Chapter 6: Thermodynamics the First Law

• Everything in the universe happens because of energy

• In this chapter, we’re going to look at the transformation of energy from one form to another. This is called Thermodynamics

• First, we’ll need to define some terms and concepts…

Systems

• To study energy, we need to make a distinction between the region/volume we want to study and the world outside it.

• The System is the volume/region we are interested in– The flask, beaker, cylinder, etc.

• The Surroundings is the volume/region immediately around the system where heat can be observed coming into or out of the system

Systems

• Systems can be open, sealed or isolated

• An Open System can take both matter and energy from the surroundings

– Engines (fuel & electricity)– Your body (food & heat)

• A Closed System can only exchange energy with the surroundings– Glow sticks– Thermal Packs

• An Isolated System exchanges nothing with the surroundings– Thermos bottles (insulated walls and sample cylinder in the middle)

Work and Energy

• Work is defined as motion against an opposing force– When we push a large block, we are doing work

against gravity– Batteries move current through a circuit

• We can use a simple test to determine if something does work:– Can the process be harnessed to raise a weight?

• Electric circuit --> Motor --> raise weight• Gas expansion --> Piston --> raise weight• Food digestion --> Muscle --> raise weight

Work

Remember from earlier discussions:

Work = Force x Distance

1J = 1 kgm2s-2

EnergyThe capacity of a system to do work is called its internal energy, U

We cannot readily compute U for a system, but we can measure the changes in U

If a system does 15J of work, then we know that the internal energy of the system has decreased by 15J or:

U = - 15 J• Always use the appropriate sign, either ‘+’ or ‘-’ when describing energy

Provided no other changes are occurring, we can set the change in internal energy equal to the work done by the system as:

U = w

Types of Work• Expansion Work: Work that changes the volume of a

system– Car engines are an example of work done by expansion

• Non-expansion Work: Work that doesn’t involve a change in volume– The electron transport chain in cells is a perfect example of

nonexpansion work

Expansion Work

Lets Look at a Cylinder…We can relate the work done by

expanding the volume of a cylinder to the pressure by:

Work = F x d

w = PA(d)

w = PV

w = -PV

But in expanding a cylinder against atmospheric pressure, F=PA

But Area x distance moved is equal to V

Now, since the system is doing work, it is losing energy, so we must use the correct sign convention.

Let’s Look at the Equation for Expansion Work

w = -PVWhat does this tell us?

1) If there is no pressure to push against, there’s no work• This is called free expansion

• Let’s check the units. Pressure is measured in Pa or 1kgm-1s-2 and volume is in m3

1 kgms-2 (m3) = 1kgm2s-2 = 1 J

ALWAYS REMEMBER YOUR UNITS!!!!!!!

Special Type of Expansion: Reversible Isothermal Expansion

If we have a piston in a water bath, the piston slowly expands against atmospheric pressure due to the constant input of thermal energy

We can reverse this process by removing heat from the cylinder

WorkP, T

Heat

In this type of setup, we have a reversible isothermal expansion

Reversible Isothermal Expansion

• The work done in this type of expansion of an ideal gas is given by :

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w = −nRT lnVfinal

Vinitial

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Heat• Heat has a special definition in Thermodynamics

Heat is the energy transferred as a result of a temperature difference

• Energy flows as heat from regions of high temperature to regions of low temperature– For hundreds of years, human thought that heat was a tangible thing like a liquid, however, despite

the terms we use it isn’t!

• Let’s think about what happens when we bring a population of high temperature gas molecules into contact with a population of low temperature gas molecules…

Heat• We’ll abbreviate the heat transferred to a system as ‘q’• When the only change to the internal energy of a system is in the form of heat:

U = q

• Conventions:– When heat enters the system from the surroundings, we us a ‘+’ sign

• This is an endothermic reaction

– When heat leaves the system and goes to the surroundings, we use a ‘-’ sign.• This is an exothermic reaction

Measuring Heat

• We can use a thermometer to measure the change in temperature caused by the transfer of heat– However, to convert this measurement to heat, we

need to know the relationship between change in temperature and heat supplied

• The Heat Capacity, C, of a system is the ratio of heat supplied and change in temperature

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C =q

ΔT

A large heat capacity means that for a given amount of heat, you’ll only see a small change in temperature

Heat Capacity

• Once we know the heat capacity, C, we can calculate the heat necessary to raise the temperature of a substance

q = CT• Heat capacity is an extensive property:

The more sample you have, the more heat would be required to raise the temperature of the sample one degree

Heat Capacity

• Because Heat Capacity is an extensive property, we can use 2 terms to describe it in a standardized manner.

1. Specific Heat Capacity:

Cs = C/m = JK-1g-1

2. Molar Heat Capacity:

Cm = C/n = JK-1mol-1

Heat Capacity

q = CT = mCsT

nCmT

Calorimetry• Practical measurements of heat

transfers are done with calorimeters

Two types:

1) Coffee cup calorimeter: Constant Pressure

2) Bomb calorimeter: Constant Volume

Measures the energy needed or Measures the energy needed or produced in a chemical reactionproduced in a chemical reaction

A A calorimeter calorimeter allows the measurement allows the measurement of this energyof this energy

A constant pressure calorimeter allows A constant pressure calorimeter allows a direct measurement of the a direct measurement of the enthalpy change during a reactionenthalpy change during a reaction

CALORIMETRYCALORIMETRY

The reaction under study The reaction under study is carried out in solutionis carried out in solution

An exothermic reaction An exothermic reaction causes heat to be causes heat to be released into the released into the solution and the solution solution and the solution temperature increasestemperature increases

We’ll measure We’ll measure ΔΔT, so if T, so if we know mass and Cwe know mass and Cp p

we can calculate we can calculate ΔΔHH

Coffee Cup CalorimeterCoffee Cup Calorimeter

System: Solution and chemicals that reactSystem: Solution and chemicals that react

Surroundings: Cup and the world around it!Surroundings: Cup and the world around it!

Assumptions: We use 2 cups to prevent energy Assumptions: We use 2 cups to prevent energy transfer to the surroundings (we assume that it transfer to the surroundings (we assume that it works as designed)works as designed)

Expected Changes:Expected Changes:

i)i) As the chemical reaction occurs, the potential As the chemical reaction occurs, the potential energy in the reactants will be released as heat energy in the reactants will be released as heat or the solution can supply heat to allow or the solution can supply heat to allow formation of a product with a higher potential formation of a product with a higher potential energyenergy

ii)ii) The solution will abosorb or release energy The solution will abosorb or release energy during the reaction. We will see this as a during the reaction. We will see this as a temperature changetemperature change

qqrr + q + qsolutionsolution = 0 = 0

Coffee Cup CalorimeterCoffee Cup Calorimeter

We place 0.05g of Mg chips in a coffee cup We place 0.05g of Mg chips in a coffee cup calorimeter and add 100 mL of 1.0M HCl, and calorimeter and add 100 mL of 1.0M HCl, and observe the temperature increase from 22.21observe the temperature increase from 22.21°C °C to 24.46°C. What is the to 24.46°C. What is the ΔΔH for the reaction?H for the reaction?

Mg(s) + 2HCl (aq) Mg(s) + 2HCl (aq) -->--> H H22(g) + MgCl(g) + MgCl22(aq)(aq)

Assume: CAssume: Cpp of the solution = 4.20 J/gK of the solution = 4.20 J/gK

Density of HCl is 1.00 g/mLDensity of HCl is 1.00 g/mL

Constant Pressure Constant Pressure Calorimetry: An ExampleCalorimetry: An Example

To solve this:To solve this:

ΔΔT = (24.46°C – 22.21°C) = (297.61K – 295.36K)=2.25KT = (24.46°C – 22.21°C) = (297.61K – 295.36K)=2.25K

Mass of solution = Mass of solution =

Now, let’s calculate qNow, let’s calculate qsolutionsolution::

qqsolutionsolution = mC = mCmmΔΔT = T = (100.05g)(4.20 J/gK)(2.25K)(100.05g)(4.20 J/gK)(2.25K)

= 945.5 J= 945.5 J

Now, let’s calculate qNow, let’s calculate qrr::

qqrr = -q = -qsolutionsolution = -945.5 J = -945.5 J

Constant Pressure Constant Pressure Calorimetry: An ExampleCalorimetry: An Example

g 100.05 0.05g ml

1g x ml100 =+⎟

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A piece of chromium metal weighing A piece of chromium metal weighing 24.26 g is heated in boiling water to a 24.26 g is heated in boiling water to a temperature of 98.3temperature of 98.3°C°C and then and then dropped into a coffee cup calorimeter dropped into a coffee cup calorimeter containing 82.3g of water at 23.3containing 82.3g of water at 23.3°C. °C. When thermal equilibrium is reached, When thermal equilibrium is reached, the final temperature is 25.6°C. the final temperature is 25.6°C. Calculate the CCalculate the Cmm of chromium. of chromium.

Constant Pressure Constant Pressure Calorimetry: Another ExampleCalorimetry: Another Example

• Technique can be used to obtain the Technique can be used to obtain the heat content of combustion of heat content of combustion of compoundscompounds

• Used in the food, fuel and Used in the food, fuel and pharmaceutical industries to know how pharmaceutical industries to know how much energy would be released by much energy would be released by completely consuming the compoundcompletely consuming the compound

• Uses a Uses a BOMB CalorimeterBOMB Calorimeter

Constant Volume Calorimetry: Using Constant Volume Calorimetry: Using Bombs in the LabBombs in the Lab

CALORIMETRYCALORIMETRYCALORIMETRYCALORIMETRY

•Place sample of known mass inside the bomb

•Place oxygen in the sample chamber and immerse bomb into water

•Ignite the bomb and measure temperature of water

•Since the volume doesn’t change, no P-V work is done, so the qr is a measurement of the ΔU

CalorimetryCalorimetry

Some heat from reaction warms waterqwater = CmH2O(water mass)(∆T)

Some heat from reaction warms “bomb”qbomb = (heat capacity, J/K)(∆T)

Total heat evolved = qtotal = qwater + qbomb

Calculate energy of combustion (∆U) of Calculate energy of combustion (∆U) of octane. octane. 2C2C88HH1818 + 25O + 25O22 -->--> 16CO16CO22 + 18 H + 18 H22OO

•• Burn 1.00 g of octaneBurn 1.00 g of octane

• Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20 ooCC

• Calorimeter contains 1200. g waterCalorimeter contains 1200. g water

• Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

Step 1Step 1 Calc. energy transferred from reaction to water. Calc. energy transferred from reaction to water.q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 Jq = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 JStep 2Step 2 Calc. energy transferred from reaction to bomb. Calc. energy transferred from reaction to bomb.q = (bomb heat capacity)(∆T)q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 JStep 3Step 3 Total energy evolved Total energy evolved 41,200 J + 6860 J = 48,060 J41,200 J + 6860 J = 48,060 J

Energy of combustion (∆U) of 1.00 g of octane Energy of combustion (∆U) of 1.00 g of octane = - 48.1 kJ= - 48.1 kJ

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

The First Law of Thermodynamics

• Up until now, we have only considered the changes in the internal energy of a system as functions of a single change: either work or heat

• However, these changes rarely occur singly, so we can describe the change in internal energy as:

U = q + w (The 1st Law)

• The change in internal energy is dependent upon the work done by the system and the heat gained or lost by the system

The First Law: Put Another Way

11stst Law of Thermodynamics Law of Thermodynamics

A A systemsystem can store energy. A change in the can store energy. A change in the energy of a system means that there must be energy of a system means that there must be a change in the heat or the work done a change in the heat or the work done BYBY or or

TOTO the system. the system.

OROR

The Total Energy of the Universe is ConstantThe Total Energy of the Universe is Constant

The First Law

• Internal energy is an example of a state function

• A state function is a property that only depends on the current state of the system and is independent of how that state was reached

• Pressure, Volume, Temperature and density are all examples of state functions

State Functions and Things that Aren’t

• Work and heat ARE NOT state functions

• The amount of work done depends on how the change was brought about

Change in Internal Energy (Implications of a State Function)

• It doesn’t matter what path we take to get to the final point, the change in internal energy is only dependent on where we started and where we finished

• Let’s think about this on a molecular level…– If we expand an ideal gas isothermally, the molecules will have

the same kinetic energy and will move at the same speed– Despite the fact that the volume has increased, the potential

energy of the system remains the same because there are no forces between molecules (KMT)

– Since neither the kinetic nor potential energy has changed, the change in internal energy is…

Zero!

U = 0 for the isothermal expansion of an ideal gas