Why does there need to be an isothermal compression in a Carnot cycle?
$begingroup$
This is the part that allows the Carnot engine to not violate the 2nd law of thermodynamics, but, hypothetically, why can't we just adiabatically compress the working substance to get it back to State A?
As in, the working substance starts at $A$,
- Isothermal expansion - work is done on the environment
- Adiabatic expansion - work is done on the environment, temperature is decreasing though so the internal energy of the working substance most now be increased in order for the process to be a cycle
- Now use an adiabatic compression to bring the internal energy of the working substance back to that of $A$ so that it can repeat the cycle, without rejecting heat to a cold reservoir.
I know it violates the 2nd law of thermodynamics, but is there explanation for as to why this is impossible other than that?
thermodynamics heat-engine carnot-cycle
$endgroup$
add a comment |
$begingroup$
This is the part that allows the Carnot engine to not violate the 2nd law of thermodynamics, but, hypothetically, why can't we just adiabatically compress the working substance to get it back to State A?
As in, the working substance starts at $A$,
- Isothermal expansion - work is done on the environment
- Adiabatic expansion - work is done on the environment, temperature is decreasing though so the internal energy of the working substance most now be increased in order for the process to be a cycle
- Now use an adiabatic compression to bring the internal energy of the working substance back to that of $A$ so that it can repeat the cycle, without rejecting heat to a cold reservoir.
I know it violates the 2nd law of thermodynamics, but is there explanation for as to why this is impossible other than that?
thermodynamics heat-engine carnot-cycle
$endgroup$
add a comment |
$begingroup$
This is the part that allows the Carnot engine to not violate the 2nd law of thermodynamics, but, hypothetically, why can't we just adiabatically compress the working substance to get it back to State A?
As in, the working substance starts at $A$,
- Isothermal expansion - work is done on the environment
- Adiabatic expansion - work is done on the environment, temperature is decreasing though so the internal energy of the working substance most now be increased in order for the process to be a cycle
- Now use an adiabatic compression to bring the internal energy of the working substance back to that of $A$ so that it can repeat the cycle, without rejecting heat to a cold reservoir.
I know it violates the 2nd law of thermodynamics, but is there explanation for as to why this is impossible other than that?
thermodynamics heat-engine carnot-cycle
$endgroup$
This is the part that allows the Carnot engine to not violate the 2nd law of thermodynamics, but, hypothetically, why can't we just adiabatically compress the working substance to get it back to State A?
As in, the working substance starts at $A$,
- Isothermal expansion - work is done on the environment
- Adiabatic expansion - work is done on the environment, temperature is decreasing though so the internal energy of the working substance most now be increased in order for the process to be a cycle
- Now use an adiabatic compression to bring the internal energy of the working substance back to that of $A$ so that it can repeat the cycle, without rejecting heat to a cold reservoir.
I know it violates the 2nd law of thermodynamics, but is there explanation for as to why this is impossible other than that?
thermodynamics heat-engine carnot-cycle
thermodynamics heat-engine carnot-cycle
asked 10 hours ago
sangstarsangstar
1,1101617
1,1101617
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because what you propose is impossible. You are essentially trying to make a cycle out of only these three steps:
1) Isothermal expansion (A to B)
2) Adiabatic expansion (B to C)
3) Adiabatic compression back to original state (C to A)
The curve going from C to A cannot be an adiabatic process. Adiabatic processes are characterized by
$$PV^n=text{const}$$
where $n$ is a property of the gas being used.
Therefore, if you want to follow an adiabatic curve during compression, you will just end up going back to state B. You can't go to state A from C using an adiabatic compression.
This is why we need the isothermal compression step after the adiabatic expansion step. This step is needed so that we can get on the correct adiabatic curve back to state A
To be a little more specific, let's say the pressure and volume at states $B$ and $C$ are $(P_B,V_B)$ and $(P_C,V_C)$ respectively. Then we know in process 2
$$P_BV_B^n=P_CV_C^n=alpha$$
Or, in other words, the entire curve is described by $$P=frac{alpha}{V^n}=frac{P_BV_B^n}{V^n}=frac{P_CV_C^n}{V^n}$$
Now we want to do adiabatic compression from state C. Well we have to follow the curve defined by $PV^n=beta$, but since we know we start in state $C$ it must be that the constant is the same one as before: $beta=alpha=P_CV_C^n$. Therefore, the curve is given by
$$P=frac{beta}{V^n}=frac{P_CV_C^n}{V^n}$$
which is the same curve we followed going from B to C.
We need the isothermal compression step in order to get to the appropriate state D such that $P_DV_D^n=P_AV_A^n$
$endgroup$
$begingroup$
This is exactly how I would have tackled the problem for ages and ages. But not only is it a mathematical pain-in-the-neck, it is limited by depending on the choice of equation of state. There is a much simpler way ...
$endgroup$
– dmckee♦
2 hours ago
$begingroup$
@dmckee I guess this is more specific to an ideal gas, but I don't think simple algebra is a pain in the neck. But yeah, it is simpler in the T-S plane here. It seemed like the OP was just learning about this, so I decided to go with what I had initially learned. The T-S plane was introduced to me while studying for the physics GRE, and is a shame that it isn't emphasized more like you say in your answer.
$endgroup$
– Aaron Stevens
29 mins ago
add a comment |
$begingroup$
Examine the problem in the $T$-$S$ plane
It is traditional for books to diagram the Carnot cycle in the $P$-$V$ plane (it's also useful because the enclosed area is the net work on a cycle).
It is much less common to diagram it in the $T$-$S$ plane (even though it's just as useful because that enclosed area is the net heat). I suspect that the main reason that diagram is rarely shown explicitly is that it is a boring diagram: it's an axis-aligned rectangle.
Think about that for a minute.
If we do one adiabatic stage (vertical movement in the $T$-$S$ plane) and one isothermal stage (horizontal movement), then there is no adiabatic path back to the starting point because adiabatic paths are vertical.
And this fact is independent of the equation of state of the working fluid: it only relies on the assumption that you are executing adiabatic and isothermal stages.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f459771%2fwhy-does-there-need-to-be-an-isothermal-compression-in-a-carnot-cycle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because what you propose is impossible. You are essentially trying to make a cycle out of only these three steps:
1) Isothermal expansion (A to B)
2) Adiabatic expansion (B to C)
3) Adiabatic compression back to original state (C to A)
The curve going from C to A cannot be an adiabatic process. Adiabatic processes are characterized by
$$PV^n=text{const}$$
where $n$ is a property of the gas being used.
Therefore, if you want to follow an adiabatic curve during compression, you will just end up going back to state B. You can't go to state A from C using an adiabatic compression.
This is why we need the isothermal compression step after the adiabatic expansion step. This step is needed so that we can get on the correct adiabatic curve back to state A
To be a little more specific, let's say the pressure and volume at states $B$ and $C$ are $(P_B,V_B)$ and $(P_C,V_C)$ respectively. Then we know in process 2
$$P_BV_B^n=P_CV_C^n=alpha$$
Or, in other words, the entire curve is described by $$P=frac{alpha}{V^n}=frac{P_BV_B^n}{V^n}=frac{P_CV_C^n}{V^n}$$
Now we want to do adiabatic compression from state C. Well we have to follow the curve defined by $PV^n=beta$, but since we know we start in state $C$ it must be that the constant is the same one as before: $beta=alpha=P_CV_C^n$. Therefore, the curve is given by
$$P=frac{beta}{V^n}=frac{P_CV_C^n}{V^n}$$
which is the same curve we followed going from B to C.
We need the isothermal compression step in order to get to the appropriate state D such that $P_DV_D^n=P_AV_A^n$
$endgroup$
$begingroup$
This is exactly how I would have tackled the problem for ages and ages. But not only is it a mathematical pain-in-the-neck, it is limited by depending on the choice of equation of state. There is a much simpler way ...
$endgroup$
– dmckee♦
2 hours ago
$begingroup$
@dmckee I guess this is more specific to an ideal gas, but I don't think simple algebra is a pain in the neck. But yeah, it is simpler in the T-S plane here. It seemed like the OP was just learning about this, so I decided to go with what I had initially learned. The T-S plane was introduced to me while studying for the physics GRE, and is a shame that it isn't emphasized more like you say in your answer.
$endgroup$
– Aaron Stevens
29 mins ago
add a comment |
$begingroup$
Because what you propose is impossible. You are essentially trying to make a cycle out of only these three steps:
1) Isothermal expansion (A to B)
2) Adiabatic expansion (B to C)
3) Adiabatic compression back to original state (C to A)
The curve going from C to A cannot be an adiabatic process. Adiabatic processes are characterized by
$$PV^n=text{const}$$
where $n$ is a property of the gas being used.
Therefore, if you want to follow an adiabatic curve during compression, you will just end up going back to state B. You can't go to state A from C using an adiabatic compression.
This is why we need the isothermal compression step after the adiabatic expansion step. This step is needed so that we can get on the correct adiabatic curve back to state A
To be a little more specific, let's say the pressure and volume at states $B$ and $C$ are $(P_B,V_B)$ and $(P_C,V_C)$ respectively. Then we know in process 2
$$P_BV_B^n=P_CV_C^n=alpha$$
Or, in other words, the entire curve is described by $$P=frac{alpha}{V^n}=frac{P_BV_B^n}{V^n}=frac{P_CV_C^n}{V^n}$$
Now we want to do adiabatic compression from state C. Well we have to follow the curve defined by $PV^n=beta$, but since we know we start in state $C$ it must be that the constant is the same one as before: $beta=alpha=P_CV_C^n$. Therefore, the curve is given by
$$P=frac{beta}{V^n}=frac{P_CV_C^n}{V^n}$$
which is the same curve we followed going from B to C.
We need the isothermal compression step in order to get to the appropriate state D such that $P_DV_D^n=P_AV_A^n$
$endgroup$
$begingroup$
This is exactly how I would have tackled the problem for ages and ages. But not only is it a mathematical pain-in-the-neck, it is limited by depending on the choice of equation of state. There is a much simpler way ...
$endgroup$
– dmckee♦
2 hours ago
$begingroup$
@dmckee I guess this is more specific to an ideal gas, but I don't think simple algebra is a pain in the neck. But yeah, it is simpler in the T-S plane here. It seemed like the OP was just learning about this, so I decided to go with what I had initially learned. The T-S plane was introduced to me while studying for the physics GRE, and is a shame that it isn't emphasized more like you say in your answer.
$endgroup$
– Aaron Stevens
29 mins ago
add a comment |
$begingroup$
Because what you propose is impossible. You are essentially trying to make a cycle out of only these three steps:
1) Isothermal expansion (A to B)
2) Adiabatic expansion (B to C)
3) Adiabatic compression back to original state (C to A)
The curve going from C to A cannot be an adiabatic process. Adiabatic processes are characterized by
$$PV^n=text{const}$$
where $n$ is a property of the gas being used.
Therefore, if you want to follow an adiabatic curve during compression, you will just end up going back to state B. You can't go to state A from C using an adiabatic compression.
This is why we need the isothermal compression step after the adiabatic expansion step. This step is needed so that we can get on the correct adiabatic curve back to state A
To be a little more specific, let's say the pressure and volume at states $B$ and $C$ are $(P_B,V_B)$ and $(P_C,V_C)$ respectively. Then we know in process 2
$$P_BV_B^n=P_CV_C^n=alpha$$
Or, in other words, the entire curve is described by $$P=frac{alpha}{V^n}=frac{P_BV_B^n}{V^n}=frac{P_CV_C^n}{V^n}$$
Now we want to do adiabatic compression from state C. Well we have to follow the curve defined by $PV^n=beta$, but since we know we start in state $C$ it must be that the constant is the same one as before: $beta=alpha=P_CV_C^n$. Therefore, the curve is given by
$$P=frac{beta}{V^n}=frac{P_CV_C^n}{V^n}$$
which is the same curve we followed going from B to C.
We need the isothermal compression step in order to get to the appropriate state D such that $P_DV_D^n=P_AV_A^n$
$endgroup$
Because what you propose is impossible. You are essentially trying to make a cycle out of only these three steps:
1) Isothermal expansion (A to B)
2) Adiabatic expansion (B to C)
3) Adiabatic compression back to original state (C to A)
The curve going from C to A cannot be an adiabatic process. Adiabatic processes are characterized by
$$PV^n=text{const}$$
where $n$ is a property of the gas being used.
Therefore, if you want to follow an adiabatic curve during compression, you will just end up going back to state B. You can't go to state A from C using an adiabatic compression.
This is why we need the isothermal compression step after the adiabatic expansion step. This step is needed so that we can get on the correct adiabatic curve back to state A
To be a little more specific, let's say the pressure and volume at states $B$ and $C$ are $(P_B,V_B)$ and $(P_C,V_C)$ respectively. Then we know in process 2
$$P_BV_B^n=P_CV_C^n=alpha$$
Or, in other words, the entire curve is described by $$P=frac{alpha}{V^n}=frac{P_BV_B^n}{V^n}=frac{P_CV_C^n}{V^n}$$
Now we want to do adiabatic compression from state C. Well we have to follow the curve defined by $PV^n=beta$, but since we know we start in state $C$ it must be that the constant is the same one as before: $beta=alpha=P_CV_C^n$. Therefore, the curve is given by
$$P=frac{beta}{V^n}=frac{P_CV_C^n}{V^n}$$
which is the same curve we followed going from B to C.
We need the isothermal compression step in order to get to the appropriate state D such that $P_DV_D^n=P_AV_A^n$
edited 9 hours ago
answered 10 hours ago
Aaron StevensAaron Stevens
10.6k31742
10.6k31742
$begingroup$
This is exactly how I would have tackled the problem for ages and ages. But not only is it a mathematical pain-in-the-neck, it is limited by depending on the choice of equation of state. There is a much simpler way ...
$endgroup$
– dmckee♦
2 hours ago
$begingroup$
@dmckee I guess this is more specific to an ideal gas, but I don't think simple algebra is a pain in the neck. But yeah, it is simpler in the T-S plane here. It seemed like the OP was just learning about this, so I decided to go with what I had initially learned. The T-S plane was introduced to me while studying for the physics GRE, and is a shame that it isn't emphasized more like you say in your answer.
$endgroup$
– Aaron Stevens
29 mins ago
add a comment |
$begingroup$
This is exactly how I would have tackled the problem for ages and ages. But not only is it a mathematical pain-in-the-neck, it is limited by depending on the choice of equation of state. There is a much simpler way ...
$endgroup$
– dmckee♦
2 hours ago
$begingroup$
@dmckee I guess this is more specific to an ideal gas, but I don't think simple algebra is a pain in the neck. But yeah, it is simpler in the T-S plane here. It seemed like the OP was just learning about this, so I decided to go with what I had initially learned. The T-S plane was introduced to me while studying for the physics GRE, and is a shame that it isn't emphasized more like you say in your answer.
$endgroup$
– Aaron Stevens
29 mins ago
$begingroup$
This is exactly how I would have tackled the problem for ages and ages. But not only is it a mathematical pain-in-the-neck, it is limited by depending on the choice of equation of state. There is a much simpler way ...
$endgroup$
– dmckee♦
2 hours ago
$begingroup$
This is exactly how I would have tackled the problem for ages and ages. But not only is it a mathematical pain-in-the-neck, it is limited by depending on the choice of equation of state. There is a much simpler way ...
$endgroup$
– dmckee♦
2 hours ago
$begingroup$
@dmckee I guess this is more specific to an ideal gas, but I don't think simple algebra is a pain in the neck. But yeah, it is simpler in the T-S plane here. It seemed like the OP was just learning about this, so I decided to go with what I had initially learned. The T-S plane was introduced to me while studying for the physics GRE, and is a shame that it isn't emphasized more like you say in your answer.
$endgroup$
– Aaron Stevens
29 mins ago
$begingroup$
@dmckee I guess this is more specific to an ideal gas, but I don't think simple algebra is a pain in the neck. But yeah, it is simpler in the T-S plane here. It seemed like the OP was just learning about this, so I decided to go with what I had initially learned. The T-S plane was introduced to me while studying for the physics GRE, and is a shame that it isn't emphasized more like you say in your answer.
$endgroup$
– Aaron Stevens
29 mins ago
add a comment |
$begingroup$
Examine the problem in the $T$-$S$ plane
It is traditional for books to diagram the Carnot cycle in the $P$-$V$ plane (it's also useful because the enclosed area is the net work on a cycle).
It is much less common to diagram it in the $T$-$S$ plane (even though it's just as useful because that enclosed area is the net heat). I suspect that the main reason that diagram is rarely shown explicitly is that it is a boring diagram: it's an axis-aligned rectangle.
Think about that for a minute.
If we do one adiabatic stage (vertical movement in the $T$-$S$ plane) and one isothermal stage (horizontal movement), then there is no adiabatic path back to the starting point because adiabatic paths are vertical.
And this fact is independent of the equation of state of the working fluid: it only relies on the assumption that you are executing adiabatic and isothermal stages.
$endgroup$
add a comment |
$begingroup$
Examine the problem in the $T$-$S$ plane
It is traditional for books to diagram the Carnot cycle in the $P$-$V$ plane (it's also useful because the enclosed area is the net work on a cycle).
It is much less common to diagram it in the $T$-$S$ plane (even though it's just as useful because that enclosed area is the net heat). I suspect that the main reason that diagram is rarely shown explicitly is that it is a boring diagram: it's an axis-aligned rectangle.
Think about that for a minute.
If we do one adiabatic stage (vertical movement in the $T$-$S$ plane) and one isothermal stage (horizontal movement), then there is no adiabatic path back to the starting point because adiabatic paths are vertical.
And this fact is independent of the equation of state of the working fluid: it only relies on the assumption that you are executing adiabatic and isothermal stages.
$endgroup$
add a comment |
$begingroup$
Examine the problem in the $T$-$S$ plane
It is traditional for books to diagram the Carnot cycle in the $P$-$V$ plane (it's also useful because the enclosed area is the net work on a cycle).
It is much less common to diagram it in the $T$-$S$ plane (even though it's just as useful because that enclosed area is the net heat). I suspect that the main reason that diagram is rarely shown explicitly is that it is a boring diagram: it's an axis-aligned rectangle.
Think about that for a minute.
If we do one adiabatic stage (vertical movement in the $T$-$S$ plane) and one isothermal stage (horizontal movement), then there is no adiabatic path back to the starting point because adiabatic paths are vertical.
And this fact is independent of the equation of state of the working fluid: it only relies on the assumption that you are executing adiabatic and isothermal stages.
$endgroup$
Examine the problem in the $T$-$S$ plane
It is traditional for books to diagram the Carnot cycle in the $P$-$V$ plane (it's also useful because the enclosed area is the net work on a cycle).
It is much less common to diagram it in the $T$-$S$ plane (even though it's just as useful because that enclosed area is the net heat). I suspect that the main reason that diagram is rarely shown explicitly is that it is a boring diagram: it's an axis-aligned rectangle.
Think about that for a minute.
If we do one adiabatic stage (vertical movement in the $T$-$S$ plane) and one isothermal stage (horizontal movement), then there is no adiabatic path back to the starting point because adiabatic paths are vertical.
And this fact is independent of the equation of state of the working fluid: it only relies on the assumption that you are executing adiabatic and isothermal stages.
edited 2 hours ago
answered 2 hours ago
dmckee♦dmckee
74.3k6134269
74.3k6134269
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f459771%2fwhy-does-there-need-to-be-an-isothermal-compression-in-a-carnot-cycle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown