Understanding the Carnot Cycle better : Exercises

Introduction

In this exercise we are going to consider how a refrigerator works. Read the text that follows that explains how a refrigerator works and then answer the questions that follow.

In a refrigerator heat is transferred from the cabinet to a heat dump by means of a coolant liquid called freon. We thus pass heat from a cooler object (the inside of the fridge) to a hotter object (the outside world), which we know means that we must do some work on this system. Remember no process is possible in which the sole result is the transfer of heat from a colder to a hotter body. In a fridge we thus have a motor that does work on the freon coolant. For the purposes of this question we will suppose that there are four stages in the operation of our refrigerator:

The freon coolant is released into a pipe that runs around the back of the fridge cabinet. During this stage the temperature of the freon increases in order to cool the cabinet. The volume of the freon is kept fixed during this stage.
The freon gas is compressed (it's volume is decreased) by a motor. During this stage the pressure of the freon gas is kept fixed.
The freon gas is released into a pipe that runs outside the fridge cabinet. The temperature of the freon decreases during this stage. Once again, however, the volume of the gas is kept fixed.
The freon gas is expanded (it's volume is increased) by a motor that ensures that the pressure of the gas is kept fixed throughout.

Once one cycle through these four stages are completed step one is repeated. For the purposes of the questions that follow you may assume that the freon gas reverts to its original thermodynamic state on completion of a cycle.

Example problems

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Problem 1

Draw a diagram illustrating the changes in the volume and pressure that take place during each of the four transitions that take place for the freon gas inside the engine.

The important things to note from the information in the question here are that:

The volume does not change in steps 1 and 2 but the temperature increases during these stages. Gibbs Phase Rule tells us that for a one component one phase system like this one there can only be two indepdenent thermodynamic variables. Consequently, because the temperature changes but the volume doesn't the pressure of the freon must change in steps 1 and 3. In particular, the pressure increases during step 1 (as the temperature goes up) and the pressure decreases during step 3 (as the temperature goes down).
In steps 2 and 4 the volume changes because we are told this above. In particular the volume decreases during step 2 and increases during step 4.

The above realisations provide sufficient information to draw the thermodynamic cycle.

In which of the four transitions is heat transferred from the freon gas into the surroundings? For the purposes of this question the heat output during these stages will be referred to as $q_1$ in the questions that follow. Will $q_1$ have a positive or negative sign?

Heat is transferred from the freon gas into the surroundings during steps 3 and 4. $q_1$ will have a negative sign as heat is being transferred away from freon gas and into the surroundings. The entropy of the freon gas is therefore lowered during these transitions.

In which of the four transitions transfer is heat transferred from the surroundings into the freon gas? For the purposes of this question the heat input during these stages will be referred to as $q_2$ in the questions that follow. Will $q_2$ have a positive or negative sign?

Heat is transferred from the surroudnigs into the freon gas during steps 1 and 2. $q_2$ will have a positive sign as heat is being transferred into the freon gas from the surroudnings. The entropy of the freon gas is therefore increased during these transitions.

Ideally we would like $q_2-q_1 \lt 0$. Why?

We want our fridge to transfer heat from its interior to the outside world. If $q_2-q_1>0$ this tells us that there is a net transfer of heat from the environment to the inside of the fridge, which would render this device for keeping food cool rather useless.

In which of the four transition is work done on the freon gas? For the purposes of this question the work done during these stages will be referred to as $w_1$ in the questions that follows. Will $w_1$ have a positive or negative sign?

Work is done on the freon gas during stage 2 when it is compressed. The volume of the gas will decrease during this stage and as such $\textrm{d}V$ will be negative. $-P\textrm{d}V$ will, therefore, be positive so $w_1$ will have a positive sign.

In which of the four transition is work done by the freon gas on its surroundings? For the purposes of this question the work done during these stages will be referred to as $w_2$ in the questions that follows. Will $w_2$ have a positive or negative sign?

Work is done by the freon gas during stage 4 when it is expanded. The volume of the gas will increase during this stage and as such $\textrm{d}V$ will be positive. $-P\textrm{d}V$ will, therefore, be negative so $w_2$ will have a negative sign.

Given that the freon reverts to its original state when a cycle completes what must $q_2 - q_1 + w_1 - w_2$ equal to? Use this result to show that $\frac{q_1 - q_2}{w_1} = \frac{w_1 - w_2}{w_1}$. Consider the Carnot cycle and explain why this quantity is important and why this expression is significant.

$q_2-q_1 + w_1 - w_2$ must be equal to zero. By the first law of thermodynamics this sum is equal to the total change in the internal energy. We know, however, that internal energy is a function of state. The energy of the system thus only depends on the state we are in. If we are in the same state at the end of the thermodynamic cycle the energy must be the same. There must, therefore, be no overall change in the internal energy on completing a cycle.

We have established that: $$ q_2-q_1 + w_1 - w_2 = 0 $$ This expression can be rearranged to give: $$ q_1 - q_2 = w_1 - w_2 $$ Dividing both sides of the above by $w_1$ gives the required result: $$ \frac{q_1 - q_2}{w_1} = \frac{w_1 - w_2}{w_1} $$ The result is significant as $w_1$ is the amount of work that is required to compress the gas. $q_1-q_2$ by contrast is a negative quantity that tells us how much heat will be transferred from the inside of the fridge to the environment. $\frac{q_1 - q_2}{w_1}$ is thus the ratio of the amount of heat transferred away to the amount of work done. It is thus a measure of the efficiency of the fridge.

We are now going to prove that the refrigerator described above is the most efficient refrigerator that can be constructed. We will do so by imagining that a second more efficient refrigerator does indeed exist. In this hypothetical fridge doing you have $q_1'-q_2'=q_1-q_2$, where $q_1$ and $q_2$ are as they were above and $q_1'$ and $q_2'$ are the heat input and output from the hypothetical fridge. The important thing about this new fridge is that it transfers the same amount of heat as our hypothetical fridge but the motor in the fridge does less work. In other words $w_1' \lt w_1$, where $w_1'$ is the amount of work done by our more efficient fridge.

Explain why $w_2' \lt w_2$ where $w_2'$ is the amount of work done by our new (more efficient refridgerator) and $w_2$ is as it is above.

Our more efficient energy must return to the same thermodynamic state on completion of a cycle, however, and it must also obey the first law of thermodynamics. As such: $$ q_2'-q_1' + w_1' - w_2' = 0 \qquad \rightarrow \qquad q_1' - q_2' = w_1' - w_2' $$ We also know from the question that: $$ q_1'-q_2'=q_1-q_2 $$ which implies that: $$ q_1-q_2 = w_1' - w_2' $$ From the previous question, however, we know that: $$ q_1 - q_2 = w_1 - w_2 $$ which in turn ensures that: $$ w_1' - w_2' = w_1 - w_2 $$ Therefore if $w_1' \lt w_1$ then $w_2' \lt w_2$ in order to ensure that the above equality holds.

Lets now suppose that the heat output from our hypothetical fridge is used to drive the fridge that was described at the start of this page in the reverse direction. Calculate the total amount of heat exchanged between the fridges during this process and the amount of work that is done during each of the constant pressure stages (steps 2 and 4) of the two fridges' operation. Use these considerations to prove that no fridge can be more efficient than the fridge we examined at the start of this question. Hint: Something similar was done for the Carnot engine in the video so use that as your guide.

HEAT TRANFER

When steps 1 and 2 are run in the forward direction for the more efficient fridge an ammount of heat, $-q_2'=-q_2$, will be transferred from the less efficient fridge to the more efficient fridge. For the purposes of this question we use a negative sign to indicate that heat is being transferred from the less efficient fridge to the more efficient fridge.
When steps 3 and 4 are run in the forward direction for the more efficient fridge an ammount of heat, $q_1'=q_1$, will be transferred from the more efficient fridge to the less efficient fridge. For the purposes of this question we use a positive sign to indicate that heat is being transferred to the less efficient fridge to the more efficient fridge.
When steps 1 and 2 are run in the backwards direction for the less efficient fridge an ammount of heat, $q_2$, will be transferred from the more efficient fridge to the less efficient fridge. For the purposes of this question we use a positive sign to indicate that heat is being transferred to the less efficient fridge
When steps 3 and 4 are run in the backwards direction for the less efficient fridge an ammount of heat, $-q_1$, will be transferred from the less efficient fridge to the more efficient fridge. For the purposes of this question we use a negative sign to indicate that heat is being transferred from the less efficient fridge to the more efficient fridge.

When we add together the total amount of heat transferred to/from the less efficient fridge during the cycle we thus find that: $$ -q_2 + q_1 + q_2 - q_1 = 0 $$ That is to say there is no net transfer of heat during the cycle.

WORK DONE

As discussed above work is only done during steps 2 and 4 in the fridges operation i.e. when the volume of the gas is changing. In particular:

An ammount of work $w_1'$ must be done during step 2 when the motor in the more efficient fridge compresses the gas as this fridge operates in the forward direction. This work is driven by the expansion of the gas in the less efficient fridge, which is running in reverse. This expansion also takes place during stage 2. For the less efficient fridge this process of expansion gives out an ammount of work $-w_1$.
An ammount of work $w_2$ must be done during step 4 when the motor in the less efficient compresses the gas as this fridge operates in the reverse direction. This work is driven by the expansion of the gas in the more efficient fridge, which is running in the fowards direction. The expansion in the more efficient fridge also takes place during stage 4. For the more efficient fridge this process of expansion gives out an ammount of work $-w_2'$.

If we think about the total amount of work done during these two steps we thus have: $$ \textrm{Stage 2:} \quad w_1' - w_1 \lt 0 \qquad \textrm{and} \qquad \textrm{Stage 4:} \quad w_2 - w_2' > 0 $$ The inequalities here hold because of the answers to the previous given in the previous question. Consider what the above result would mean, however. The pressure on the gas is higher during stage 2 than it is in step 4. The above inequalities are telling us that work is output from the combined pair of fridges during stage 2 as $w_1' - w_1 \lt 0$ and that work is dnoe on the combined pair of fridges during stage 4 as $w_2 - w_2' > 0$. In other words, the motor that is applying a lower pressure to the freon is making its volume contract, while the motor that is applying the lower pressure is making its volume expand. In other words, there is a net transfer of volume across a pressure gradient. This combination of fridges is therefore doing something that is at odds with our intuitive feelings about how the world operates.

The point with this question is that if you look at it carefully and compare the answers above with the answers we obtain for the Carnot cycle you see that all we have done is swap round the roles played by the heat $q$ and the work $w$ in the two derivations.

Problem 2

Consider now a fictitious heating device that stores and releases energy by dissolving a salt in a solution. The charging and heating stages for this device together constitute a thermodynamic cycle involving the following four (isochoric) stages.

The number of atoms in the solution is increased. During this stage the temperature of the solution is kept fixed at a value of $T_2$.

The solution is thermally lagged and the number of atoms in the solution is increased further.

The thermal lagging is removed and the number of atoms in the solution is decreased. During this stage the temperature of the solution is kept fixed at a value of $T_1$.

The solution is thermally lagged one further time and the number of atoms in the solution is decreased still further.

Use the fact that on completion of the four stages of the devices operation the solution returns to its original thermodynamic state and the fact that that $T_2>T_1$ to determine whether the reaction in which the salt dissolves in the solution is exothermic or endothermic. Justify your answer.

The problem here is once again a variant on the Carnot cycle. As such you should start by thinking about which of the stages in the devices operation output heat/work and which stages in the devices operation take heat/work in. Remember that work must be done on the system to increase the number of atoms in the solution as $\textrm{d}w = \mu \textrm{d}N$.

Understanding the Carnot Cycle better : Exercises

Introduction

Example problems

Problem 1

Problem 2

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