Thermodynamics

Thermodynamics Systems and Their Surroundings Thermodynamics is the branch of physics that is built upon the fundamental laws that heat and work obey. In thermodynamics the collection of objects on which attention is being focused is called the system, while everything else in the environment is called the surroundings. For example, the system in an automobile engine could be the burning gasoline, while the surroundings would then include the pistons, the exhaust system, the radiator, and the outside air. The system and its surroundings are separated by walls of some kind. Walls that permit heat to flow through them, such as those of engine block, are called diathermal walls. Perfectly insulating walls that do not permit heat to flow between the system and its surroundings are known as adiabatic walls.

The Laws of Thermodynamics

Thermodynamics literally means “moving or evolving heat”. The science of thermodynamics is concerned with heat and its transformation to mechanical energy.Internal energy refers to the sum of the kinetic energies of the molecules of a body and the potential energy due to intermolecular.

Reversible process is one in which the system and its surroundings can be returned to their initial state before the process occurs. The opposite, of course, is irreversible. Growing old, breaking a glass, and burning pieces of papers are irreversible processes. Fusion, vaporization, sublimation, expansion, and contraction of substances when heat is added or released are examples of reversible process.There are four basic laws in thermodynamics: zeroth law, first law, second law, and the third law.

Zeroth Law of Thermodynamics

The zeroth law was first formulated by R.H. Fowler in 1931 after the first law and second law had been in use for some time. Thermal equilibrium is the main concern of the zeroth law. Two bodies are said to be in thermal equilibrium if they have the same temperature. The zeroth law of thermodynamics states that if object A is in thermal equilibrium with object B and object A is in thermal equilibrium with a third object C, then object B must be in equilibrium with object C.

Let the temperatures of objects A, B, and C be TA, TB, and TC, respectively. If TA = TB and TA = TC, then TB = TC. In effect, the zeroth law of thermodynamics is very similar to the transitive law in elementary algebra.

The zeroth law of thermodynamics deals with the concept of thermal equilibrium. Two systems are said to be in thermal equilibrium if there is no net flow of heat between them when they are brought into thermal contact.

Example: Consider 2 beakers of water, in one beaker, the temperature of water is above room temperature, and the other is below room temperature. They are left on a table (they are not in contact with each other), after some time, equilibrium is reached. Both beakers of water are at the same temperature. The two beakers become in thermal equilibrium with the surroundings, thus they are in thermal equilibrium with each other, and they are at the same temperature.

First Law of Thermodynamics

The first law of thermodynamics relates heat, mechanical work, and internal energy of a system. It states that when heat is added to a system, some of it remains in the system increasing its internal energy, while the rest leaves the system as the system does work. The first law may be written in symbols as: H=∆U + W

Where H is the heat, ∆U is the internal energy, and W is the work. Heat, change in internal energy, and work expressed in joules. In using this equation, the following sign convention must be applied. H is + when added to the system and – when removed from the system. W is + if done by the system and – if done on the system.

The first law of thermodynamics may be considered as a statement of conservation of energy. Example of this is a car engine, part of the energy stored in gasoline transformed into useful work in moving the car. Some of the energy heats up the car, while the rest is given off as heat in exhaust gases.

Problem 1: Figure 1 illustrates a system and its surroundings. In part a, the system gains 1500 J of heat from its surroundings, and 2200 J of work is done by the system on the surroundings. In part b, the system also gains 1500 J of heat, but 2200 J of work is done on the system by the surroundings. In each case, determine the change in the internal energy of the system.

Reasoning: In figure 1a the system loss more energy in doing work than it gains in the form of heat, so the internal energy of the system decreases. Thus, we expect the change in the internal energy, ∆U= U f – Ui, to be negative. In part b of the drawing, the system gains energy in the form of heat and work. The internal energy of the system increases, and we expect ∆U to be positive.

Solution:a.) The heat is positive, Q = +1500 J, since it is gained by the system. The work is positive, W =

+2200 J, since it is done by the system. According to the first law of thermodynamics ∆U = Q – W = (1500J) – (+2200 J) = -700 J The minus sign for ∆U indicates that the internal energy has decreased, as expected.

b.) The heat is positive, Q = +1500 J, since it is gained by the system. But the work is negative, W = -2200 J, since it is done on the system. Thus,

∆U = Q – W = (+1500 J) – (-2200 J) = + 3700 J The plus sign for ∆U indicates that the internal energy has increased, as expected.

Problem 2: The internal energy decreases by 700 J when it absorbs 2 000 J of heat. How much work is done during the process? Is the work done by the system or on the system?

Solution: Given: ∆U= -700 J H= 2000 J Using the first law of thermodynamics, H= ∆U + W 2 000 J=-700 J + W W= 2700 J Since W is +, it is the system that does the work.

The Second Law of Thermodynamics

There are three versions of the second law of thermodynamics.a.) Kelvin-Planck Statement. No heat engine can completely convert heat energy to work. In other

words, there is no 100% efficient heat engine.b.) Clausius Statement. Heat flows naturally from hot to cold objects.c.) Entropy Statement. When a reversible process occurs, the total entropy of the universe remains

the same. When an irreversible process occurs, the total entropy of the universe increases.

But what is entropy? Before formally defining entropy, let us first discuss order and disorder. It is very easy to create disorder out of order. Let us suppose that a child has 15 red balls and 15 white balls. The child arranges these balls in an “orderly manner” in a box by placing the white balls together on one side of the box and all the red ones on the other side. Shaking the box will mix the balls, causing the “ordered” state of the balls to disordered.

The thermodynamic measure of disorder is entropy. Entropy represented by S. Some examples where entropy increases are (a) when heat is added to an object because the molecules tend to move faster, (b) when gas flows from a container under high pressure to a space under low pressure, just like spraying air freshener in a big room because the gas molecules have a bigger space to move around freely, and (c) when ice melts. In ice, the water molecules are free to move around throughout the liquid. In general, solids are more orderly than liquids and gases.

The change in entropy ∆S for a process occurring at constant temperature is defined as the heat (H) added or released during the process divided by the temperature (T) in Kelvin.

∆S=HT

When heat is added or removed from a solid or liquid of mass m and specific heat c and its temperature changes from T1 to T2, the change in entropy is

∆S=(mc) ln T2

T1

The SI unit of entropy is obviously J/K.

The entropy statement of the second law may be written as: ∆S of universe = 0 for a reversible process ∆S of universe > for an irreversible process

Problem 3: An Automobile Engine An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine?

Qc = QH – W = Wl - W = (2510 J) __1___ - 1

0.220 = 8900 J

The Third Law of Thermodynamics

The third law of thermodynamics simply states that it is impossible to attain the absolute zero temperature. The laws of thermodynamics may be stated as “you cannot win, you cannot break even, nor can you get out of the game.”

The third law of thermodynamics is sometimes stated as follows:

The entropy of a perfect crystal at absolute zero is exactly equal to zero.

At zero kelvin the system must be in a state with the minimum possible energy, and this statement

of the third law holds true if the perfect crystal has only one minimum energy state. Entropy is related to

the number of possible microstates, and with only one microstate available at zero kelvin, the entropy is

exactly zero.[1]

A more general form of the third law applies to systems such as glasses that may have more than

one minimum energy state:

The entropy of a system approaches a constant value as the temperature approaches zero.

Nernst-Simon statement follows:

The entropy change associated with any condensed system undergoing through a reversible isothermal process approaches zero as temperature approaches 0K. Where condensed system refers to liquids and solids.

Another simple formulation of the third law can be:

It is impossible for any process, no matter how idealized, to reduce the entropy of a system to its zero point value in a finite number of operations.

The constant value (not necessarily zero) is called the residual entropy of the system.

Problem 4:

The sublimation of zinc

The sublimation of zinc (mass per mole = 0.0654 kg/mol) occurs at a temperature of 6.00x102 K, and the

latent heat of sublimation is 1.99x106 J/Kg. The pressure remains constant during the sublimation.

Assume that the zinc vapour can be treated as a monatomic ideal gas and that the volume of solid zinc is

negligible compared to the corresponding vapour. What is the change in the internal energy of the zinc

when 1.50 Kg of zinc sublimates?

Q = ∆U + W Q = ∆U + W or mLs = ∆U + nRT

Q = mLs ∆U = mLs - nRT

W = nRT ∆U = mLs – nRT

(1.50kg) (1.99x106 J/Kg) – _1.50 Kg__ (8.31 J/mol•K) (6.00x102)

0.0654 Kg/mol

= 2.87x106 J