Dr. Cooper's Comments: Be sure to label ALL figures, pictures etc. Not only call them Figure 1, 2 etc... but give a short description of what the figure shows.

**(Edited Version) By Joan Rozario: In BLack**

**The second law: the concepts**

The second law of thermodynamics focuses on the spontaneous and non spontaneous processes. Some things occur naturally like, gas expanding to fill available volume; a hot object cools down to the temperature of its surroundings etc. But we don’t observe a can of coke cooling all by itself form room temperature, until we put it in the refrigerator. All these natural observations take us to the two different types of change: spontaneous and non-spontaneous.

A **spontaneous change** is a change that has a tendency to occur without work being done on the object or system. A spontaneous change naturally occurs. A **non-spontaneous change** however is a change that can be brought about only by doing work on the system. A non-spontaneous change has no natural tendency to occur.

**An overview of spontaneous and non-spontaneous change:**

For example, we know that H_{2}O molecules are always in constant motion in random direction, but what are the odds that H_{2}O molecules suspended in a bottle will all move in the same direction at the same time and the molecular motion will make the bottle jump.

The answer is never. Then the question arises why not? If the H_{2}O molecules are in random motion sooner or later they should all go in the same direction? It’s because of probability. That one event to occur may take longer than the life of the universe.

Probability is the essence what entropy is all about. Water bottles don’t just jump up by themselves as probability says that the chance of that happening is so incredibly small that in likelihood it’s not going to occur. Theoretically, this is possible. A spontaneous change may occur once in thousands, or millions or billions of years which we won’t experience in the span of our life time.

A second example given in class is about the glass being shattered in pieces. What are the probabilities for the glass pieces to get back together and reform the glass? Our answer would be not at all. But what physical conditions restrict it to reform? If we saw the movie of the glass breaking and then we saw the movie backwards when the glass reforms we see the direction of time. In one of them, time has to go backwards, because the glass reforming is impossible. This takes us to an important concept: what determines the direction of time? Why don’t we experience everything backwards, or why don’t we experience glasses reforming? The reason is the second law of thermodynamics and entropy.

**Figure1: broken glass analogy**

**The concept of spontaneity**:

When we have gas suspended in a small space attached to a larger container with a permeable boundary,

why do the gas molecules spontaneously want to go through to the other side?

**Figure2: gas expansion to additional space and lower pressure**

** **

When we have Ice water in a glass, sitting in a room temperature at 20 degree Celsius,

we see the ice melt but why does not the water freeze instead?

**Figure 3: the ice melting vs. H20 freezing example**

There is some aspect of the universe that determines the direction of change. Ice melting, gas expanding are all examples of change. When the gas is expanding to the vacuum, or the ice is melting we are not doing work on them, which tells us that, the direction of change is not dependent on Work, because we can observe things changing without doing work on them. Thus, spontaneous direction of changes does not require work to be done to happen.

**Spontaneity vs. Reaction time:**

Spontaneity and thermodynamics don’t tell us how quickly stuff happens. Even though something is spontaneous in reality it may not happen, in reasonable amount of time. The classic example of this spontaneity vs. reaction time is diamonds. Diamonds are the most stable form of Carbon; they are tremendously strong in bond which gives them their rigidity. It has been around us for a very long period of time. The process of diamond to form takes a long amount of time as the crystal lattice of its backbone is extremely complex. People say that diamonds are forver. But does a diamond always remain in its structural conformation? The answe is not. Probability says, diamonds will turn into graphite, after a long period of time as its bonds weaken and it takes a more primary state of it's conformation 'graphite.( last sentence revised)

**Figure 4: energy barrier that keeps diamond to go back to graphite state**

This process will take much later than sooner and the reason behind it is the huge energy barrier for the reorientation of all the molecules. For diamonds tetrahedral strong structure to break and reform back to graphite sheets, takes a lot of energy. But there is no indication for how long this would take to actually happen. Which tells us, Spontaneity and Thermodynamics does not provide information on the rate of change.

** **

**Spontaneity:**

We have still not discussed about the causes of the spontaneous changes like ice melting to H_{2}O, and gas moving to lower pressure. What determines the process of spontaneous change? In the first law of thermodynamics we have learned that for an isolated system, energy is conserved.

The equation is:

∆U= q+w.

What if we have a spontaneous change in an isolated system, where energy cannot be going out in the surrounding? It can’t be energy because we know the energy in an isolated system is conserver. But it is possible to have a spontaneous change in a system without changing the energy of the system. It is the effect of dispersible energy. We hear about it when said, universe is going towards disorder. So, to say that everything tend towards a state of the lowest energy is not true.

So, spontaneous change is all about taking some localized amount of energy and spreading it out. And that really is the essence of the argument about the water bottle. If spreading the energy out is the criteria of the spontaneity the reverse must be the opposite. If we are taking the random distribution of the motion of the molecules, that energy is spread out all across the molecules. To bring them together and make them move all in the same direction we have to order the energy rather than have it dispersed.

**Therefore, direction of spontaneous change is the direction where energy gets more dispersed.**

**Dispersal of energy:**

Here is an example of the same argument with the demonstration of a ball. Molecules in the ball would need all the random motion of the ball to go in the same direction in order for it to

**Figure 5: the example of arrangement of atoms in oder for a ball to jump by itself**

If we go back to chapter 2 and think about the moveable piston we read about that enables us to compress and expand gas, it helps us understand how it extracts work out of the system. The amount of work to get to drive the wheel to run a machine is governed by the rules of compression and expansion. In the time of the industrial revolution merchants were interested in the type of machine that would be the most efficient. The rules that we are concentrating on were basically built to solve these problems.

**Figure 6: the expansion and compression of gas in the engine Figure 7: steam engine (industrial revolution)**

**Heat engine:**

Heat engine takes energy from some source (like for steam engine it is water vapor), and through some process turns the energy into work.

**Figure 8: heat engine that shows 100% efficiency**

In this diagram it shows us that all the energy input is coming out as work. But in reality that is not the case as no engine functions 100% efficiently. The diagram is missing the amount of energy that is being lost. It is physically impossible to get all the energy out and get it transferred all to work. And the reason for this is entropy.

So, one way of saying the first statement of the second law of thermodynamics is: it’s impossible to convert 100% energy to 100% work. The second statement is, for a spontaneous change for an isolated system, the total entropy has to increase. So, when we calculate entropy if it a positive (+) value than change is spontaneous and if a negative(-) value than the change is is non- spontaneous.

**End of Joan's Section.**

--beginning of Hamna's section--

__I. __**Entropy**

Definition:

As mentioned in the Second Law of Thermodynamics, not all of the heat put into a system will be converted to work . What causes this to occur?

The answer is *entropy**,* which is the measure of the amount of heat that dispersed within a sytem and therefore not converted into work. The Thermodynamic definition of entropy is:

(eq 1)

Where dS is the change in entropy, T is the temperature in Kelvins and dq_{rev} is the reverse path of the heat put into the system. As demonstrated by the broken glass demonstration (Figure 1), once changes occur in a system the original state of the system cannot be restored unless additional work or heat is put into it. As the Second Law states, not all of the heat put into a system is converted into work, however if the movement of the heat is observed from the reversed direction, all of the heat put into the system can be accounted for in an infinitesimally small amount.

Equation 1 can be integrated to find measurable changes made in a system as a result of entropy:

(eq 2)

Example Problem:

**Calculate the entropy change of 1 mol of an ideal gas when it expands isothermally from 1.0 L to 2.0 L.**

Equation 2 can be applied to solve this problem.

Since 1/T is a coefficient, it can be moved in front of the integral (this is a property of integrals)

The integral of dq_{rev} is simply q_{rev} _{} , and the remaining equation is:

In order to Solve for q_{rev}, knowledge from previous chapters must be applied

For Example, from chapter two it is known that

and that

The change in internal energy ( ∆U) is zero because this system is expanding *isothermally* (at a constant temperature) therefore

w = -q and q can also be defined as:

Entropy can finally be solved for by plugging in the values to the equation above:

∆S = (1.00mol)(8.314J/Kmol)(ln2) = **+5.76J/K**

(side note: the units for entropy is J/K)

__II. Entropy In Terms of the Surroundings__

When heat from the surroundings is transferred into the system, a change in entropy will result. The equation used to solve for entropy in respect to the surroundings is the same, however the subscript “sur” is used to acknowledge the affect the surroundings can have on the system:

dS_{sur} = (dq_{sur, rev})/ T = dq_{sur} / T (eq. 3)

In chapter 2, it was established that internal energy change q is a state function. Since it is a state function the path taken to undergo a change is not important, only the current condition or "state" of the change is considered. For this reason, it does not matter what direction the heat of the system proceeds (forward or reverse) therefore equation 3 can be simplified to dq_{sur}/T. Equation 3 can be simplified further because chapter 2 also established that the temperature of the surrounding is constant despite heat exchange with the system. For this reason

∆S_{sur} = q_{sur}/T (eq. 4)

because there are no changes in the enthalpy of the surroundings.

In an Adiabatic system, the change in q_{sur} is 0 by definition because adiabatic systems are isolated from the surrounding. For this reason:

∆S_{sur} = 0 (eq. 5 only in adiabatic change)

__III. Molecular Aspect of a System__

Each phase of a compound contains differing amounts of energy. The instructor gives an example of a glass of ice water. The energy in the molecules of the ice is at a lower energy state than the molecules of the water. This is demonstrated in figure 3.4 of the text

*(fig 9) The Boltzmann Distribution*

Graph (a) in Figure 9 represents the energy distribution of the molecules in ice whereas graph (b) shows the distribution for water. Keep in mind that it is not possible to account for the individual states of energy of each molecule at one instant. For this reason the energy of the population, or the molecules of the entire system, is averaged. so

Based on this figure, it could also be said that increasing the energy in a system in which the population already occupies higher energy levels will not lead to a drastic total change in comparison to increasing the energy through of a system with a population at a lower energy level s. After all, systems with lower energy have more access to higher energy levels than systems that already occupy those levels.

It was previously established that the surroundings of a system are a reservoir of energy (Ch. 2 of Atkins) and for this reason the surroundings remain at a constant state despite changes made in the system. This is demonstrated in the example of an ice cube melting in an environment at room temperature.

**Ice Melting:**

∆S_{syst} = q/273K

This equation represents changes made in the ice cube.

This value will be much greater than changes made in the surroundings:

∆S_{surr} = -q/298K

Side note: the sign of the entropy in this equation is negative to indicate

that heat is flowing from the warmer surroundings into the colder system.

Now to calculate the total change in entropy, we find the difference between the change in energy of the system and the change in energy of the surroundings:

∆S_{total} = ∆S_{sys} – (-∆S_{surr} )

∆S_{total} = **∆S**_{sys} + **∆S**_{surr}

∆S_{total} > 0

Since the change in entropy of the system is much greater than the change in entropy of the surroundings (which remains relatively constant), the total change in entropy is still positive.

__IV. Heat Engines Redefined__

Recall the structure of an engine from Figure 8. In the type of engine depicted in figure 8, the heat that is put into a system is not completely converted to work. Since entropy is a state function that is independent of the path taken to bring the system to its current condition, this engine will have to be redesigned in such a way that the change between the initial and final states should be zero.

**Fig 10: The Thermodynamic Cycle in terms of state functions**

(eq. 6)

The contour integral symbol of Equation 6 (the integral symbol with the circle in the center) represents the fact that the final and initial states of the function are the same. Keep in mind, that this concept is just a description of entropy on paper, it is not actually possible in real life.

The cycle described in both figure 10 and in equation 6 is called the *Carnot cyle* which will be elaborated on in the

next section

--End of Hamna's Section--

-------------------------------------

Carlos Rodriguez

**THE HEAT ENGINE**

**Figure 11: The Heat Eng****ine. A heat engine is a device that can convert heat energy to mechanical work. There are numerous thermodynamic cycles, and in the case of a heat engine the conversion of heat to work manifests from the exploitation of a temperature difference between and a hot source and a "cold sink." **

In a heat engine the heat flows from the source (in this case a "hot" source), through some sort of working body, to the sink (in this case a "cold" sink); in such a process some of the heat is converted into work by exploiting the properties of a working substance such as a gas. Note that a heat engine can be used to demonstrate that entropy is a state function; that the integral of dS around a closed path is equal to zero serves as proof of this (eq. 7 below). An integral of this nature assures that the initial and final states of the given thermodynamic cycle are equal (hence figure 10 above).

eq. 7 (p. 82 Atkins)

**THE CARNOT CYCLE**

The Carnot cycle was derived from Nicolas L. Sadi Carnot's work regarding the efficiency of heat engines. While it has been shown that an engine in which there is a 100% transfer of heat energy to work energy is not possible, Carnot believed that the efficiency of a heat engine was based on the difference between the highest and lowest temperatures achieved during the heat cycle. Described below is a brief description of the four general steps that comprise the Carnot cycle and form the basis for the mathematics that can be used to describe the amount of work produced from this cycle.

**Figure 12: The Carnot Cycle. A graphical representation of the volume changes and accompanying pressure changes that occur during the "four-step" Carnot cycle.**

Step 1: The cycle begins with a piston moving in a given direction while an expanding gas absorbs heat from a hot source. The portion of the image from point A to point B represents this behavior. Although heat has been introduced into the system, since the gas is expanding there is no net increase in temperature and thus we call this an isothermal expansion.

Step 2: From point B to point C, what has occurred is a removal of the hot source, though the piston continues to move as the gas continues to expand (note that it experiences a lowering in temperature since the hot source has been removed). Though the expansion persists, since no energy has been added or removed we call this an adiabatic expansion.

Step 3: In this stage the expansion has ended and compression is initiated as a cold sink is attached to the system. Point C to point D represents this decrease in volume and increase in pressure. In this stage heat is released to the cold source and thus energy is released into the environment. Though the system is compressing the gas molecules, since the heat is transferring to the cold source the temperature remains the same and thus this is termed an isothermal compression.

Stage 4: Lastly, the piston continues to move towards compressing the now cool gas; point C to point D demonstrates this shrinking volume (though less volume loss then in step 3). The temperature of the gas is back to its temperature (pre contact with hot source) until it is once again exposed to the hot source. Once exposed to the hot source this allows the gas to act as a sort of heat transfer agent to start the cycle anew as the heat causes the piston to once again expand and do work.

This is a great link that helped me conceptualize this process; it may help to open it in another window and view it while reading over the Carnot cycle steps.

http://www.cs.sbcc.net/%7Ephysics/flash/heatengines/Carnot%20cycle.html

**CARNOT CYCLE MATHEMATICS**

In Step 1:

(eq. 8):

In step 2:

(eq. 9):

In step 3:

(eq. 10):

In step 4:

(eq. 11):

Thus,

(eq.12),

-Remember that this closed path integral that was shown to be equal to be 0. Using this formula it can be shown (using justification 3.1 on p. 83 Atkins) that:

(eq. 13)

This equation is very important: it relates the hot/cold fractions for both the heat transfer and temperature in a carnot cycle involving a perfect gas. We will see shortly how important this is in deriving work efficiency values from temperature values.

Now what we'll do is take the formula for the efficiency of a heat engine, where epsilon = efficiency,

eq. 14

and if we remember that this is a system with no change in internal energy (because as we've shown it is a state function), then we will realize that the absolute value of work (W) in this system is equal to heat (Q) through the formula we already know, U = W + Q, where U = 0. Thus we get:

eq. 15

Now we will use the all important equation 13 by plugging in temperature values of the (hot source) and (cold sink) to replace the transferred heat values from equation 15. The final equation will thus give us the efficiency of the heat engine from only the temperature values. We can also use the equation to obtain the value of W or derive W directly by adding the qc and qh values that are easily derived from the temperatures used in this heat transfer cycle.

eq. 16

**Figure 13: The Heat Engine (see definition above).**

Thus the work done by the heat transfer using an ideal gas can be realized simply by using the hot and cold temperatures in the cycle.

**REFRIGERATION**

**Figure 14: Refrigeration. Note that work is being put into the system so as to remove heat energy from a cold source (Qc) to a hot sink.**

Refrigeration is not the mere opposite of the Carnot cycle, though similarities with a "reverse carnot cycle" can be drawn. In this cycle, work is used to move heat from a cold source to a hot sink. This process is not spontaneous but the work required to transfer the heat (though not to maintain the low temperature, see p. 85 for more details) can be found by simply plugging in temperatures to an equation.

In refrigeration, the efficiency, defined as "c," is higher when less work is required to pump heat out of the cold source. Note that c can be found using just temperatures. In the figure below it is also important to realize that the absolute value of w is equal to the absolute value of Qc subtracted from Qh. This results from the fact that in this device heat is flowing from a cold source, work is incorporated into the system and thus is "added to the energy stream." Thus the energy deposited as heat in the hot sink is abs(Qh) = abs(Qc) + abs(W), where abs(X) is equal to the absolute value of X.

## Comments (2)

## hriaz3@gmu.edu said

at 11:15 pm on Sep 25, 2009

hey guys,

this is a good site for LaTeX codes for more complicated equations.

http://www.math.uiuc.edu/~hildebr/tex/displays.html

it helped me out, hope it helps you too

## hriaz3@gmu.edu said

at 11:33 pm on Sep 26, 2009

why cant i see my section anymore???

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