Chapter 6: Thermodynamics the First Law

1. 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…

2. 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

3. 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)

4. 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

5. Work Remember from earlier discussions: Work = Force x Distance 1J = 1 kgm2s-2

6. Energy The 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

7. 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

8. 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 = PV w = -PV 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.

9. Let’s Look at the Equation for Expansion Work w = -PV What does this tell us? • 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 1kgm-1s-2 and volume is in m3 1 kgms-2 (m3) = 1kgm2s-2 = 1 J ALWAYS REMEMBER YOUR UNITS!!!!!!! 

10. Work P, T Heat 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 In this type of setup, we have a reversible isothermal expansion

11. Reversible Isothermal Expansion • The work done in this type of expansion of an ideal gas is given by : 

12. Heat • Heat has a special definition inThermodynamics 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…

13. 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

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

15. Heat Capacity • Once we know the heat capacity, C, we can calculate the heat necessary to raise the temperature of a substance q = CT • 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

16. Heat Capacity • Because Heat Capacity is an extensive property, we can use 2 terms to describe it in a standardized manner. • Specific Heat Capacity: Cs = C/m = JK-1g-1 • Molar Heat Capacity: Cm = C/n = JK-1mol-1

17. Heat Capacity q = CT = mCsT nCmT

18. Calorimetry • Practical measurements of heat transfers are done with calorimeters Two types: Coffee cup calorimeter: Constant Pressure Bomb calorimeter: Constant Volume

19. CALORIMETRY Measures the energy needed or produced in a chemical reaction A calorimeter allows the measurement of this energy A constant pressure calorimeter allows a direct measurement of the enthalpy change during a reaction

20. Coffee Cup Calorimeter The reaction under study is carried out in solution An exothermic reaction causes heat to be released into the solution and the solution temperature increases We’ll measure ΔT, so if we know mass and Cp we can calculate ΔH

21. Coffee Cup Calorimeter System: Solution and chemicals that react Surroundings: Cup and the world around it! Assumptions: We use 2 cups to prevent energy transfer to the surroundings (we assume that it works as designed) Expected Changes: • As the chemical reaction occurs, the potential energy in the reactants will be released as heat or the solution can supply heat to allow formation of a product with a higher potential energy • The solution will abosorb or release energy during the reaction. We will see this as a temperature change qr + qsolution = 0

22. Constant Pressure Calorimetry: An Example We place 0.05g of Mg chips in a coffee cup calorimeter and add 100 mL of 1.0M HCl, and observe the temperature increase from 22.21°C to 24.46°C. What is the ΔH for the reaction? Mg(s) + 2HCl (aq) --> H2(g) + MgCl2(aq) Assume: Cp of the solution = 4.20 J/gK Density of HCl is 1.00 g/mL

23. Constant Pressure Calorimetry: An Example To solve this: ΔT = (24.46°C – 22.21°C) = (297.61K – 295.36K)=2.25K Mass of solution = Now, let’s calculate qsolution: qsolution = mCmΔT = (100.05g)(4.20 J/gK)(2.25K) = 945.5 J Now, let’s calculate qr: qr = -qsolution = -945.5 J

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

25. Constant Volume Calorimetry: Using Bombs in the Lab • Technique can be used to obtain the heat content of combustion of compounds • Used in the food, fuel and pharmaceutical industries to know how much energy would be released by completely consuming the compound • Uses a BOMB Calorimeter

26. CALORIMETRY • 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

27. Some heat from reaction warms water qwater = CmH2O(water mass)(∆T) Some heat from reaction warms “bomb” qbomb = (heat capacity, J/K)(∆T) Calorimetry Total heat evolved = qtotal = qwater + qbomb

28. Measuring Heats of ReactionCALORIMETRY Calculate energy of combustion (∆U) of octane. 2C8H18 + 25O2-->16CO2 + 18 H2O • Burn 1.00 g of octane • Temp rises from 25.00 to 33.20 oC • Calorimeter contains 1200. g water • Heat capacity of bomb = 837 J/K

29. Measuring Heats of ReactionCALORIMETRY Step 1 Calc. energy transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. energy transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J Step 3 Total energy evolved 41,200 J + 6860 J = 48,060 J Energy of combustion (∆U) of 1.00 g of octane = - 48.1 kJ

30. 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

31. The First Law: Put Another Way 1st Law of Thermodynamics A system can store energy. A change in the energy of a system means that there must be a change in the heat or the work done BY or TO the system. OR The Total Energy of the Universe is Constant

32. 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

33. 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

34. 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…

35. Zero! U = 0 for the isothermal expansion of an ideal gas