Why does there need to be an isothermal compression in a Carnot cycle?
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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
asked 5 hours ago
sangstarsangstar
1,0951617
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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
asked 5 hours ago
sangstarsangstar
1,0951617
$endgroup$
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
asked 5 hours ago
sangstarsangstar
1,0951617
$endgroup$
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 5 hours ago
sangstarsangstar
1,0951617
asked 5 hours ago
sangstarsangstar
1,0951617
asked 5 hours ago
sangstarsangstar
1,0951617
asked 5 hours ago
sangstarsangstar
1,0951617
asked 5 hours ago
sangstarsangstar
1,0951617
1,0951617
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
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 $P_DV_D^n=P_AV_A^n$
answered 5 hours ago
Aaron StevensAaron Stevens
10.6k31742
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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$begingroup$
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 $P_DV_D^n=P_AV_A^n$
answered 5 hours ago
Aaron StevensAaron Stevens
10.6k31742
$endgroup$
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 $P_DV_D^n=P_AV_A^n$
answered 5 hours ago
Aaron StevensAaron Stevens
10.6k31742
$endgroup$
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 $P_DV_D^n=P_AV_A^n$
answered 5 hours ago
Aaron StevensAaron Stevens
10.6k31742
$endgroup$
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 $P_DV_D^n=P_AV_A^n$
answered 5 hours ago
Aaron StevensAaron Stevens
10.6k31742
edited 5 hours ago
answered 5 hours ago
Aaron StevensAaron Stevens
10.6k31742
answered 5 hours ago
Aaron StevensAaron Stevens
10.6k31742
answered 5 hours ago
Aaron StevensAaron Stevens
10.6k31742
10.6k31742
add a comment |
add a comment |
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