Pressure Volume Work Heats of reaction can be measured either at constant pressure, giving q P or ∆H values, or at constant volume, giving q V or ∆U values

Pressure Volume Work• Heats of reaction can be measured either at

constant pressure, giving qP or ∆H values, or at constant volume, giving qV or ∆U values. ∆H and ∆U values can be appreciably different if there are gaseous reactants or products. The difference arises from gases doing work on the surroundings or the surroundings doing work on a system (normally compressing it). We need to consider PV (pressure- volume) work.

Work and Motion• In work calculations we need usually to

consider both a force(s) and motion. For gas expansion or compression there is obviously motion. One can easily show that, for expansion of a gas in a regular cylinder, the product P∆V has work units. We’ll do this as a class exercise. The easy P∆V calculations have volume in L and pressure in kPa.

• 1 joule = 1 J = 1 L.kPa = 1 m3.Pa.

System Expanding Does PV Work• An expanding system does work on its

surroundings. A system can expand because of a chemical reaction or due to being heated. Heating can cause gas expansion or a phase change.

• He(l) → He(g) VSystem increases. Why?

• 2 KClO3(s) →2 KCl(s) + 3 O2(g) In this 2nd case VSystem also increases. Why?

Slide 4 of 57

Work During a Chemical Reaction

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7

FIGURE 7-7 •Illustrating work (expansion) during the chemical reaction •2 KClO3(s) 2 KCl(s) + 3 O2(g)

In addition to heat effects chemical reactions may also do work.

Gas formed pushes against the atmosphere.The volume changes.

Pressure-volume work.

PV Work With No Chemistry• The next slide shows a compressed gas. The

system is at mechanical equilibrium unless the pressure on the gas is reduced by removing weights. The expanding gas does work on the surroundings and we write

• wSystem = - PExternal∆VSystem or w = -P∆V

• Aside: If Pexternal varies we will need an integration (not in Chemistry 1050).

Slide 6 of 57

Pressure-volume workFIGURE 7-8

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

(m x g)

= PV

w = -PextV

A h= x A

x hw = F x d

= (m x g)

PV Calculation Using Previous Slide• The previous slide can be used to do a simple

PV “work term” calculation. This “work term” will be one of two considered for First Law of Thermodynamics calculations. The text provides an example problem (next slide) which we can also tackle slightly differently (Class example).

Copyright 2011 Pearson Canada Inc. 7 - 8

First Law & Kinetic Theory• During the discussion of ideal gases it was

suggested that gas molecules are point masses whose kinetic energy increased with T due to increasing translational velocity. Real molecules are not point masses and polyatomic gaseous molecules can also rotate and vibrate. Average rotational and vibrational energies increase with T. Condensed phases are more complicated. Why?

Slide 10 of 57

The First Law of Thermodynamics


FIGURE 7-9

•Some contributions to the internal energy of a system

Internal Energy, U.

Total energy (potential and kinetic) in a system.

• Translational kinetic energy.

• Molecular rotation.• Bond vibration.• Intermolecular attractions.• Chemical bonds.

• Electrons.

Slide 11 of 57

The First Law of Thermodynamics• A system contains only internal energy.– A system does not contain heat or work.– These only occur during a change in the system.

• Law of Conservation of Energy– The energy of an isolated system is constant


U = q + w

Slide 12 of 57

The First Law of Thermodynamics•An isolated system is unable to exchange either heat or work with its surroundings, so that DUisolated system = 0, and we can say:


The energy of an isolated system is constant.

Thermodynamic Sign Conventions

• All of us are familiar with the fact that our bank balances increase when we make a deposit (a “positive” event) and decrease when we make a withdrawal (a “negative” event). Similarly, in thermodynamics, we need to keep track of whether our system is losing heat to the surroundings or gaining heat from the surroundings when a physical or chemical change takes place. Work done on or by a system must also be accounted for carefully.

Slide 14 of 57

Illustration of sign conventions used in thermodynamicsFIGURE 7-10


Slide 15 of 57

Functions of State


•Any property that has a unique value for a specified state of a system is said to be a function of state or a state function.

•Water at 293.15 K and 1.00 atm is in a specified state. •d = 0.99820 g/mL•This density is a unique function of the state.•It does not matter how the state was established.

Slide 16 of 57

Functions of State• U is a function of

state.– Not easily measured.

• U has a unique value between two states.– Is easily measured.


Slide 17 of 57

Path Dependent Functions


•Changes in heat and work are not functions of state.–Remember example 7-5, w = -1.24 x 102 J in a one step expansion of gas:

let us consider a two step process from 2.40 atm to 1.80 atm, and finally to 1.20 atm.

Slide 18 of 57

Path Dependent Functions


FIGURE 7-11

•A two-step expansion for the gas shown in Figure 7-8

w = (-1.80 atm)(1.36-1.02)L – (1.30 atm)(2.04-

1.36)L

= -0.61 L atm – 0.82 L atm

= -1.43 L atm= -1.44 x 102 JCompared -1.24 102 J for the one

stage process

Copyright 2011 Pearson Canada Inc. 7 - 19

First Law Example• Ex. A system undergoes a series of four

changes in which (a) the system absorbs 4.88 kJ of heat from its surroundings, (b) the system does 1.25 kJ of electrical work on its surroundings, (c) the system gives up 210 J of heat to the surroundings and (d) the surroundings do 1.00 kJ of pressure volume do work on the system. Evaluate ΔUSystem for the series of four changes outlined.

Hydrocarbon Combustion Example

• Ex. The heat (enthalpy) of combustion of CH4(g) is • -890 kJ mol∙ -1 at 298 K . How much heat is released

when 9.00 g of methane is burned at constant pressure at 298 K? “Asides”: This is a highly exothermic reaction. The balanced chemical reaction for the combustion is

• CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)• Combustion of hydrocarbons drives much of our

industry and our transportation systems. The amount of CO2(g) being produced is enormous and is thought to contribute to climate change/global warming.

Heat Capacity Units

• Ex. The specific heat capacity of Hg(l) is 0.140 J K∙ -1 g∙ -1. Find the molar heat capacity of Hg(l). Hint: The problem is easily solved once the units for molar heat capacity are written.

PV Work and Mole Counting• For the reactions below no PV work is done in only

one case (at constant P and T). Which one?

• (a) CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(aq)

• (b) K2CO3(s) + 2 HCl(aq) → 2 KCl(aq) + H2O(l) + CO2(g)

• (c) H2(g) + Cl2(g) → 2 HCl(g)

• (d) H2(g) + ½ O2(g) → H2O(l)

• (e) SO2Cl2(g) + 2 H2O(l) → H2SO4(aq) + 2 HCl(g)

Reaction System Expands ?

wSystem ΔnGas

(a) CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(aq) Yes -ve +1

(b) K2CO3(s) + 2 HCl(aq) → 2 KCl(aq) + H2O(l) +CO2(g) Yes -ve +1

(c)H2(g) + Cl2(g) → 2 HCl(g) No 0 0

Contracts +ve -1.5

(e) SO2Cl2(l) + 2 H2O(l) → H2SO4(aq) + 2 HCl(g) Yes -ve +2

Heats of Combustion

• Given time, we will practice writing thermochemical equations corresponding to combustion processes for common organic compounds.

Documents

Pressure Volume Work Heats of reaction can be measured either at constant pressure, giving q P or ∆H values, or at constant volume, giving q V or ∆U values