Mixing
Work done by a gas
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In the expression for work done by a gas integral(Pdv) aren't we supposed to use internal pressure?
Moreover work done by gas is the work done by the force exerted by th gas,but everywhere i find people using external pressure instead of internal pressure
thermodynamics
asked 36 mins ago
Alfred Mathew
211
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Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
4
down vote
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In the expression for work done by a gas integral(Pdv) aren't we supposed to use internal pressure?
Moreover work done by gas is the work done by the force exerted by th gas,but everywhere i find people using external pressure instead of internal pressure
thermodynamics
asked 36 mins ago
Alfred Mathew
211
New contributor
Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In equilibrium, the pressure exerted by a gas on the walls ("internal pressure") is equal to the pressure exerted by the walls on the gas ("external pressure"). Thermodynamics deals with transitions between a series of equilibrium states.
– probably_someone
30 mins ago
@probably_someone The issue might be that if the pressure is changing, then it can't always be equal to the external pressure.
– Aaron Stevens
28 mins ago
@probably_someone external pressure is equal to internal pressure if the piston is massless..but that is not true if the piston has sufficient mass
– Alfred Mathew
25 mins ago
@probably_someone moreover in a thermodynamic process equilibrium need not be there in the intermediate stages
– Alfred Mathew
23 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In the expression for work done by a gas integral(Pdv) aren't we supposed to use internal pressure?
Moreover work done by gas is the work done by the force exerted by th gas,but everywhere i find people using external pressure instead of internal pressure
thermodynamics
asked 36 mins ago
Alfred Mathew
211
New contributor
Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In the expression for work done by a gas integral(Pdv) aren't we supposed to use internal pressure?
Moreover work done by gas is the work done by the force exerted by th gas,but everywhere i find people using external pressure instead of internal pressure
thermodynamics
thermodynamics
asked 36 mins ago
Alfred Mathew
211
New contributor
Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 36 mins ago
Alfred Mathew
211
New contributor
Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 36 mins ago
Alfred Mathew
211
New contributor
Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 36 mins ago
Alfred Mathew
211
asked 36 mins ago
Alfred Mathew
211
211
New contributor
Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alfred Mathew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In equilibrium, the pressure exerted by a gas on the walls ("internal pressure") is equal to the pressure exerted by the walls on the gas ("external pressure"). Thermodynamics deals with transitions between a series of equilibrium states.
– probably_someone
30 mins ago
@probably_someone The issue might be that if the pressure is changing, then it can't always be equal to the external pressure.
– Aaron Stevens
28 mins ago
@probably_someone external pressure is equal to internal pressure if the piston is massless..but that is not true if the piston has sufficient mass
– Alfred Mathew
25 mins ago
@probably_someone moreover in a thermodynamic process equilibrium need not be there in the intermediate stages
– Alfred Mathew
23 mins ago
add a comment |Â
In equilibrium, the pressure exerted by a gas on the walls ("internal pressure") is equal to the pressure exerted by the walls on the gas ("external pressure"). Thermodynamics deals with transitions between a series of equilibrium states.
– probably_someone
30 mins ago
@probably_someone The issue might be that if the pressure is changing, then it can't always be equal to the external pressure.
– Aaron Stevens
28 mins ago
@probably_someone external pressure is equal to internal pressure if the piston is massless..but that is not true if the piston has sufficient mass
– Alfred Mathew
25 mins ago
@probably_someone moreover in a thermodynamic process equilibrium need not be there in the intermediate stages
– Alfred Mathew
23 mins ago
In equilibrium, the pressure exerted by a gas on the walls ("internal pressure") is equal to the pressure exerted by the walls on the gas ("external pressure"). Thermodynamics deals with transitions between a series of equilibrium states.
– probably_someone
30 mins ago
In equilibrium, the pressure exerted by a gas on the walls ("internal pressure") is equal to the pressure exerted by the walls on the gas ("external pressure"). Thermodynamics deals with transitions between a series of equilibrium states.
– probably_someone
30 mins ago
@probably_someone The issue might be that if the pressure is changing, then it can't always be equal to the external pressure.
– Aaron Stevens
28 mins ago
@probably_someone The issue might be that if the pressure is changing, then it can't always be equal to the external pressure.
– Aaron Stevens
28 mins ago
@probably_someone external pressure is equal to internal pressure if the piston is massless..but that is not true if the piston has sufficient mass
– Alfred Mathew
25 mins ago
@probably_someone external pressure is equal to internal pressure if the piston is massless..but that is not true if the piston has sufficient mass
– Alfred Mathew
25 mins ago
@probably_someone moreover in a thermodynamic process equilibrium need not be there in the intermediate stages
– Alfred Mathew
23 mins ago
@probably_someone moreover in a thermodynamic process equilibrium need not be there in the intermediate stages
– Alfred Mathew
23 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
The work done by an expanding gas is the energy transferred to its surroundings. In effect, as the gas expands it is compressing its surroundings so the work done is the force exerted on the surroundings (i.e. the pressure of the surroundings times the area) times the distance moved.
The extreme case of this is a Joule expansion where a gas expands into a vacuum i.e. the pressure of the surroundings is zero. In this case the expanding gas does no work regardless of the initial pressure of the gas.
answered 26 mins ago
John Rennie
265k41518766
force exerted on the surroundings is pressure exerted by the gas times the area,not external pressure
– Alfred Mathew
21 mins ago
1
@AlfredMathew imagine you're compressing a spring with your hand. No matter how hard (how much force) you push with your hand, it's the $kx^2$ force from the spring that determines the work/energy that gets stored in it.
– John Forkosh
17 mins ago
@John Forkosh That is because the spring is massless..and by newtons second law there must be no net force,hence the force we apply is equal to the force exerted by the spring
– Alfred Mathew
5 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The work done by an expanding gas is the energy transferred to its surroundings. In effect, as the gas expands it is compressing its surroundings so the work done is the force exerted on the surroundings (i.e. the pressure of the surroundings times the area) times the distance moved.
The extreme case of this is a Joule expansion where a gas expands into a vacuum i.e. the pressure of the surroundings is zero. In this case the expanding gas does no work regardless of the initial pressure of the gas.
answered 26 mins ago
John Rennie
265k41518766
force exerted on the surroundings is pressure exerted by the gas times the area,not external pressure
– Alfred Mathew
21 mins ago
1
@AlfredMathew imagine you're compressing a spring with your hand. No matter how hard (how much force) you push with your hand, it's the $kx^2$ force from the spring that determines the work/energy that gets stored in it.
– John Forkosh
17 mins ago
@John Forkosh That is because the spring is massless..and by newtons second law there must be no net force,hence the force we apply is equal to the force exerted by the spring
– Alfred Mathew
5 mins ago
add a comment |Â
up vote
5
down vote
The work done by an expanding gas is the energy transferred to its surroundings. In effect, as the gas expands it is compressing its surroundings so the work done is the force exerted on the surroundings (i.e. the pressure of the surroundings times the area) times the distance moved.
The extreme case of this is a Joule expansion where a gas expands into a vacuum i.e. the pressure of the surroundings is zero. In this case the expanding gas does no work regardless of the initial pressure of the gas.
answered 26 mins ago
John Rennie
265k41518766
force exerted on the surroundings is pressure exerted by the gas times the area,not external pressure
– Alfred Mathew
21 mins ago
1
@AlfredMathew imagine you're compressing a spring with your hand. No matter how hard (how much force) you push with your hand, it's the $kx^2$ force from the spring that determines the work/energy that gets stored in it.
– John Forkosh
17 mins ago
@John Forkosh That is because the spring is massless..and by newtons second law there must be no net force,hence the force we apply is equal to the force exerted by the spring
– Alfred Mathew
5 mins ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The work done by an expanding gas is the energy transferred to its surroundings. In effect, as the gas expands it is compressing its surroundings so the work done is the force exerted on the surroundings (i.e. the pressure of the surroundings times the area) times the distance moved.
The extreme case of this is a Joule expansion where a gas expands into a vacuum i.e. the pressure of the surroundings is zero. In this case the expanding gas does no work regardless of the initial pressure of the gas.
answered 26 mins ago
John Rennie
265k41518766
The work done by an expanding gas is the energy transferred to its surroundings. In effect, as the gas expands it is compressing its surroundings so the work done is the force exerted on the surroundings (i.e. the pressure of the surroundings times the area) times the distance moved.
The extreme case of this is a Joule expansion where a gas expands into a vacuum i.e. the pressure of the surroundings is zero. In this case the expanding gas does no work regardless of the initial pressure of the gas.
answered 26 mins ago
John Rennie
265k41518766
answered 26 mins ago
John Rennie
265k41518766
answered 26 mins ago
John Rennie
265k41518766
answered 26 mins ago
John Rennie
265k41518766
265k41518766
force exerted on the surroundings is pressure exerted by the gas times the area,not external pressure
– Alfred Mathew
21 mins ago
1
@AlfredMathew imagine you're compressing a spring with your hand. No matter how hard (how much force) you push with your hand, it's the $kx^2$ force from the spring that determines the work/energy that gets stored in it.
– John Forkosh
17 mins ago
@John Forkosh That is because the spring is massless..and by newtons second law there must be no net force,hence the force we apply is equal to the force exerted by the spring
– Alfred Mathew
5 mins ago
add a comment |Â
force exerted on the surroundings is pressure exerted by the gas times the area,not external pressure
– Alfred Mathew
21 mins ago
1
@AlfredMathew imagine you're compressing a spring with your hand. No matter how hard (how much force) you push with your hand, it's the $kx^2$ force from the spring that determines the work/energy that gets stored in it.
– John Forkosh
17 mins ago
@John Forkosh That is because the spring is massless..and by newtons second law there must be no net force,hence the force we apply is equal to the force exerted by the spring
– Alfred Mathew
5 mins ago
force exerted on the surroundings is pressure exerted by the gas times the area,not external pressure
– Alfred Mathew
21 mins ago
force exerted on the surroundings is pressure exerted by the gas times the area,not external pressure
– Alfred Mathew
21 mins ago
1
1
@AlfredMathew imagine you're compressing a spring with your hand. No matter how hard (how much force) you push with your hand, it's the $kx^2$ force from the spring that determines the work/energy that gets stored in it.
– John Forkosh
17 mins ago
@AlfredMathew imagine you're compressing a spring with your hand. No matter how hard (how much force) you push with your hand, it's the $kx^2$ force from the spring that determines the work/energy that gets stored in it.
– John Forkosh
17 mins ago
@John Forkosh That is because the spring is massless..and by newtons second law there must be no net force,hence the force we apply is equal to the force exerted by the spring
– Alfred Mathew
5 mins ago
@John Forkosh That is because the spring is massless..and by newtons second law there must be no net force,hence the force we apply is equal to the force exerted by the spring
– Alfred Mathew
5 mins ago
add a comment |Â
Alfred Mathew is a new contributor. Be nice, and check out our Code of Conduct.
Alfred Mathew is a new contributor. Be nice, and check out our Code of Conduct.
Alfred Mathew is a new contributor. Be nice, and check out our Code of Conduct.
Alfred Mathew is a new contributor. Be nice, and check out our Code of Conduct.
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In equilibrium, the pressure exerted by a gas on the walls ("internal pressure") is equal to the pressure exerted by the walls on the gas ("external pressure"). Thermodynamics deals with transitions between a series of equilibrium states.
– probably_someone
30 mins ago
@probably_someone The issue might be that if the pressure is changing, then it can't always be equal to the external pressure.
– Aaron Stevens
28 mins ago
@probably_someone external pressure is equal to internal pressure if the piston is massless..but that is not true if the piston has sufficient mass
– Alfred Mathew
25 mins ago
@probably_someone moreover in a thermodynamic process equilibrium need not be there in the intermediate stages
– Alfred Mathew
23 mins ago