The Laws of Thermodynamics Chapter 18 The Laws of Thermodynamics

Thermodynamics The study of heat and its transformation into mechanical work.

18-1 The Zeroth Law of Thermodynamics We have already discussed the zeroth law, and include it here for completeness: If object A is in thermal equilibrium with object C, and object B is separately in thermal equilibrium with object C, then objects A and B will be in thermal equilibrium if they are placed in thermal contact.

Work done by a gas - Suppose you had a piston filled with a specific amount of gas. As you add heat, the temperature rises and thus the volume of the gas expands. The gas then applies a force on the piston wall pushing it a specific displacement. Thus it can be said that a gas can do WORK.

Work done BY gas

Work done ON gas

Sample Problem: An engine cylinder compresses a volume of gas by 0 Sample Problem: An engine cylinder compresses a volume of gas by 0.04 m3. How much work is done by the cylinder if it exerts a constant pressure of 7.5 x 105 Pa?

Sample problem: Calculate the work done by a gas that expands from 0 Sample problem: Calculate the work done by a gas that expands from 0.020 m3 to 0.80 m3 at constant atmospheric pressure. How much work is done by the environment when the gas expands this much?

Work is the AREA of a P vs. V graph The “negative” sign in the equation for WORK is often misunderstood. Since work done BY a gas has a positive volume change we must understand that the gas itself is USING UP ENERGY or in other words, it is losing energy, thus the negative sign. When work is done ON a gas the change in volume is negative. This cancels out the negative sign in the equation. This makes sense as some EXTERNAL agent is ADDING energy to the gas.

The “negative” sign in the equation for WORK is often misunderstood The “negative” sign in the equation for WORK is often misunderstood. Since work done BY a gas has a positive volume change we must understand that the gas itself is USING UP ENERGY or in other words, it is losing energy, thus the negative sign.   When work is done ON a gas the change in volume is negative. This cancels out the negative sign in the equation. This makes sense as some EXTERNAL agent is ADDING energy to the gas.

Example Sketch a PV diagram and find the work done by the gas during the following stages. A gas is expanded from a volume of 1.0 L to 3.0 L at a constant pressure of 3.0 atm. The gas is then cooled at a constant volume until the pressure falls to 2.0 atm 600 J

Example continued The gas is then compressed at a constant pressure of 2.0 atm from a volume of 3.0 L to 1.0 L. The gas is then heated until its pressure increases from 2.0 atm to 3.0 atm at a constant volume. -400 J

Example continued What is the NET WORK? 600 J + -400 J = 200 J NET work is the area inside the shape. 600 J + -400 J = 200 J Rule of thumb: If the system rotates CW, the NET work is positive. If the system rotates CCW, the NET work is negative.

Sketch a PV diagram and find the work done by the gas during the following stages.

(a) A gas is expanded from a volume of 1. 0 L to 3 (a) A gas is expanded from a volume of 1.0 L to 3.0 L at a constant pressure of 3.0 atm.   (b) The gas is then cooled at a constant volume until the pressure falls to 2.0 atm. (c) What is the net work?

Rule of thumb: If the system rotates CW, the NET work is positive. If the system rotates CCW, the NET work is negative.

Sample problem: Consider the cycle ABCDA, where State A: 200 kPa, 1.0 m3 State B: 200 kPa, 1.5 m3 State C: 100 kPa, 1.5 m3 State D: 100 kPa, 1.0 m3 Sketch the cycle.

B) Graphically estimate the work done by the gas in one cycle. C) Is the work from ABC (more, less, or the same) as the work from ADC. Work is path dependent.

Work done by a cycle

Internal Energy (DU) All of the energy inside a system is called INTERNAL ENERGY, DU. When you add HEAT(Q), you are adding energy and the internal energy INCREASES. Both are measured in joules. But when you add heat, there is usually an increase in temperature associated with the change.

The First Law of Thermodynamics The first law of thermodynamics is a statement of the conservation of energy. If a system’s volume is constant, and heat is added, its internal energy increases.

The First Law of Thermodynamics If a system does work on the external world, and no heat is added, its internal energy decreases.

First Law of Thermodynamics “The internal energy of a system tend to increase when HEAT is added and work is done ON the system.” Suggests a CHANGE or subtraction You are really adding a negative here! The bottom line is that if you ADD heat then transfer work TO the gas, the internal energy must obviously go up as you have MORE than what you started with.

The First Law of Thermodynamics Combining these gives the first law of thermodynamics. The change in a system’s internal energy is related to the heat Q and the work W as follows: It is vital to keep track of the signs of Q and W.

DU: change in internal energy of system(J) DU = Q + W DU: change in internal energy of system(J) Q: heat added to the system (J). This heat exchange is driven by temperature difference. W: work done on the system (J). Work will be related to the change in the system’s volume. This law is sometimes paraphrased as “you can’t win”.

Sample Problem: When 5000 J of heat is added to an engine, the engine does 1250 J of work. What is the change in the internal energy of engine?

Sample Problem: A system absorbs 200 J of energy from the environment and does 100 J of work on the environment. What is its change in internal energy?

Sample Problem: How much work does the environment do on a system if its internal energy changes from 40,000 J to 45,000 J without the addition of heat?

The thermodynamic state of a gas is defined by pressure, volume, and temperature. A gas process describes how gas gets from one state to another state. Processes depend on the behavior of the boundary and the environment more than they depend on the behavior of the gas.

Isothermal Processes This is an idealized reversible process. The gas is compressed; the temperature is constant, so heat leaves the gas. As the gas expands, it draws heat from the reservoir, returning the gas and the reservoir to their initial states. The piston is assumed frictionless.

To keep the temperature constant both the pressure and volume change to compensate. (Volume goes up, pressure goes down) “BOYLES’ LAW”

Isobaric Process Heat is added to the gas which increases the Internal Energy (U) Work is done by the gas as it changes in volume. The path of an isobaric process is a horizontal line called an isobar. ∆U = Q - W can be used since the WORK is POSITIVE in this case

Isovolumetric or Isometric Process

Thermodynamic Processes - Adiabatic ADIABATIC- (GREEK- adiabatos- "impassable") In other words, NO HEAT can leave or enter the system.

An adiabatic process is one in which no heat flows into or out of the system. The adiabatic P-V curve is similar to the isothermal one, but is steeper. One way to ensure that a process is adiabatic is to insulate the system.

18-3 Thermal Processes Another way to ensure that a process is effectively adiabatic is to have the volume change occur very quickly. In this case, heat has no time to flow in or out of the system.

Isobaric Work Question: How does the pressure of an isobarically expanding gas affect the amount of work it does? Work done by a cycle: What is DU for the gas in a complete cycle? What is changed when a gas undergoes a complete cycle?

18-3 Thermal Processes Here is a summary of the different types of thermal processes:

In Summary

Second Law of Thermodynamics “No process is possible whose sole result is the complete conversion of heat from a hot reservoir into mechanical work. (Kelvin-Planck statement.) No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. (Clausius statement.) The bottom line: Heat always flows from a hot body to a cold body Nothing is 100% efficient

Heat engines can convert heat into useful work. According to the 2nd Law of Thermodynamics. Heat engines always produce some waste heat. Efficiency can be used to tell how much heat is needed to produce a given amount of work. NOTE: A heat engine is not something that produces heat. A heat engine transfers heat from hot to cold, and does mechanical work in the process.

Heat Transfer

Heat Engine

Engines Heat flows from a HOT reservoir to a COLD reservoir QH = remove from, absorbs = hot QC= exhausts to, expels = cold

Engine Efficiency In order to determine the thermal efficiency of an engine you have to look at how much ENERGY you get OUT based on how much you energy you take IN. In other words:

In order to reverse heat flow work must be done. Spontaneous flow of heat from a cold area to a hot area would constitute a perfect refrigerator, forbidden by the second law.

Sample Problem: A piston absorbs 3600 J of heat and dumps 1500 J of heat during a complete cycle. How much work does it do during the cycle? What is the efficiency?

Carnot Cycle

Carnot Efficiency Carnot a believed that there was an absolute zero of temperature, from which he figured out that on being cooled to absolute zero, the fluid would give up all its heat energy.  Therefore, if it falls only half way to absolute zero from its beginning temperature, it will give up half its heat, and an engine taking in heat at T and shedding it at ½T will be utilizing half the possible heat, and be 50% efficient.  Picture a water wheel that takes in water at the top of a waterfall, but lets it out halfway down.  So, the efficiency of an ideal engine operating between two temperatures will be equal to the fraction of the temperature drop towards absolute zero that the heat undergoes.

Efficiency of a Carnot Cycle

Sample Problem: Calculate the Carnot efficiency of a heat engine operating between the temperatures of 60 and 1500 oC.

Entropy