Lecture GJ 1-1 – Thermodynamics: The First Law Page 1

THERMODYNAMICS: THE FIRST LAW

Chapter 7

Thermochemistry

General Chemistry 10th Ed.

R. H. Petrucci, F. G. Herring, J. D. Madura & C. Bissonnette

Lecture GJ 1-1 – Thermodynamics: The First Law Page 2

INTRODUCTION

Energy maintains life (human body temperature control, thinking, etc.)

Access to energy is one of the major problems facing humanity

Thermodynamics: the study of the transformations of energy from one form into another

The first law of thermodynamics: it is concerned with keeping track of energy changes

The second law of thermodynamics: it explains why some chemical reactions take place but others do not

Statistical thermodynamics: the interpretation of the laws of thermodynamics in terms of the average behavior of the large numbers of atoms and molecules that make up a typical (bulk) sample – it relates atoms/molecules to bulk properties

Lecture GJ 1-1 – Thermodynamics: The First Law Page 3

SYSTEMS, STATES, AND ENERGY

Heat and work: (i) fundamental concepts of thermodynamics; (ii) two ways in which energy can be transferred

7.1 Systems Thermodynamics studies how energy is transformed from one form into another and transferred from one place to another

System: the region in which we are interested – place where a process occurs

Surroundings: everything else

The system and surroundings make up the universe

The only part of the universe significantly affected by a process is the sample we measure changes in the sample and its immediate surroundings

Lecture GJ 1-1 – Thermodynamics: The First Law Page 4

Open system: it can exchange both matter and energy with the surroundings

Closed system: it has a fixed amount of matter, but can exchange energy with the surroundings

Isolated system: it has no contact with the surroundings

7.2 Work and Energy Work, w: the process of achieving motion against an opposing force

Work required to move an object a distance, d, against an opposing force, F:

dFw =

Energy: the capacity of a system to do work

SI unit of work and energy: the joule, J; 1 J = 1 kg·m2·s−2

Lecture GJ 1-1 – Thermodynamics: The First Law Page 5

Important: (i) If a system can do a lot of work (on the surroundings), it possesses a lot of energy; (ii) when a system does work on the surroundings, its capacity to do work is reduced, so its energy has fallen (e.g. a hot compressed gas can do more work than the same gas after it has expanded and cooled it possesses more energy initially); (iii) when work is done on a system, then its capacity to do work and its energy has increased

Internal energy, U: the total store of energy in a system

Important: (i) we cannot measure the absolute value of U of a system – it includes the energies of all the atoms, their electrons, and the components of their nuclei; (ii) we can measure changes in internal energy, ∆U

Change of property X, ∆X:

initialfinal XXX −=∆

A negative value of ∆X: the value of X has decreased

A positive value of ∆X: the value of X has increased

Important: We always show the sign of ∆X; we write ∆X = +15 J or ∆X = –15 J

Lecture GJ 1-1 – Thermodynamics: The First Law Page 6

7.3 Expansion Work Two kinds of work: (i) expansion work – the work arising from a change in the volume of a system (e.g. a gas expanding in a cylinder fitted with a piston pushes out against the atmosphere); (ii) non-expansion work – the work that does not involve a change in volume of the system (e.g. a chemical reaction in a battery can do non-expansion work by causing an electrical current to flow)

Table: Varieties of Work

Lecture GJ 1-1 – Thermodynamics: The First Law Page 7

Analysis of expansion work: The work done by a system consisting of a gas in a cylinder; (ii) the external pressure, Pex, provides the force opposing expansion; (iii) Pex = const when the piston is pressed on by the atmosphere

Work done upon expansion:

w = Pex A d = Pex ∆V

When a system expands, it loses energy ∆V is positive and w is negative

VPw ∆−= ex

Important: (i) Equation applies to all system (liquids, solids); (ii) it applies only when Pex = const

Observation: The internal energy of the system decreases because some energy

is lost as work when the system expands (hence a negative sign!)

SI unit: 1 Pa m3 = 1 kg m–1 s–2 × 1 m3 = 1 kg m2 s–2 = 1 J

Free expansion: expansion against Pex = 0 w = 0; a system does no expansion work when it expands into a vacuum – no opposing force

Lecture GJ 1-1 – Thermodynamics: The First Law Page 8

Example: Water expands when it freezes. How much work does 100. g of water do when it freezes at 0°C and pushes back the metal wall of a pipe that exerts an opposing pressure of 1070 atm? The densities of water and ice at 0°C are 1.00 g·cm−3 and 0.92 g·cm−3, respectively.

Solution:

w = –Pex ∆V = –Pex (Vi – Vw)

Volume change: ∆V = Vi – Vw = mi/di – mw/dw = m(1/di – 1/dw) mi = mw = m

∆V = 100 (1/0.92 – 1/1.00) = 100 × (1.087 – 1.00) = 8.7 cm3

∆V = 8.7 cm3 = 8.7 × 10–6 m3

Pex = 1070 × 1.01325 × 105 Pa

w = –Pex ∆V = –(1070 × 1.01325 × 105) × 8.7 × 10–6 m3 = –940 J

Answer: w = –0.94 kJ

Lecture GJ 1-1 – Thermodynamics: The First Law Page 9

Reversible process: (i) a process that can take place in either direction (the system is very close to equilibrium); (ii) a process that can be reversed by an infinitely small change in a variable

Reversible expansion: (i) expansion of an ideal gas against a changing external pressure at T = const; (ii) Pex differs by a finite (measurable) amount from the pressure of the system Pex ≅ P but we accept that Pex = P; (iii) at T = const, P falls as it expands to achieve reversible expansion, Pex must be reduced in step: at every stage Pex is the same as P (see Figure); (iv) to calculate w, we have to take into account the gradual reduction in Pex (the external opposing force changes)

For an infinitesimal change in volume, dV:

VPw dd ex−=

Pex = P PV = nRT P = nRT/V at each stage of expansion

VV

nRTw dd −=

Lecture GJ 1-1 – Thermodynamics: The First Law Page 10

The total work done is the sum (integral) of these infinitesimal contributions as the volume changes from Vi to Vf:

∫−=f

i

dV

V VVTRnw

i

flnVVTRnw −=

The work done by the system as it expands is equal to the area beneath the ideal gas isotherm lying between Vi and Vf (see Figure)

Important: The work is done at the expense of the internal energy of the system (U decreases by ∆U = w)

Lecture GJ 1-1 – Thermodynamics: The First Law Page 11

Example: A cylinder of volume 2.00 L contains 0.100 mol He(g) at 30°C. Which process does more work on the system, compressing the gas isothermally to 1.00 L with a constant external pressure of 1.00 atm or compressing it reversibly and isothermally to the same final volume?

Solution:

Irreversible compression

w = –Pex ∆V = –1.00 × 1.01325 × 105 ×(–1.0 × 10–3)

w = +101 J

Reversible compression

w = –n R T ln (Vf/Vi) = 0.1 ×8.314 × 303 ×ln(1/2)

w = +174 J

Answer: The reversible compression does more work on the system (U increases).

Lecture GJ 1-1 – Thermodynamics: The First Law Page 12

7.4 Heat Heat: the energy transferred as a result of a temperature difference

The internal energy of a system (its capacity to do work) can be changed by (i) doing work on the surroundings (or the surroundings do work on the system) or (ii) transferring energy to or from the surroundings as heat

Observation: Energy flows as heat from a high T region to a low T region (the system cannot be thermally insulating for energy flow to occur)

Heat is not a substance and does not flow what is heat (what is transferred)?

The “thermal energy” of a system is the sum of Ep and Ekin arising from the chaotic thermal motion of atoms, ions, and molecules

The “flowing” of energy due to a T difference is the transfer of the energy of thermal motion, the energy associated with the random motion of molecules

q: the energy transferred to/from a system as heat

Lecture GJ 1-1 – Thermodynamics: The First Law Page 13

When U of the system is changed by transferring energy as heat ∆U = q

Observation: (i) If energy enters a system as heat, U of the system increases and q is positive; (ii) if energy leaves the system as heat, U of the system decreases and q is negative

Energy transferred as heat (like any energy) is measured in joules, J

Biochemistry: a unit still widely used is the calorie, cal; 1 cal is the energy needed to raise the temperature of 1 g of water by 1°C

The modern definition: 1 cal = 4.184 J

Nutrition: the calorie, Cal, is 1 kilocalorie (kcal)

Lecture GJ 1-1 – Thermodynamics: The First Law Page 14

7.5 The Measurement of Heat Change in U is monitored by measuring heat produced/absorbed by a process

Analysis: System and surroundings are separated by a wall

Adiabatic wall: the wall is thermally insulating (e.g. a vacuum flask with a silvered surface) – heat cannot be transferred across it even if ∆T = Tsys – Tsurr ≠ 0

Important: An adiabatic system is not isolated – energy can be transferred to/from a system as work in a closed adiabatic system ∆U = w

Diathermic wall: the wall permits the transfer of energy as heat

Diathermic system does not lose energy as work: an influx of energy through the walls raises Tsys analysis of Tsys is a way to monitor the heat transferred and to infer the change in U

Heat capacity, C: it is used to convert ∆T into energy

producedriseetemperatur

suppliedheat=

∆=

TqC

Lecture GJ 1-1 – Thermodynamics: The First Law Page 15

Important: Heat capacity is an extensive property: the larger the sample, the more heat is required to raise its T by a given amount

Specific heat capacity, Cs: the heat capacity divided by the mass of the sample

mCC =s

Molar heat capacity, Cm: the heat capacity divided by the amount (in moles) of the sample

nCC =m

Water: The specific heat capacity of liquid water at room temperature is 4.184 J·(°C)−1·g−1, or 4.184 J·K−1·g−1, and its molar heat capacity is 75 J·K−1·mol−1

Lecture GJ 1-1 – Thermodynamics: The First Law Page 16

Important: We can relate the energy supplied to the system to: (i) its mass, its specific heat capacity and the temperature rise or (ii) its amount (in moles), its molar heat capacity and the temperature

TCmq ∆= s

TCnq ∆= m

Example: Potassium perchlorate, KClO4, is used as an oxidizer in fireworks. Calculate the heat required to raise the temperature of 10.0 g of KClO4 from 25°C to an ignition temperature of 900°C. The specific heat capacity of KClO4 is 0.8111 J·K−1·g−1.

Solution:

q = m Cs ∆T = 10 ×0.8111 × (900 – 25) = 7097 J

Answer: q = 7.1 kJ

Lecture GJ 1-1 – Thermodynamics: The First Law Page 17

Calorimeter: (i) A device used to measure transfers of energy as heat; (ii) the heat transfer is monitored by recording the change in T that it produces; (iii) the heat capacity of the calorimeter (calorimeter constant, Ccal) needs to be known; (iv) Ccal is determined by “calibration”: the calorimeter’s heat capacity is measured by supplying a known quantity of heat and noting the resulting rise in T (Ccal = q/∆T)

Figure: A bomb calorimeter

Example: A small piece of calcium carbonate was placed in a calorimeter whose Ccal = 488 J °C–1 (it was determined with 0.100 L of another aqueous solution), and 0.100 L of dilute hydrochloric acid was poured over it. The temperature of the calorimeter rose by 3.57°C. What is the value of ∆U for the reaction of the hydrochloric acid with the calcium carbonate?

Solution: q = Ccal ∆T = 488 × 3.57 = 1742 J

Heat is released at the expense of internal energy, thus U = –q

Answer: U = –1.74 kJ

Lecture GJ 1-1 – Thermodynamics: The First Law Page 18

7.6 The First Law The internal energy changes as a result of both work and heat (e.g. a spark ignites the mixture of gasoline vapor and air in the internal combustion engine of a moving automobile, the vapor burns and expands, transferring energy to its surroundings as both work and heat)

The change in U of a closed system is the net result of both kinds of transfers

wqU +=∆ (*)

Observation: (i) Both heat and work are means of transferring energy and thereby changing U of a system; (ii) the fundamental molecular difference between work and heat: (a) when energy is transferred as work, the system moves molecules in the surroundings in a definite direction; (b) during the transfer of energy as heat, the molecules of the surroundings are moved in random directions but more vigorously (T rises νrms increases)

Lecture GJ 1-1 – Thermodynamics: The First Law Page 19

Example: An automobile engine does 520 kJ of work and loses 220 kJ of energy as heat. What is the change in the internal energy of the engine? Treat the engine, fuel, and exhaust gases as a closed system.

Solution:

System does work on the surroundings: ∆U1 = w = –520 kJ

It also loses heat to the surroundings: ∆U2 = q = –220 kJ

Answer: Total internal energy change: ∆U = ∆U1 + ∆U2 = –740 kJ

Eq. (*) is a complete statement: the only way to change U of such a system is to transfer energy into/from it as heat or as work

If the system is isolated (no contact with the surroundings), then the internal energy cannot change at all

First law of thermodynamics:

The internal energy of an isolated system is constant

Lecture GJ 1-1 – Thermodynamics: The First Law Page 20

State function: a property that depends only on the current state of the system and is independent of how that state was prepared

U is a state function

P, V, T and d of a system are also state functions

Figure: If we took 100 g of water at 25°C and raised its temperature to 60°C, its internal energy would change by a certain amount. If we took the same mass of water at 25°C and heated it to boiling, vaporized it, condensed the vapor, and then allowed the water to cool to 60°C, the net change in the internal energy of the water would be exactly the same as if we had heated it to 60°C in one step.

Important: Work done by a system is not a state function – it depends on the path

Analysis: A gas expands at T = 25°C through 100 cm3 by two different paths: (i) against an external force (Pex ≠ 0), so w = –Pex∆V and (ii) against vacuum (Pex = 0), so w = 0 the change in the state of the gas is the same in each case, but the work done by the system is different

Lecture GJ 1-1 – Thermodynamics: The First Law Page 21

Important: Heat is not a state function

Analysis: We raise the temperature of 100 g of water from 25°C to 30°C. (i) The energy is supplied as heat by using an electric heater; the heat required is: q = (4.18 J·°C−1·g−1) × (100 g) × (5°C) = +2 kJ; (ii) the temperature is raised by stirring the water vigorously with paddles until 2 kJ of work has been supplied – all the energy required is transferred as work; none is supplied as heat, so q = 0

Example: Suppose that 2.00 mol CO2 at 2.00 atm and 300 K is compressed isothermally and reversibly to half its original volume before being used to produce soda water. Calculate w, q, and ∆U by treating the CO2 as an ideal gas.

Solution:

w = –n R T ln (Vf/Vi) = 2.0 ×8.314 × 300 ×ln(1/2) = –3460 J = –3.46 kJ

Isothermal process (T = const) ∆U = 0

∆U = w + q q = –w = + 3.46 kJ

Answer: ∆U = 0; w = –3.46 kJ; q = +3.46 kJ