity of the metal?
Heating/Cooling Curve for Water
Note that T is
constant as ice melts
or water boils
Thermochemical Equations
• Thermochemical equations are a balanced chemical reaction plus the H value for
the reaction.
– For example, this is a thermochemical equation.
C5 H12( )  8 O 2(g)  5 CO2(g)  6 H 2 O (  )  3523 kJ
1 mole
8 moles
5 moles
6 moles
1 mole
• The stoichiometric coefficients in thermochemical equations must be interpreted as
numbers of moles.
• 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of H2O, and
releasing 3523 kJ is referred to as one mole of reactions.
Thermochemical Equations
•
This is an equivalent method of writing thermochemical equations.
C5H12()  8 O2(g)  5 CO2(g)  6 H2O() H
•
•
H < 0 designates an exothermic reaction.
H > 0 designates an endothermic reaction
o
rxn
 - 3523 kJ
Thermochemical Equations
• Write the thermochemical equation for the reaction for
CuSO4(aq) + 2NaOH(aq)
Cu(OH)2(s) + Na2SO4(aq)
50.0mL of 0.400 M CuSO4 at 23.35 oC
50.0mL of 0.600 M NaOH at 23.35 oC
Density final solution = 1.02 g/mL
Tfinal 25.23oC
CH2O = 4.184 J/goC
Gas Laws
At the macroscopic level, a complete physical description of a sample of a
gas requires four quantities:
1. Temperature (expressed in K)
2. Volume (expressed in liters)
3. Amount (expressed in moles)
4. Pressure (given in atmospheres)
• These variables are not independent — if the values of any three of these
quantities are known, the fourth can be calculated.
Standard Temperature and Pressure
• Standard temperature and pressure is given the symbol STP.
– It is a reference point for some gas calculations.
• Standard P  1.00000 atm or 101.3 kPa
o
K = C
• Standard T  273.15 K or 0.00oC
– Gas laws must use the Kelvin scale to be correct.
• Relationship between Kelvin and centigrade.
+ 273
Boyle’s Law: The Volume-Pressure Relationship
•
•
•
•
•
•
V  1/P or V= k (1/P) or PV = k
P1V1 = k1 for one sample of a gas.
P2V2 = k2 for a second sample of a gas.
k1 = k2 for the same sample of a gas at the same T.
Mathematically we write Boyle’s Law as P1V1 = P2V2
This relationship between pressure and volume is
known as Boyle’s law which states that at constant
temperature, the volume of a fixed amount of a gas is inversely
proportional to its pressure.
Robert Boyle (1627-1691). Son
of Earl of Cork, Ireland.
Charles’ Law: The Volume-Temperature Relationship;
The Absolute Temperature Scale
Charles’s and Gay-Lussac’s findings can be stated as: At
constant pressure, the volume of a fixed amount of a gas is
directly proportional to its absolute temperature
(in K).
This relationship is referred to as Charles’s law and is
stated mathematically as
V = (constant) [T (in K)] or V  T (in K, at constant P).
Jacques Charles (1746-1823).
Isolated boron and studied
gases. Balloonist.
Charles’ Law: The Volume-Temperature Relationship;
The Absolute Temperature Scale
Volume (L)
35
30
25
20
15
10
5
0
Gases liquefy
before reaching 0K
0
50
100
150 200 250
Temperature (K)
300
350
400
absolute zero = -273.15 0C
• Gay-Lussac showed that a plot of V versus T was a straight line that could be extrapolated to –
273.15ºC at zero volume, a theoretical state.
• The slope of the plot of V versus T varies for the same gas at different pressures, but the intercept
remains constant at –273.15ºC.
• Significance of the invariant T intercept in plots of V versus T was recognized by Thomson (Lord
Kelvin), who postulated that –273.15ºC was the lowest possible temperature that could theoretically be
achieved, and he called it absolute zero (0 K).
The Combined Gas Law Equation
• Boyle’s and Charles’ Laws combined into one statement is
called the combined gas law equation.
– Useful when the V, T, and P of a gas are changing.
Boyle' s Law
P1V1  P2 V2
Charles' Law
V1 V2

T1 T2
For a given sample of gas : The combined gas law is :
PV
P1 V1 P2 V2
k

T
T1
T2
Avogadro’s Law and
the Standard Molar Volume
• Avogadro’s Law states that at the same temperature and pressure, equal
volumes of two gases contain the same number of molecules (or moles) of
gas.
• If we set the temperature and pressure for any gas to be STP, then one mole
of that gas has a volume called the standard molar volume.
• Stated mathematically: V = (constant) (n) or V  n (at constant T and P)
• The standard molar volume is 22.4 L at STP.
– This is another way to measure moles.
– For gases, the volume is proportional to the number of moles.
11.2 L of a gas at STP = 0.500 mole
44.8 L of a gas at STP = ? Moles
Boyle’s Law:
The Volume-Pressure Relationship
• At 25oC a sample of He has a volume of 4.00 x 102 mL under a
pressure of 7.60 x 102 torr. What volume would it occupy under a
pressure of 2.00 atm at the same T?
Charles’ Law:
The Volume-Temperature Relationship;
The Absolute Temperature Scale
• A sample of hydrogen, H2, occupies 1.00 x 102 mL at 25.0oC and 1.00
atm. What volume would it occupy at 50.0oC under the same pressure?
The Combined Gas Law Equation
• A sample of nitrogen gas, N2, occupies 7.50 x 102 mL at 75.00C under
a pressure of 8.10 x 102 torr. What volume would it occupy at STP?
Avogadro’s Law and the
Standard Molar Volume
• One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given
temperature and pressure. (a) What is its molar mass? (b) What is its
density at STP?
Summary of Gas Laws:
The Ideal Gas Law
•
What volume would 50.0 g of ethane, C2H6, occupy at 1.40 x 102 oC
under a pressure of 1.82 x 103 torr?
Summary of Gas Laws:
The Ideal Gas Law
• Calculate the number of moles in, and the mass of, an 8.96 L sample of
methane, CH4, measured at standard conditions?
Dalton’s Law of Partial Pressures
1. The ideal gas law assumes that all gases behave identically. If V, T, and n for
each gas in a mixture are known, the pressure of each gas, its partial pressure,
can be calculated. Holding V and T constant, P is directly proportional to n:
P = n(RT/V) = n(constant)
2. Generally, for a mixture of i components, the total pressure is given by
Pt = (n1 + n2 + n3 + - - - +ni) (RT/V).
3. The above equation makes it clear that, at constant T and V, P depends on
only the total number of moles of gas present, whether the gas is a single
chemical species or a mixture of gaseous species. Nothing in the equation
depends on the nature of the gas, only on the quantity.
4. The total pressure exerted by a mixture of gases is the sum of the partial pressures of
component gases. This law is known as Dalton’s law of partial pressures and can be
written mathematically as
Pt = P1 + P2 + P3 - - - + Pi
where Pt is the total pressure and the other terms are the partial pressures of
the individual gases.
• Vapor Pressure is the pressure exerted by a substance’s
vapor over the substance’s liquid at equilibrium.
• Partial pressure is the pressure the gas would exert if it
were the only one present (at the same temperature and
volume).
John Dalton 1766-1844
Dalton’s Law of Partial Pressures
• If 1.00 x 102 mL of hydrogen, measured at 25.0 oC and 3.00 atm pressure,
and 1.00 x 102 mL of oxygen, measured at 25.0 oC and 2.00 atm pressure,
were forced into one of the containers at 25.0 oC, what would be the
pressure of the mixture of gases?
Dalton’s Law of Partial Pressures
• A sample of hydrogen was collected by displacement of water at 25.0 oC.
The atmospheric pressure was 748 torr. What pressure would the dry
hydrogen exert in the same container?
Mass-Volume Relationships
in Reactions Involving Gases
•In this section we are looking at reaction stoichiometry, like in Chapter 3,
just including gases in the calculations.
MnO 2 &
2 KClO 3(s) 
 2 KCl(s) + 3 O 2 ( g)
2 mol KClO3
2(122.6g)
yields
yields
2 mol KCl and 3 mol O2
2 (74.6g)
and 3 (32.0g)
Those 3 moles of O2 can also be thought of as:
3(22.4L)
or
67.2 L at STP
Mass-Volume Relationships in
Reactions Involving Gases
• What volume of oxygen measured at STP, can be produced by the
thermal decomposition of 120.0 g of KClO3?
Real Gases: Deviations from Ideality
The Ideal Gas Law ignores both the volume occupied by
the molecules of a gas and all interactions between
molecules, whether attractive or repulsive.
In reality, all gases have a volume and the molecules of
real gases interact with one another.
For an ideal gas, a plot of PV/nRT versus P gives a
horizontal line with an intercept of 1 on the PV/nRT axis.
Real gases behave ideally at ordinary temperatures
and pressures. At low temperatures and high
pressures real gases do not behave ideally.
The reasons for the deviations from ideality are:
• The molecules are very close to one another, thus
their volume is important.
• The molecular interactions also become important.
J. van der Waals, 1837-1923, Professor of
Physics, Amsterdam. Nobel Prize 1910.
Real Gases: Deviations from Ideality
The van der Waals’ equation
(P + an2/V2) (V – nb) = nRT
accounts for the behavior of real gases at low temperatures and high pressures.
The van der Waals constants a and b are empirical constants that differ for
each gas that take into account two things:
Pressure term, P + (an2/V2)
a corrects for intermolecular attractive forces that tend to reduce
the pressure from that predicted by the ideal gas law
For nonpolar gases the attractive forces are London Forces
For polar gases the attractive forces are dipole-dipole attractions
or hydrogen bonds.
n2/V2 represents the concentration of the gas (n/V) squared because
it takes two particles to engage in the pairwise intermolecular
interactions
Volume term, V – nb, corrects for the volume occupied by the gaseous
molecules
b accounts for volume of gas molecules
At large volumes a and b are relatively small and the van der Waal’s
equation reduces to the ideal gas law at high temperatures and low
pressures.
Real Gases:
Deviations from Ideality
• Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L
container at 200. oC using the ideal gas law.
PV = nRT
P = nRT/V
n = 84.0g * 1mol/17 g T = 200 + 273
P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K)
(5 L)
P = 38.3 atm
Real Gases: Deviations from Ideality
• Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC
using the van der Waal’s equation. The van der Waal's constants for ammonia are: a = 4.17
atm L2 mol-2 b =3.71x10-2 L mol-1

n 2a 
V  nb  nRT
P +
2 
V 

nRT n 2a
P
- 2
V - nb V
n = 84.0g * 1mol/17 g T = 200 + 273
P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K)
5 L – (4.94 mol*3.71E-2 L mol-1)
P = 39.81 atm – 4.07 atm = 35.74
P = 38.3 atm
7% error
(4.94 mol)2*4.17 atm L2 mol-2
(5 L)2