Why is there Net Work Done on a Pressure/Volume Cycle?Net work done for rubber bandsWhy are volume and pressure inversely proportional to each other?Net work output of an engine performing Carnot cycle?Volume quotient in Carnot-cyclePressure vs Volume Calculation for airWork done by pressurePressure Volume graphs and work donePressure-Volume DiagramWhich fundamental reversible thermodynamic process has maximum work associated with it?Which direction will this object move in at the end of this process?Doubt in thermodynamics pressure volume work done
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Why is there Net Work Done on a Pressure/Volume Cycle?
Net work done for rubber bandsWhy are volume and pressure inversely proportional to each other?Net work output of an engine performing Carnot cycle?Volume quotient in Carnot-cyclePressure vs Volume Calculation for airWork done by pressurePressure Volume graphs and work donePressure-Volume DiagramWhich fundamental reversible thermodynamic process has maximum work associated with it?Which direction will this object move in at the end of this process?Doubt in thermodynamics pressure volume work done
$begingroup$
The area enclosed by the rectangle is said to be the Net Work Done. My question is why would there be any net workdone if it returns to its initial position (W).
Shouldn't it act like a force-extension graph, where the net work done of a sample (eg.rubber), is the Area under the loading curve substracted by the Area under the unloading graph, and the area would be present if the sample undergoes plastic deformation.
I'm currently still in highschool, so any simple concise real-world example might be very helpful.
thermodynamics energy pressure volume carnot-cycle
asked Apr 19 at 11:51
KEVIN IOPKEVIN IOP
183
$endgroup$
add a comment |
$begingroup$
The area enclosed by the rectangle is said to be the Net Work Done. My question is why would there be any net workdone if it returns to its initial position (W).
Shouldn't it act like a force-extension graph, where the net work done of a sample (eg.rubber), is the Area under the loading curve substracted by the Area under the unloading graph, and the area would be present if the sample undergoes plastic deformation.
I'm currently still in highschool, so any simple concise real-world example might be very helpful.
thermodynamics energy pressure volume carnot-cycle
asked Apr 19 at 11:51
KEVIN IOPKEVIN IOP
183
$endgroup$
1
$begingroup$
Work is a path function and hence depends on the path followed rather than the initial and final state. So even though it returns to its original position there is a net work done
$endgroup$
– Harsh Wasnik
Apr 19 at 12:12
$begingroup$
Suppose your spring constant depends on temperature, and you stretch the spring at a low temperature and then let it return at a higher temperature. Then, you have done a net amount of work. After letting it return, you change the temperature back to the original temperature and start over.
$endgroup$
– Chet Miller
Apr 19 at 12:12
1
$begingroup$
You "apply" energy to increase the pressure. Not to decrease it again. That happens by itself (that energy is lost or absorbed elsewhere).
$endgroup$
– Steeven
Apr 19 at 13:05
1
$begingroup$
Work = force x distance. The distance that the piston moves is the same in either direction. But what the answers below are telling you, in various different ways, is that the force acting on the piston when it moves in one direction is different from the force that acts on it when it moves in the other direction.
$endgroup$
– Solomon Slow
Apr 19 at 14:46
add a comment |
$begingroup$
The area enclosed by the rectangle is said to be the Net Work Done. My question is why would there be any net workdone if it returns to its initial position (W).
Shouldn't it act like a force-extension graph, where the net work done of a sample (eg.rubber), is the Area under the loading curve substracted by the Area under the unloading graph, and the area would be present if the sample undergoes plastic deformation.
I'm currently still in highschool, so any simple concise real-world example might be very helpful.
thermodynamics energy pressure volume carnot-cycle
asked Apr 19 at 11:51
KEVIN IOPKEVIN IOP
183
$endgroup$
The area enclosed by the rectangle is said to be the Net Work Done. My question is why would there be any net workdone if it returns to its initial position (W).
Shouldn't it act like a force-extension graph, where the net work done of a sample (eg.rubber), is the Area under the loading curve substracted by the Area under the unloading graph, and the area would be present if the sample undergoes plastic deformation.
I'm currently still in highschool, so any simple concise real-world example might be very helpful.
thermodynamics energy pressure volume carnot-cycle
thermodynamics energy pressure volume carnot-cycle
asked Apr 19 at 11:51
KEVIN IOPKEVIN IOP
183
asked Apr 19 at 11:51
KEVIN IOPKEVIN IOP
183
asked Apr 19 at 11:51
KEVIN IOPKEVIN IOP
183
asked Apr 19 at 11:51
KEVIN IOPKEVIN IOP
183
asked Apr 19 at 11:51
KEVIN IOPKEVIN IOP
183
183
1
$begingroup$
Work is a path function and hence depends on the path followed rather than the initial and final state. So even though it returns to its original position there is a net work done
$endgroup$
– Harsh Wasnik
Apr 19 at 12:12
$begingroup$
Suppose your spring constant depends on temperature, and you stretch the spring at a low temperature and then let it return at a higher temperature. Then, you have done a net amount of work. After letting it return, you change the temperature back to the original temperature and start over.
$endgroup$
– Chet Miller
Apr 19 at 12:12
1
$begingroup$
You "apply" energy to increase the pressure. Not to decrease it again. That happens by itself (that energy is lost or absorbed elsewhere).
$endgroup$
– Steeven
Apr 19 at 13:05
1
$begingroup$
Work = force x distance. The distance that the piston moves is the same in either direction. But what the answers below are telling you, in various different ways, is that the force acting on the piston when it moves in one direction is different from the force that acts on it when it moves in the other direction.
$endgroup$
– Solomon Slow
Apr 19 at 14:46
add a comment |
1
$begingroup$
Work is a path function and hence depends on the path followed rather than the initial and final state. So even though it returns to its original position there is a net work done
$endgroup$
– Harsh Wasnik
Apr 19 at 12:12
$begingroup$
Suppose your spring constant depends on temperature, and you stretch the spring at a low temperature and then let it return at a higher temperature. Then, you have done a net amount of work. After letting it return, you change the temperature back to the original temperature and start over.
$endgroup$
– Chet Miller
Apr 19 at 12:12
1
$begingroup$
You "apply" energy to increase the pressure. Not to decrease it again. That happens by itself (that energy is lost or absorbed elsewhere).
$endgroup$
– Steeven
Apr 19 at 13:05
1
$begingroup$
Work = force x distance. The distance that the piston moves is the same in either direction. But what the answers below are telling you, in various different ways, is that the force acting on the piston when it moves in one direction is different from the force that acts on it when it moves in the other direction.
$endgroup$
– Solomon Slow
Apr 19 at 14:46
1
1
$begingroup$
Work is a path function and hence depends on the path followed rather than the initial and final state. So even though it returns to its original position there is a net work done
$endgroup$
– Harsh Wasnik
Apr 19 at 12:12
$begingroup$
Work is a path function and hence depends on the path followed rather than the initial and final state. So even though it returns to its original position there is a net work done
$endgroup$
– Harsh Wasnik
Apr 19 at 12:12
$begingroup$
Suppose your spring constant depends on temperature, and you stretch the spring at a low temperature and then let it return at a higher temperature. Then, you have done a net amount of work. After letting it return, you change the temperature back to the original temperature and start over.
$endgroup$
– Chet Miller
Apr 19 at 12:12
$begingroup$
Suppose your spring constant depends on temperature, and you stretch the spring at a low temperature and then let it return at a higher temperature. Then, you have done a net amount of work. After letting it return, you change the temperature back to the original temperature and start over.
$endgroup$
– Chet Miller
Apr 19 at 12:12
1
1
$begingroup$
You "apply" energy to increase the pressure. Not to decrease it again. That happens by itself (that energy is lost or absorbed elsewhere).
$endgroup$
– Steeven
Apr 19 at 13:05
$begingroup$
You "apply" energy to increase the pressure. Not to decrease it again. That happens by itself (that energy is lost or absorbed elsewhere).
$endgroup$
– Steeven
Apr 19 at 13:05
1
1
$begingroup$
Work = force x distance. The distance that the piston moves is the same in either direction. But what the answers below are telling you, in various different ways, is that the force acting on the piston when it moves in one direction is different from the force that acts on it when it moves in the other direction.
$endgroup$
– Solomon Slow
Apr 19 at 14:46
$begingroup$
Work = force x distance. The distance that the piston moves is the same in either direction. But what the answers below are telling you, in various different ways, is that the force acting on the piston when it moves in one direction is different from the force that acts on it when it moves in the other direction.
$endgroup$
– Solomon Slow
Apr 19 at 14:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Think about going from X to Y on the diagram.
The volume is constant, but the pressure increases.
Something must be happening to the gas to cause that. One way to make it happen would to add heat energy to the gas and raise its temperature.
The rectangular shape of the pressure-volume diagram doesn't correspond to anything that is physically simple, but you should be able to see that as you go from Y to Z and the volume increases, the gas pressure would decrease unless you continue to add more heat energy to the system and keep the pressure constant as the gas expands.
For the other two sides, Z to W and back to X, the gas is losing heat energy. Don't try to over-think this (because as the previous paragraph said, this diagram doesn't correspond to any simple physical device), but going round the complete cycle, the amount of heat added is not the same as the amount of heat removed.
The net amount of heat added to the system is converted into mechanical work when the gas pressure changes the volume of the system, moving from Y to Z and then from W back to X.
Since the starting and ending conditions at point X are the same, the amount of heat energy added to the system during the cycle is the same as the amount of mechanical work done by the gas.
Similar diagrams (but with curved sides, not straight) are used to understand the behaviour of different types of "heat engines" such as steam engines, gasoline and diesel car engines, jet aircraft engines, etc. Devices where mechanical work is used to remove heat from a system, for example refrigerators and air conditioning systems, can be understood in the same way, moving round a loop on the pressure-volume diagram in the opposite direction.
answered Apr 19 at 12:30
alephzeroalephzero
5,98421122
$endgroup$
add a comment |
$begingroup$
In effect a “piston” moves the same distance during the volume changes along paths $WX$ and $YZ$ but a greater force needs to be exerted along path $YZ$ than along path $WX$ and so the work done (force $times$ distance) along path $YZ$ is greater than along path $WZ$.
Your example of the force-extension curve for rubber is a good one but in that case the effect is called hysteresis and it is unlike plastic deformation in that there is no permanent deformation of the rubber after it is unloaded.
answered Apr 19 at 12:12
FarcherFarcher
52.5k340112
$endgroup$
$begingroup$
I see. physics.stackexchange.com/a/149122/226894 . So the small section between the loading & unloading curve is the net work done, right? And if that is so what exactly is meant by NET work done?
$endgroup$
– KEVIN IOP
Apr 19 at 12:46
$begingroup$
@KEVINIOP For an elastic spring the work done expanding the spring by an external force is exactly equal to the work done by the spring when it is contracting. The net work done is zero. In your case the work done by the gas in expanding is less than the work done by an external force compressing the gas,. The difference between these two quantities is the net work done.
$endgroup$
– Farcher
Apr 22 at 12:29
add a comment |
$begingroup$
If you come back to initial position by same path, the work done would be obviously zero.
The PV work is path dependant; not state dependant. So even if you come back to same state , the work done may not be zero. The same physics is used here also, like in a spring the work done is zero if returned back to initial position by same force, but what do you expect when the force applied on backward path is higher? . Think of it like this: The work done is $PDelta V$, (assuming pressure =constant). By referencing to your graph, you can easily understand, that in one path, the pressure is higher than other, i.e., the returning path is not exactly the same. The pressure and forces are different. So in a loop enclosing a non-zero area, a non-zero work is done by gas in cycle.
Hope you got something out of my explanation:)
answered Apr 19 at 13:08
HS SinghHS Singh
562
$endgroup$
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Think about going from X to Y on the diagram.
The volume is constant, but the pressure increases.
Something must be happening to the gas to cause that. One way to make it happen would to add heat energy to the gas and raise its temperature.
The rectangular shape of the pressure-volume diagram doesn't correspond to anything that is physically simple, but you should be able to see that as you go from Y to Z and the volume increases, the gas pressure would decrease unless you continue to add more heat energy to the system and keep the pressure constant as the gas expands.
For the other two sides, Z to W and back to X, the gas is losing heat energy. Don't try to over-think this (because as the previous paragraph said, this diagram doesn't correspond to any simple physical device), but going round the complete cycle, the amount of heat added is not the same as the amount of heat removed.
The net amount of heat added to the system is converted into mechanical work when the gas pressure changes the volume of the system, moving from Y to Z and then from W back to X.
Since the starting and ending conditions at point X are the same, the amount of heat energy added to the system during the cycle is the same as the amount of mechanical work done by the gas.
Similar diagrams (but with curved sides, not straight) are used to understand the behaviour of different types of "heat engines" such as steam engines, gasoline and diesel car engines, jet aircraft engines, etc. Devices where mechanical work is used to remove heat from a system, for example refrigerators and air conditioning systems, can be understood in the same way, moving round a loop on the pressure-volume diagram in the opposite direction.
answered Apr 19 at 12:30
alephzeroalephzero
5,98421122
$endgroup$
add a comment |
$begingroup$
Think about going from X to Y on the diagram.
The volume is constant, but the pressure increases.
Something must be happening to the gas to cause that. One way to make it happen would to add heat energy to the gas and raise its temperature.
The rectangular shape of the pressure-volume diagram doesn't correspond to anything that is physically simple, but you should be able to see that as you go from Y to Z and the volume increases, the gas pressure would decrease unless you continue to add more heat energy to the system and keep the pressure constant as the gas expands.
For the other two sides, Z to W and back to X, the gas is losing heat energy. Don't try to over-think this (because as the previous paragraph said, this diagram doesn't correspond to any simple physical device), but going round the complete cycle, the amount of heat added is not the same as the amount of heat removed.
The net amount of heat added to the system is converted into mechanical work when the gas pressure changes the volume of the system, moving from Y to Z and then from W back to X.
Since the starting and ending conditions at point X are the same, the amount of heat energy added to the system during the cycle is the same as the amount of mechanical work done by the gas.
Similar diagrams (but with curved sides, not straight) are used to understand the behaviour of different types of "heat engines" such as steam engines, gasoline and diesel car engines, jet aircraft engines, etc. Devices where mechanical work is used to remove heat from a system, for example refrigerators and air conditioning systems, can be understood in the same way, moving round a loop on the pressure-volume diagram in the opposite direction.
answered Apr 19 at 12:30
alephzeroalephzero
5,98421122
$endgroup$
add a comment |
$begingroup$
Think about going from X to Y on the diagram.
The volume is constant, but the pressure increases.
Something must be happening to the gas to cause that. One way to make it happen would to add heat energy to the gas and raise its temperature.
The rectangular shape of the pressure-volume diagram doesn't correspond to anything that is physically simple, but you should be able to see that as you go from Y to Z and the volume increases, the gas pressure would decrease unless you continue to add more heat energy to the system and keep the pressure constant as the gas expands.
For the other two sides, Z to W and back to X, the gas is losing heat energy. Don't try to over-think this (because as the previous paragraph said, this diagram doesn't correspond to any simple physical device), but going round the complete cycle, the amount of heat added is not the same as the amount of heat removed.
The net amount of heat added to the system is converted into mechanical work when the gas pressure changes the volume of the system, moving from Y to Z and then from W back to X.
Since the starting and ending conditions at point X are the same, the amount of heat energy added to the system during the cycle is the same as the amount of mechanical work done by the gas.
Similar diagrams (but with curved sides, not straight) are used to understand the behaviour of different types of "heat engines" such as steam engines, gasoline and diesel car engines, jet aircraft engines, etc. Devices where mechanical work is used to remove heat from a system, for example refrigerators and air conditioning systems, can be understood in the same way, moving round a loop on the pressure-volume diagram in the opposite direction.
answered Apr 19 at 12:30
alephzeroalephzero
5,98421122
$endgroup$
Think about going from X to Y on the diagram.
The volume is constant, but the pressure increases.
Something must be happening to the gas to cause that. One way to make it happen would to add heat energy to the gas and raise its temperature.
The rectangular shape of the pressure-volume diagram doesn't correspond to anything that is physically simple, but you should be able to see that as you go from Y to Z and the volume increases, the gas pressure would decrease unless you continue to add more heat energy to the system and keep the pressure constant as the gas expands.
For the other two sides, Z to W and back to X, the gas is losing heat energy. Don't try to over-think this (because as the previous paragraph said, this diagram doesn't correspond to any simple physical device), but going round the complete cycle, the amount of heat added is not the same as the amount of heat removed.
The net amount of heat added to the system is converted into mechanical work when the gas pressure changes the volume of the system, moving from Y to Z and then from W back to X.
Since the starting and ending conditions at point X are the same, the amount of heat energy added to the system during the cycle is the same as the amount of mechanical work done by the gas.
Similar diagrams (but with curved sides, not straight) are used to understand the behaviour of different types of "heat engines" such as steam engines, gasoline and diesel car engines, jet aircraft engines, etc. Devices where mechanical work is used to remove heat from a system, for example refrigerators and air conditioning systems, can be understood in the same way, moving round a loop on the pressure-volume diagram in the opposite direction.
answered Apr 19 at 12:30
alephzeroalephzero
5,98421122
edited Apr 19 at 12:36
answered Apr 19 at 12:30
alephzeroalephzero
5,98421122
answered Apr 19 at 12:30
alephzeroalephzero
5,98421122
answered Apr 19 at 12:30
alephzeroalephzero
5,98421122
5,98421122
add a comment |
add a comment |
$begingroup$
In effect a “piston” moves the same distance during the volume changes along paths $WX$ and $YZ$ but a greater force needs to be exerted along path $YZ$ than along path $WX$ and so the work done (force $times$ distance) along path $YZ$ is greater than along path $WZ$.
Your example of the force-extension curve for rubber is a good one but in that case the effect is called hysteresis and it is unlike plastic deformation in that there is no permanent deformation of the rubber after it is unloaded.
answered Apr 19 at 12:12
FarcherFarcher
52.5k340112
$endgroup$
$begingroup$
I see. physics.stackexchange.com/a/149122/226894 . So the small section between the loading & unloading curve is the net work done, right? And if that is so what exactly is meant by NET work done?
$endgroup$
– KEVIN IOP
Apr 19 at 12:46
$begingroup$
@KEVINIOP For an elastic spring the work done expanding the spring by an external force is exactly equal to the work done by the spring when it is contracting. The net work done is zero. In your case the work done by the gas in expanding is less than the work done by an external force compressing the gas,. The difference between these two quantities is the net work done.
$endgroup$
– Farcher
Apr 22 at 12:29
add a comment |
$begingroup$
In effect a “piston” moves the same distance during the volume changes along paths $WX$ and $YZ$ but a greater force needs to be exerted along path $YZ$ than along path $WX$ and so the work done (force $times$ distance) along path $YZ$ is greater than along path $WZ$.
Your example of the force-extension curve for rubber is a good one but in that case the effect is called hysteresis and it is unlike plastic deformation in that there is no permanent deformation of the rubber after it is unloaded.
answered Apr 19 at 12:12
FarcherFarcher
52.5k340112
$endgroup$
$begingroup$
I see. physics.stackexchange.com/a/149122/226894 . So the small section between the loading & unloading curve is the net work done, right? And if that is so what exactly is meant by NET work done?
$endgroup$
– KEVIN IOP
Apr 19 at 12:46
$begingroup$
@KEVINIOP For an elastic spring the work done expanding the spring by an external force is exactly equal to the work done by the spring when it is contracting. The net work done is zero. In your case the work done by the gas in expanding is less than the work done by an external force compressing the gas,. The difference between these two quantities is the net work done.
$endgroup$
– Farcher
Apr 22 at 12:29
add a comment |
$begingroup$
In effect a “piston” moves the same distance during the volume changes along paths $WX$ and $YZ$ but a greater force needs to be exerted along path $YZ$ than along path $WX$ and so the work done (force $times$ distance) along path $YZ$ is greater than along path $WZ$.
Your example of the force-extension curve for rubber is a good one but in that case the effect is called hysteresis and it is unlike plastic deformation in that there is no permanent deformation of the rubber after it is unloaded.
answered Apr 19 at 12:12
FarcherFarcher
52.5k340112
$endgroup$
In effect a “piston” moves the same distance during the volume changes along paths $WX$ and $YZ$ but a greater force needs to be exerted along path $YZ$ than along path $WX$ and so the work done (force $times$ distance) along path $YZ$ is greater than along path $WZ$.
Your example of the force-extension curve for rubber is a good one but in that case the effect is called hysteresis and it is unlike plastic deformation in that there is no permanent deformation of the rubber after it is unloaded.
answered Apr 19 at 12:12
FarcherFarcher
52.5k340112
answered Apr 19 at 12:12
FarcherFarcher
52.5k340112
answered Apr 19 at 12:12
FarcherFarcher
52.5k340112
answered Apr 19 at 12:12
FarcherFarcher
52.5k340112
52.5k340112
$begingroup$
I see. physics.stackexchange.com/a/149122/226894 . So the small section between the loading & unloading curve is the net work done, right? And if that is so what exactly is meant by NET work done?
$endgroup$
– KEVIN IOP
Apr 19 at 12:46
$begingroup$
@KEVINIOP For an elastic spring the work done expanding the spring by an external force is exactly equal to the work done by the spring when it is contracting. The net work done is zero. In your case the work done by the gas in expanding is less than the work done by an external force compressing the gas,. The difference between these two quantities is the net work done.
$endgroup$
– Farcher
Apr 22 at 12:29
add a comment |
$begingroup$
I see. physics.stackexchange.com/a/149122/226894 . So the small section between the loading & unloading curve is the net work done, right? And if that is so what exactly is meant by NET work done?
$endgroup$
– KEVIN IOP
Apr 19 at 12:46
$begingroup$
@KEVINIOP For an elastic spring the work done expanding the spring by an external force is exactly equal to the work done by the spring when it is contracting. The net work done is zero. In your case the work done by the gas in expanding is less than the work done by an external force compressing the gas,. The difference between these two quantities is the net work done.
$endgroup$
– Farcher
Apr 22 at 12:29
$begingroup$
I see. physics.stackexchange.com/a/149122/226894 . So the small section between the loading & unloading curve is the net work done, right? And if that is so what exactly is meant by NET work done?
$endgroup$
– KEVIN IOP
Apr 19 at 12:46
$begingroup$
I see. physics.stackexchange.com/a/149122/226894 . So the small section between the loading & unloading curve is the net work done, right? And if that is so what exactly is meant by NET work done?
$endgroup$
– KEVIN IOP
Apr 19 at 12:46
$begingroup$
@KEVINIOP For an elastic spring the work done expanding the spring by an external force is exactly equal to the work done by the spring when it is contracting. The net work done is zero. In your case the work done by the gas in expanding is less than the work done by an external force compressing the gas,. The difference between these two quantities is the net work done.
$endgroup$
– Farcher
Apr 22 at 12:29
$begingroup$
@KEVINIOP For an elastic spring the work done expanding the spring by an external force is exactly equal to the work done by the spring when it is contracting. The net work done is zero. In your case the work done by the gas in expanding is less than the work done by an external force compressing the gas,. The difference between these two quantities is the net work done.
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– Farcher
Apr 22 at 12:29
add a comment |
$begingroup$
If you come back to initial position by same path, the work done would be obviously zero.
The PV work is path dependant; not state dependant. So even if you come back to same state , the work done may not be zero. The same physics is used here also, like in a spring the work done is zero if returned back to initial position by same force, but what do you expect when the force applied on backward path is higher? . Think of it like this: The work done is $PDelta V$, (assuming pressure =constant). By referencing to your graph, you can easily understand, that in one path, the pressure is higher than other, i.e., the returning path is not exactly the same. The pressure and forces are different. So in a loop enclosing a non-zero area, a non-zero work is done by gas in cycle.
Hope you got something out of my explanation:)
answered Apr 19 at 13:08
HS SinghHS Singh
562
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add a comment |
$begingroup$
If you come back to initial position by same path, the work done would be obviously zero.
The PV work is path dependant; not state dependant. So even if you come back to same state , the work done may not be zero. The same physics is used here also, like in a spring the work done is zero if returned back to initial position by same force, but what do you expect when the force applied on backward path is higher? . Think of it like this: The work done is $PDelta V$, (assuming pressure =constant). By referencing to your graph, you can easily understand, that in one path, the pressure is higher than other, i.e., the returning path is not exactly the same. The pressure and forces are different. So in a loop enclosing a non-zero area, a non-zero work is done by gas in cycle.
Hope you got something out of my explanation:)
answered Apr 19 at 13:08
HS SinghHS Singh
562
$endgroup$
add a comment |
$begingroup$
If you come back to initial position by same path, the work done would be obviously zero.
The PV work is path dependant; not state dependant. So even if you come back to same state , the work done may not be zero. The same physics is used here also, like in a spring the work done is zero if returned back to initial position by same force, but what do you expect when the force applied on backward path is higher? . Think of it like this: The work done is $PDelta V$, (assuming pressure =constant). By referencing to your graph, you can easily understand, that in one path, the pressure is higher than other, i.e., the returning path is not exactly the same. The pressure and forces are different. So in a loop enclosing a non-zero area, a non-zero work is done by gas in cycle.
Hope you got something out of my explanation:)
answered Apr 19 at 13:08
HS SinghHS Singh
562
$endgroup$
If you come back to initial position by same path, the work done would be obviously zero.
The PV work is path dependant; not state dependant. So even if you come back to same state , the work done may not be zero. The same physics is used here also, like in a spring the work done is zero if returned back to initial position by same force, but what do you expect when the force applied on backward path is higher? . Think of it like this: The work done is $PDelta V$, (assuming pressure =constant). By referencing to your graph, you can easily understand, that in one path, the pressure is higher than other, i.e., the returning path is not exactly the same. The pressure and forces are different. So in a loop enclosing a non-zero area, a non-zero work is done by gas in cycle.
Hope you got something out of my explanation:)
answered Apr 19 at 13:08
HS SinghHS Singh
562
answered Apr 19 at 13:08
HS SinghHS Singh
562
answered Apr 19 at 13:08
HS SinghHS Singh
562
answered Apr 19 at 13:08
HS SinghHS Singh
562
562
add a comment |
add a comment |
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Work is a path function and hence depends on the path followed rather than the initial and final state. So even though it returns to its original position there is a net work done
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– Harsh Wasnik
Apr 19 at 12:12
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Suppose your spring constant depends on temperature, and you stretch the spring at a low temperature and then let it return at a higher temperature. Then, you have done a net amount of work. After letting it return, you change the temperature back to the original temperature and start over.
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– Chet Miller
Apr 19 at 12:12
1
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You "apply" energy to increase the pressure. Not to decrease it again. That happens by itself (that energy is lost or absorbed elsewhere).
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– Steeven
Apr 19 at 13:05
1
$begingroup$
Work = force x distance. The distance that the piston moves is the same in either direction. But what the answers below are telling you, in various different ways, is that the force acting on the piston when it moves in one direction is different from the force that acts on it when it moves in the other direction.
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– Solomon Slow
Apr 19 at 14:46