Basic concept:1. Work, heat and energy2. The internal energy3. Expansion work4. Heat transaction5. Adiabatic changes

Thermochemistry:1. Standard enthalpy changes2. Standard enthalpies of

formation3. The temperature dependence

of reaction enthalpies.

State of function and exact differentials:1. Exact and inexact differentials2. Changes in internal energy3. The Joule – Thomson effect

First Law Thermodynamic

Work• Work (w) is defined as the force (F) that produces the movement of an object through a distance (d):

Work = force × distancew = F x d

• Work also has units of J, kJ, cal, kcal, Cal, etc.• The two most important types of chemical work are:

– the electrical work done by moving charged particles.– the expansion work done as a result of a volume change in a system,

particularly from an expanding or contracting gas. This is also known as pressure-volume work, or PV work.

• PV work occurs when the force is the result of a volume change against anexternal pressure.

The example of PV work – in the cylinder of an automobile engine

The combustion of the gasoline causes gases within the cylinder to expand,pushing the piston outward and ultimately moving the wheel of the car.

The relationship between a volume change (∆V) and work (w):

W= -P ∆V

Where P is external pressure

The units of PV work are L·atm; 1 L·atm = 101.3 J.

• If the gas expands, ΔV is positive, and the work termwill have a negative sign (work energy is leaving thesystem).

• If the gas contracts, ΔV is negative, and the work termwill have a positive sign (work energy is entering thesystem).

• If there is no change in volume, ΔV = 0, and there isno work done. (This occurs in reactions in whichthere is no change in the number of moles of gas.)

• If the gas expands, ΔV is positive, and the work term will have a negative sign (work energy is leaving the system).

• If the gas contracts, ΔV is negative, and the work term will have a positive sign (work energy is entering the system).

• If there is no change in volume, ΔV = 0, and there is no work done. (This occurs in reactions in which there is no change in the number of moles of gas.)

Expansion Work

b) Free expansion is expansion against zero opposing force. W=0

No work is done when a system expand freely. Expansion of this kind occurs when a gasexpands into a vacuum.

c) Expansion against constant pressure – now suppose that the external pressure is constant throughout expansion.

Therefore, if we write the change in volume as ∆V=Vf -Vi

d) Reversible ExpansionA reverse change in thermodynamic is a change that can be reversed by aninfinitesimal modification of a variable.The key “infinitesimal” sharpen the everyday meaning of the word “reversible” assomething that can change direction.

d) Isothermal reversible expansion

• When the volume is greater than the initial volume, as in expansion, the logarithmin above equation is positive and hence w<0. In this case the system has done workon the surroundings and there is a corresponding reduction in its internal energy.

• From the equation also show that more work is done for a given change of volumewhen temperature is increased:

at higher temperature the greater pressure of the confined gas needs a higheropposing pressure to ensure reversibility and the work done iscorrespondingly greater.

Table 1 Type of work

Type of work Variables Conventional unit

Volume expansion Pressure (P), volume (V) Pa m3 = J

Stretching Tension (ɤ), length (l) N m= J

Surface expansion Surface tension (ɤ), area (σ) (Nm-1)(m2) = J

Electrical Electrical potential(φ),Electrical change (q)

V C = J

Example 1:

Inflating balloon requires the inflator to do pressure-work on the surroundings. If balloons is inflated from a volume of 0.100L to 1.85L against an external pressure of 1.00 atm, how much work is done (in joules)?

Sort: you are given the initial and final volumes of the volumes of the balloon and the pressure against which its expands. The balloon and its contents are the system

Given: V1 = 0.100L, V2 = 1.85L, P=1.00 atmFind: W

Strategy: the equation W= -P ∆V specifies the amount of work done during a volume Change against an external pressure.

Conceptual Plan:P,∆V → WW= -P ∆V

Solve: To solve the problem, compute the value of ∆V and substitute it, together with P, into the equation.

The units of the answer (L.atm) can be converted to joules using 101.3J = 1L.atm

Solution:∆V = V1 - V2

= 1.85L - 0.100L= 1.75L

W= -P ∆V= - 1.00 atm x 1.75L= -1.75L.atm

-1.75 L.atm x 101.3 J = -177 J1L.atm

Check : The units (J) are correct for work. The sign it should be for an expansion: work is doneon the surroundings by the expanding balloon.

Example 2:A sample of argon of mass 6.56 g occupies 18.5 dm3 at 305K.

(a)Calculate the work done when the gas expands isothermally against a constant external pressure of 7.7 kPa until its volume has increased by 2.5 dm3.

(b) Calculate the work that would be done if the same expansion occurred reversibly.

(a)

(b)

Heat is a exchange of thermal energy between a system and its surroundings causedby temperature difference.

Notice the distinction between heat and temperature. Temperature is a measure ofthe thermal energy of a sample matter. Heat is transfer of the thermal energy.

HEAT

Heat (q)

System∆T

Temperature Change and Heat Capacity – when system absorbs heat (q) its temperature changes by ∆T:

Experimental measurements demonstrate that the heat absorbed by a system and itscorresponding temperature change are directly proportional : q α ∆T. The constant of

proportionality between q and ∆T is called the heat capacity.

• Heat capacity (C) is the amount of heat (q) a substance must absorb to raise its temperature(∆T) by 1 °C.

- Heat capacity has units of J/°C (or J/K), and is an extensive property, depending on the sample size.

• Two type of heat capacity:1) Heat capacity at constant volume:

2) Heat capacity at pressure constant

Heat Capacity

• The specific heat (c, or specific heat capacity, Cs) of an object, is thequantity of heat required to change the temperature of 1 gram of asubstance by 1 °C (or K):

- Specific heat has units of J / g °C, and is an intensive property, which isindependent of the sample size.

• The molar heat capacity (Cm), is the quantity of heat required to changethe temperature of 1 mole of a substance by 1 °C (or K):

Specific Heat

Heat Transactions

In general the change in internal energy of a system:

dwe is work in addition(e for “extra”) to the expansion work

dwexp = 0, when a system kept a constant volume cannot do any expansion work.

If the system also incapable of doing any other kind of work (if it is not, for instance, an electrochemical cell connected to an electric motor), then dwe = 0 too. Under thesecircumstances :

Heat transfer at volume

We express this relationship by writing, dU=dqv where the subscript implies a change at constant volume. For measurement change,

The amount of heat absorbed by the water inthe calorimeter is equal to the energy that isreleased by the reaction (but opposite insign).

A bomb Calorimeter

qcal : heat absorbed by the entire calorimeter assemblyCcal: heat capacity of entire calorimeter assembly

If no heat escape from the container:

The high heat capacity of water surroundingSan Francisco results in relatively cool summertemperatures.

During the summer, you have experienced theeffect of water’s high specific heat capacity.

Example:Sacramento (an inland city) and San Francisco(a coastal city) can be 18oC (30oF).San Francisco enjoy cool 20oC (68oF) whileSacramento bakes at nearly 38oC (100oF).

Why the large temperature difference?

San Francisco sit on peninsula, surrounded by thePacific Ocean. Water – absorbed much the sun’sheat without undergoing a large increase intemperature.

While the Sacramento, with its low heat capacityundergoes a large increase in temperature as itabsorbs a similar amount of heat.

heat = mass x specific heat x temperature

q= m x Cs x ∆T

Example 2:

Suppose you find a copper penny (minted pre-1982when pennies were almost entirely were almostentirely copper) in the snow. How much heat isabsorbed, by the penny as it warms, from the suntemperature of the snow, which is -8.0oC, to thetemperature of your body, 37oC? Assume the penny ispure copper and has a mass of 3.10g.

The specific heat can be used to quantify the relationship between the amount of heat added to a given amount of the substance and the corresponding temperature increase.

The equation is: heat = mass x specific heat x temperature

q= m x Cs x ∆T

Sort: You are given the mass of copper as well as its initial and final temperature. You are asked to find the heat required for the given temperature change

Given:m = 3.10 g copperTi = -8.0 oCTf = 37.0 oCFind : q

Strategy: The equation q = m x Cs x ∆T give the relationship between the amount of heat (q) and the temperature change (∆T )

Conceptual Plan:Cs , m, ∆T → qq = m x Cs x ∆T

Relationship Used:q = m x Cs x ∆T Cs = 0.385 J/g.oC

Solve: Gather the necessary quantities for the equation in the correct units and Substitute these into the equation to compute q.

Solution:∆T = Tf – Ti

= 37.0oC – (-8.0oC) = 45.0oC q = m x Cs x ∆T

= 3.10 g x 0.385 J x 45.0oC = 53.7Jg. oC

Check : The unit (J) are correct for heat. The sign of q is positive, as it should be since the penny absorbed heat from surrounding.

EnergyEnergy is define as the ability to do work.Work is done when a force is exerted through a distance.

Force through distance; work is done.

Energy is measured in Joules (J) or Calories (cal).

1 J = 1 kg m2 s-2

Energy may be converted from one to another, but it is neither created nor destroyed (conversion of energy).

In generally, system tend to move from situations of high potential energy (less stable) to situations having lower energy (more stable).

• A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1°C.

1 cal = 4.184 J

• The nutritional unit Calorie (Cal) is actually a kilocalorie (kcal):

1 Cal = 1000 cal = 1 kCal = 4184 J

Unit of Energy

• Kinetic energy (EK) is the energy due to the motion of an object with mass m and velocity v:

EK = ½ mv2

– Thermal energy, the energy associated with the temperature of an object, is a form of kinetic energy, because it arises from the vibrations of the atoms and molecules within the object.

• Potential energy (EP) is energy due to position, or any other form of “stored” energy. There are several forms of potential energy:– Gravitational potential energy– Mechanical potential energy– Chemical potential energy (stored in chemical bonds)

Potential and Kinetic Energy

Internal Energy

•The internal energy, U of a system is the sum of the kinetic and potentialenergies of all the particles that compose the system or the total energy of asystem.

•Internal energy is the state function, which means that its value depends onlythe state of the system, not the how the system arrive at the state.

•Some examples include energy (and many other thermodynamic terms),pressure, volume, altitude, distance, etc.

• An energy change in a system can occur by many different combinations ofheat (q) and work (w), but no matter what the combination, ΔU is always thesame — the amount of the energy change does not depend on how the changetakes place.

∆U = q + w

Altitude is a state function. The change in altitude during climbing depends only onthe difference between the final and initial altitudes.

In this reaction, energy is released from the system to the surroundings.

– The reactants have a higher E than the products, so ΔU for the system isnegative.

– This energy is gained by the surroundings, where ΔU is positive.(ΔUsystem = -ΔUsurroundings)

Example: reaction between carbon and oxygen to form carbon dioxide

C(s) + O2 (g) CO2(g)

ΔU < 0 (negative)

System SurroundingsEnergy flow

∆Usys < 0 (negative) ∆Usurr > 0 (positive)

If the reaction is reversed, energy is absorbed by the system from the surroundings.

– The reactants have a lower E than the products, so ΔU for the system is positive.

– This energy is lost by the surroundings, where ΔU is negative.

ΔU > 0 (positive)

The difference , ∆U is positive and energy glows into the system and out of surroundings

Summarizing:

if the reactants have a higher internal energy than a products, ∆Usys is negativeand energy flows out of the system into the surroundings.

if the reactants have a lower internal energy than a products, ∆Usys is positive and energy flows into the system from the surroundings.

System SurroundingsEnergy flow

∆Usys > 0 (positive) ∆Usurr < 0 (negative)

A system can exchange energy with its surroundings through heat and work:

System

Heat (q)

Work (w)Surroundings

According to the first law thermodynamic, the change in the internal energy of the system(∆U) must be the sum of the heat transferred (q) and the work done (w):

∆U = q +w

Sign of conventions for q, w, and ∆U

q (heat) + system gain thermal energy - System loses thermal energy

w (work) + work done on the system - Work done by the system

∆U(change in

internal energy)

+ energy flows into the system - Energy flows out of the system

• For an isolated system, with no energy flowing in or out of the system, the internal energy is a constant.

– First Law of Thermodynamics (restated): The total internal energy of an isolated system is constant.

• It is impossible to completely isolate a reaction from its surroundings, but it is possible to measure the change in the internal energy of the system, ΔU, asenergy flows into the system from the surroundings or flows from the system into the surroundings.

∆U = Uf - Ui

•Most reactions are not done in sealed containers: they are carried out in open vessels at constant pressure, with the volume capable of changing freely, especially if the reactants or products of the reaction involve gases.

• In these cases, ΔV ≠ 0, and the energy change may be due to both heat transfer and PV work.

• In order to eliminate the contribution from PV work, a quantity called enthalpy, H, is defined as internal energy (U) plus the product of pressure and volume:

p: pressure, V:volumeBecause p,V,U are all state function, the enthalpy is a state function too.

• The change in enthalpy (ΔH) is:

Enthalpy

H = U + pV

∆H = qP

The Measurement of Enthalpy Change

1) CalorimeterMonitoring the temperature change that accompanies a physical or chemical change occurring at constant pressure.Example: thermally insulated vessel open to the atmosphere: the heat released in the reaction is monitor by measuring the change in temperature of the content.

2) Bomb CalorimeterMeasuring the internal energy change.

3) Differential Scanning CalorimeterThe most sophisticated way to measure enthalpy changes.

Hm = Um +pVm ≈ Um

Example 3:

When 2.0 mol CO2 is heated at the constant pressure of 1.25 atm, it’s temperature increase from 250 K to 277K. Given the molar heat capacity of CO2 at constant pressure is 37.11JK-1, calculate q, ∆H and ∆U.

Adiabatic ChangesThe change in internal change of a perfect gaswhen the temperature is change from Vi to Vf.

Step 1:The volume changes and the temperature isconstant at its initial value. The system expandsat the constant temperature; there is no changein internal energy if the system consist of aperfect gas.

Step 2:The temperature of the system is reduced atconstant volume.

Because the expansion is adiabatic, we know that q = 0; because ∆U=q + w , it then follows that ∆U = wad. Therefore, by equating the two expansion we have obtained for ∆U we obtained

Wad = Cv ∆T

Renungan:

“Termodinamik merupakan satu subjek yang aneh. Kali pertama anda mempelajarinya,anda tidak faham langsung. Kali kedua anda mempelajarinya, anda fikir anda

memahaminyakecuali satu dua perkara. Kali ketiga anda mempelajarinya, anda tahu bahawa anda tidakmemahaminya, akan tetapi anda sudah terbiasa dengan subjek itu maka subjek itu tidak

akan menjadi masalah lagi kepada anda”

Arnorld Sommerfeld (1868-1951)