CHEM Thermodynamics. Work. There are two ways to change the internal energy of a system:

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There are two ways to change the internal energy of a system: Thermodynamics Work 1. By flow of heat, q Heat is the transfer of thermal energy between and the surroundings 2.

2 Heat and Work are interconvertible Joule s pparatus a result of a chemical reaction System: The chemical reaction Surroundings: Everything that interacts with The amount of mechanical Energy lost =

1 There are two ways to change the internal energy of a system: Thermodynamics Work 1. By flow of heat, q Heat is the transfer of thermal energy between and the surroundings 2. By doing work, w Work can be converted into heat and vice versa. q and w are process dependent, and are not state functions. 2 Heat and Work are interconvertible Joule s pparatus a result of a chemical reaction System: The chemical reaction Surroundings: Everything that interacts with The amount of mechanical Energy lost = the amount of thermal energy gained. 1 calorie = joules ll matter has chemical energy. Chemical bonds are a source of energy (PE). The movement of molecules in space is a source of energy (KE). The vibrations and rotations of molecules is another source of chemical energy (KE). ORDERS of Magnitudes of Energy ll of these forms of chemical energy contribute in one way or another to chemical reactions. 3 4 a result of a chemical reaction a result of electrical work System: The chemical reaction Surroundings: Everything that interacts with In a chemical reaction, when bonds are broken and new bonds are formed, the internal energy of changes. Exothermic reaction heat flows out of Endothermic reaction heat flows into External load Chemical reactions can do work on their surroundings by driving an electric current through an external load

a result of work of expansion Chemical reactions can do work on their surroundings when the volume of expands during the course of the reaction.

gif Mechanical work is defined as the force exerted on an object multiplied by the distance which the object moves resulting from the applied force.

For infinitesimal change in position, dx, dw = f dx 7 8 Define a system consisting of an ideal gas enclosed in a container with a moveable piston.

f is the force P ext is the external pressure is the cross-sectional area of the piston V 2 V 2 9 10 w = f x w = P ext V Substitute f = P ext For an infinitesimal

2 a result of work of expansion Chemical reactions can do work on their surroundings when the volume of expands during the course of the reaction. Internal combustion engine burns a fuel with air and uses the hot gases for generating power. Image credit: Mechanical work is defined as the force exerted on an object multiplied by the distance which the object moves resulting from the applied force. w = f x If there is no movement (x=0), then w = 0. For infinitesimal change in position, dx, dw = f dx 7 8 Define a system consisting of an ideal gas enclosed in a container with a moveable piston. Let s evaluate the work of compression. Can we express the work of compression in terms of P and V? f is the force P ext is the external pressure is the cross-sectional area of the piston V 2 V w = f x w = P ext V Substitute f = P ext For an infinitesimal change In volume, x 1 x 2 V2 w = P ext x For a change in the position of the piston, w = P ext x x 1 x 2 V2 dw = P ext dv Integrate to get total work done. x =x 2 -x 1 V = V 2 - = (x 2 -x 1 ) x 1 is the initial piston position x 2 is the final piston position is the initial volume V w = P ext V 11 x =x 2 -x 1 x 1 is the initial piston position x 2 is the final piston position is the initial volume 12 2

Sign consideration Similarly, for work of expansion, (gas) does work. The system loses energy. Work should be a negative quantity. x 1 For work of compression, the system gains energy.

V 2 x 1 is the initial piston position x 2 is the final piston position is the initial volume dd negative sign! 13 is the initial volume Work is negative for work of expansion!

3 Sign consideration Similarly, for work of expansion, (gas) does work. The system loses energy. Work should be a negative quantity. x 1 For work of compression, the system gains energy. Work should be a positive quantity. External pressure, P ext V > 0 x 2 Since > V 2 V = V 2 < 0 x =x 2 -x 1 V2 We modify the equation. V 2 x 1 is the initial piston position x 2 is the final piston position is the initial volume dd negative sign! 13 is the initial volume Work is negative for work of expansion! System loses energy 14 Summary of work and heat w = f x q < 0 system releases heat q > 0 system absorbs heat w < 0 work done by on the environment w > 0 work done on by the environment Mechanical work PV work done by or on a gas confined in a piston and cylinder configuration. WORK w < 0; Work is done by (expansion) E < 0 E < 0 System E > 0 E > 0 w > 0; Work is done on (compression) Surroundings HET q < 0; Heat flows out of. System loses heat q > 0; Heat flows into. System gains heat rea under a PV curve! 1. n isobaric process is a constant pressure process (dp=0). The PV diagram is a useful visualization of a process. The area under the curve of a process is the amount of work done by or on during that process. s a gas is being heated slowly, the volume of the cylinder expands to maintain constant pressure. s a gas is being cooled slowly, the volume of the cylinder decreases to maintain constant pressure. Compression is work done on Expansion is work done by w < 0 w > 0 V 2 V 2 is the initial volume;

4 1. n isobaric process is a constant pressure process. 2. n isochoric process is a constant volume process. (dv=0) The gas is being heated in a rigid container, such as a chemical reaction being carried out in a bomb calorimeter. Since dv=0, the reaction does no work. This is the PV work equation for an isobaric process. This is the PV work equation for an isochoric process n isothermal process is a constant temperature process. (dt=0) nrt is a constant for an isothermal process where n = # of moles of gas R = gas constant T = temperature of the gas When the temperature stays constant, the pressure and volume are inversely proportional to one another. 3. n isothermal process is a constant temperature process. The piston is slowly moved so that the gas expands. s the gas expands, a temperature drop results. To maintain isothermal condition, heat will flow into. In fact, the amount of heat, q is equal to the magnitude of w. q w < 0 q n adiabatic process is when no heat is added or removed from. (q=0) Compare the area under the PV curve for the work of expansion done by the gas by the following processes: 1. n isobaric process. dp=0 2. n isothermal process. dt=0 3. n adiabatic. q=0 When gas expands very quickly that no heat can be transferred. Eg CO 2 fire extinguisher w w w V 2 V 2 V 2 For each of the above processes, the gas starts with a volume of and end with a volume of V 2. Work of expansion is process (or path) dependent. Work is not a state function

What is the work done as a result of the free expansion of an ideal gas? Free expansion is process where a gas expands into an insulated evacuated chamber.

H 2 O (s) H 2 O (l) During free expansion, no work is done by the gas. P ext = 0 R = 0.08206 L atm mole -1 K -1 R = 8.314 J mole -1 K -1 0.08206 L atm mole -1 K -1 = 8.

5 What is the work done as a result of the free expansion of an ideal gas? Free expansion is process where a gas expands into an insulated evacuated chamber. Calculate the work of expansion that accompanies the fusion of 1 mole of ice to form 1 mole of liquid water at 1 atm and 0 o C. decrease in volume of 1.49 ml is observed. H 2 O (s) H 2 O (l) During free expansion, no work is done by the gas. P ext = 0 R = L atm mole -1 K -1 R = J mole -1 K L atm mole -1 K -1 = J mole -1 K -1 1 L atm = J w = - (1 atm)(-1.49 x 10-3 L) w = 1.49 x 10-3 L atm Energy unit. w = J w > 0 because dv < Calculate the work of expansion that accompanies the vapourization of water. The molar volume of liquid water at 100. C is 18.8 cc, while the molar volume of water vapour at 100 C and 1 atm is 30.2 L. H 2 O (l) H 2 O (g) Calculate the work done due to the thermal expansion of 1.00 mole of ideal gas heated at a constant pressure of 1.00 atm from 0.00 C to 100. C. = 1.00 atm Expect w < 0 w = - (1 atm)(30.2l L) w = L atm Energy unit. V 2 = L = L R = L atm mole -1 K -1 R = J mole -1 K L atm mole -1 K -1 = J mole -1 K -1 1 L atm = J w =? J w < 0; Work of expansion for the vapourization of water is much bigger than the previous example. 27 T 1 = 0.00 C T 2 = 100. C n = 1.00 mole w = -199 calories 28 Calculate the work done due to the thermal expansion of 1.00 mole of ideal gas heated at a constant pressure of 1.00 atm from 0. o C to 100. C. = 1.00 atm Expect w < 0 Calculate the work compression of 2.00 moles of an ideal gas from 1.00 bar to bar at 25.0 C. (1 bar = atm) Expect w > 0 P 1 = 1.00 bar P 2 = 100. bar T 1 = 0 C T 2 = 100. C n = 1 mole 29 T = 298 K n = 2.00 moles P 1 = atm P 2 = 98.7 atm w = 23.0 kj Use this equation with pressure, and you should get the same answer, w = 23.0 kj 30 5

6 Example: The combustion of octane yields the following reaction. Calculate the work done by under standard temperature and pressure condition (i.e. STP condition is 0 C and 1 atm) 2 C 8 H 18 (l) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O (g) n = = 9 moles of gas Under STP condition, the volume that is occupied by 1 mole of gas is 22.4 Lmole -1. V 2 - = 9 moles 22.4 Lmole -1 = L w = - 1 atm L = L atm = - (201.6 L atm)( joule L -1 atm -1 ) or = kj w = - 9 moles J mole -1 K K = kj Summary: Internal energy, E: ll the energy of (chemical, potential, kinetic, etc.) Thermal energy: The part of the internal energy that changes temperature Gases: Possess translational, vibrational, rotational motions, all contributing to thermal energy Ways to change internal energy of gas system: Heat, q Both heat and work State: The values of P, V, T, E of Change of State: By altering P, V, T, and E The system does 20.4 kj of work

CHEM What is Energy? Terminology: E = KE + PE. Thermodynamics. Thermodynamics

CHEM What is Energy? Terminology: E = KE + PE. Thermodynamics. Thermodynamics Thermodynamics 2 Thermodynamics The study of energy changes accompanying physical and chemical processes. From the laws of thermodynamics, one can: 1. Predict the results of chemical reactions 2. Ascertain