 # Chapter Thermodynamics

## Presentation on theme: "Chapter Thermodynamics"— Presentation transcript:

Chapter 11 Thermodynamics

11-1 Heat and Work and Internal Energy
Heat = Work and therefore can be converted back and forth Work ----- heat if work is done on something, it can generate heat. Increasing the KE of the molecules causes movement that can be turned into work.

Thermodynamic Systems
Thermodynamics: Study of the effects of work, heat and energy on a system System – region under study which is separated by a boundary (real or imagined) Work, Heat, and Matter can go either way across the boundary. Example: flask, balloon, water, and steam that was heated. Surroundings - environment outside the system. Example: the classroom Steam does Work on the Balloon The steam’s internal energy has decreased

Recall from Chapter 5 that W = F x d units: (N x m)
Remember that P = F/A and Volume = area x displacement So, W = P x ∆V Work = pressure x change in volume Units: (N/m2 x m3) (N x m) or Joules (J)

Work done on (+) or by (-) a Gas
Question: The cross-sectional area of the piston is 0.20 m2. A N weight pushes the piston down 0.15 m and compresses the gas in the cylinder. How much pressure is exerted on the gas? Formula: P= F/A Answer: 2.0 x 103 Pa Question: How much did the gas volume decrease? Formula: W = P ●∆V Answer: m3

3 Types of Systems “Isolated” “Closed” “Open”

Internal Energy (U) The Energy contained in the system is ∆U = Q + W Internal Energy = heat + work) I am defining work by: Work done on the system is positive + (gains energy) Work done by the system is negative - (loses energy) Question: What type of “system” is the balloon and flask? Answer: Closed System 1. Heat is added to the system 2. Work is done by the system (-) W = P X ∆V As pressure inside the balloon increases – volume increases. When heat is removed, the SYSTEM returns to its original state of thermodynamic equilibrium

Example: A balloon is placed in a vacuum chamber and the air is removed. What kind of system is this? Answer: (isolated system) Work is done BY the system (expansion) As pressure on the outside of the the balloon decreases, it is allowed to expand (adiabatically). The volume of the balloon increases. According to the combined gas law: Then, T1 > T2 When pressure is restored then Work is Done ON the System (compression) T2 > T1 This increases the internal temperature in the balloon and it returns to its original state of thermodynamic equilibrium

Example 1: A system undergoes a process in which it
absorbs 800 J of heat while doing 500 J of work. the system is doing the work so W = neg What is the system’s change in internal energy? ∆U = Q + W Q = 800 J (endothermic +) and W = -500 J ∆U = 300 J Example 2: A cylinder with a movable piston at one end is filled with gas. 450 J of heat are removed from the system, while 175 J of work is done on the system. What is the change in internal energy of the system? Q = -450 (exothermic) and W = 175 J ∆U= -275 J

Isovolumetric – constant volume – no work done.
Example car tires, heart valves, calorimeters No work is done on or by the system W = 0 ∆U = Q

Isothermal Process – constant temperature
Internal energy remains unchanged ∆U = 0 ∆ T = 0

Adiabatic - work is done on or by the system but no energy is transferred to or from the system as heat. Q = 0 No heat gained or lost by the system

The first law of thermodynamics = conservation of energy.
U is the internal energy of the gas in question. Q is the heat added to the gas, and W is the work done on the gas when the gas is compressed. The signs of each term in the first law are extremely important. If heat is added to the gas, Q is positive, and if heat is taken from the gas, Q is negative. If the gas's internal energy (and, thus, its temperature) goes up, then ΔU is positive; if the gas's temperature goes down, then ΔU is negative. The sign of W is more difficult to conceptualize. Positive W represents work done on a gas. However, if a gas's volume expands, then work was not done on the gas, work was done by the gas. In the case of work done by a gas, W is given a negative sign.

Notes 11.1 Thermodynamic Processes
Internal Energy (U) = Heat (Q) + Work (W) 5 Main Thermodynamic Processes 1. Isobaric process = constant pressure 2. isometric process = constant volume (also called isovolumetric no work done Isothermal = constant temperature internal energy constant 4. Isentropic process = constant entropy 5. Adiabatic process = no change in energy (no ∆Q) energy is not transferred to or from the system as heat

An Isobaric Process – constant pressure
Draw graph

Isometric (isovolumetric) – constant volume
No work can be done, so ∆U = Q In this case, a system is heated or cooled and the gas would expand or contract if it could but the rigid container keeps the volume constant. Draw Graph

Isothermal process – constant temperature
Draw Diagram

Adiabatic Q = 0 No heat gained or lost by the system

11-2 Thermodynamic Processes
1st Law of Thermodynamics – energy conservation a systems internal energy can be changed by transferring energy as either work, heat, or a combination of the two. ∆U = ∆Q + ∆W

A Cyclic Process A thermodynamic process in which a system returns to the same conditions under which it started. Final internal energy = initial internal energy The change in internal energy of a system is ZERO in a cyclic process ∆ U = 0 Machines use heat to do work.

A refrigerator is an example of a cyclic process
A refrigerator performs work to create a temperature difference between its closed interior and its environment. Refrigeration systems are heat engines, except running in reverse. The basic principle used for refrigeration is: Expansion of a gas, reduces its temperature. Transfers energy from a body at low temperature to one at a high temperature. Draw Energy Diagram Cyclic Process - Refrigerator

Heat Engines use heat to do work
A heat engine takes heat from a hot body, converts a part of it into work and rejects the rest to a cold body. Uses Heat to do Mechanical work – opposite of a refrigerator. There are two types of heat engines: external combustion (Steam Engine) engines and internal combustion engines (gasoline engines and diesel engines) Draw Energy Diagram

Steam Engines – external combustion
Larger the difference in temperature the more work done Example of a heat pump

2nd law of Thermodynamics - Heat flows naturally from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot object

11-3 Efficiency of Heat Engines 2nd Law of Thermodynamics
Greater amount of work is gained by taking in more energy at a higher temperature and giving up less energy at a lower temperature. 2nd Law of Thermodynamics – no cyclic process that converts heat entirely into work is possible W ≠ Qh in a cyclic process Some energy must always be transferred as heat to the system’s surroundings. A measure of how well an engine operates is given by the engines’ efficiency (eff) - Useful energy Efficiency = net work done by engine/ energy added to engine as heat Found by 2 methods Eff = Wout /Qin Eff = Qin – Qout Qin Can be expressed as a decimal or a % Qin = hot Qout = cold

Nuclear power plants use controlled fission to produce energy (heat) then convert it to electricity.
Generator converts mechanical energy into electrical energy

11-4 Entropy SECOND LAW ( entropy )
The total entropy of any isolated thermodynamic system tends to increase over time. Increasing disorder reduces the energy available to do work

∆ S = Q / T Entropy ∆ S = the measure of disorder within a System. The Universe will become more disordered no matter what you do, and ENTROPY will increase.

Phase changes – major changes in entropy – isothermal
To Change the entropy in a system - Add or remove heat energy (Q). So if, ∆S = - , then Q is negative – (heat energy is lost) ∆S = +, then Q is positive (heat energy is added) Phase changes – major changes in entropy – isothermal Melting = Q +, S = + Freezing = Q - , S = - Evaporating = (adding heat energy to the system) = Q+ , S + Condensing (removing heat from the system ) = Q - , S - The system’s entropy increases. The environment (room, beaker, test tube etc. has less entropy

The 3rd LAW of THERMODYNAMICS
No Thermodynamic process can ever be 100% efficient. This would require that ALL the HEAT becomes WORK (no Qout), and that the exhaust would have to be at ABSOLUTE ZERO – impossible!