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Thread: concept of work and reversible gas expansion

  1. #1 concept of work and reversible gas expansion 
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    Hello

    I have been reviewing the physics concept of work last night in order to be able to understand the concept of the expansion of a gas in thermodynamics. Something odd came up though. Work - The Physics Hypertextbook This website mentions that positive work is done in raising a textbook up in the air at a constant velocity. I thought that the force applying work had to be the net force on the system?!?! Or are they taking into account the force of gravity and the force of your hand both doing work on it ? Then wouldn't the total work be zero? I'm confused by this.

    This leads me to adiabatic gas expansion. An example I see quite often is the case of a cylinder filled with gas with a piston loaded with weight. The system is in equilibrium. However when calculations are done to determine the work done by the gas on the surroundings during an expansion...they use the force exerted by the surroundings on the cylinder system. For me that calculation only makes sense for a reversible reaction. The reason I bring it up is because when all of the weights are taken off at once (irreversible expansion). The work that it can do is at a minimum and it is calculated by also using the external pressure of the surroundings. This doesn't make sense to me. It's not a reversible expansion so shouldn't the pressure in the work energy formula be the internal pressure of the gas? The gas itself is doing work on the surroundings.

    Thank you for your help.


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  3. #2  
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    actually now that i think about it. Wouldn't you still have maximum work applied too if you still removed all of the weights but used the internal pressure in the work formula? The gas would be applying it's force on the surroundings as it expands until it reaches equilibrium.


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  4. #3  
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    Work is the energy transferred by a force acting through a distance in the direction of the force. Since the gravitational force is acting opposite to the direction of motion, we do not say it does work. However, the total energy is zero (conservation of energy).

    In an adiabatic gas expansion, we calculate the energy of a system. The only way the system can gain or lose energy is by interacting with its surroundings, in this case by doing work since there is no loss of heat in an adiabatic process. Therefore you can't use the internal pressure of the gas to calculate work done on the outside world.
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    Thank you for your reply. So that website is wrong? It has the system doing positive work on the book...the person's hand that is.

    I still don't get the explanation of why you can't use the internal pressure to calculate work done on the surroundings.
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  6. #5  
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    Quote Originally Posted by Marwyn View Post
    Hello

    I have been reviewing the physics concept of work last night in order to be able to understand the concept of the expansion of a gas in thermodynamics. Something odd came up though. Work - The Physics Hypertextbook This website mentions that positive work is done in raising a textbook up in the air at a constant velocity. I thought that the force applying work had to be the net force on the system?!?! Or are they taking into account the force of gravity and the force of your hand both doing work on it ? Then wouldn't the total work be zero? I'm confused by this.
    Gravity is what is known as a conservative force. The energy of an object in a gravity field is equal to the objects position in the field (potential energy) and its energy of movement (kinetic energy). If you drop a weight, it changes its position while at the same time gaining velocity. Potential energy is converted to kinetic energy and the total energy of the weight remains unchanged.

    Now assume that you are raising the weight with your hand at a constant speed. In order for the energy of the weight to keep its energy unchanged, it would have to lose velocity as it climbs exchanging kinetic energy for potential energy. The upward force of your hand prevents this and the kinetic energy of the weight remains constant. It is still climbing however and gaining potential energy in the process, so your hand is imparting energy to it. This is revealed if your hand were to, after lifting the weight, to drop it. As the weight falls, it gains kinetic energy, and when it returns to its original starting point it will have more kinetic energy than what it started with but will have the same potential energy that started with for a net gain of energy. (which was provided by the hand when it lifted the object.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


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  7. #6  
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    oooh ok. I understand it. Then why do they define work as using the net force and thusly the system undergoing acceleration if that is the case?

    Thank you guys so much for your help....Can anyone help me with the gas expansion part!??!
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    Quote Originally Posted by Marwyn View Post
    Thank you for your reply. So that website is wrong? It has the system doing positive work on the book...the person's hand that is.
    The website is not wrong. They said that it was negative work if the motion is opposite the force, which is different than what I said. I have no problem calling it negative work.
    I still don't get the explanation of why you can't use the internal pressure to calculate work done on the surroundings.
    Think of it like this. If you have a cylinder pressurized to 1000 pounds per square inch, and it weighs 100 pounds, would it be any more work for you to lift than an unpressurized cylinder that weighs 100 pounds? No, because the internal pressure does not affect the surroundings (i.e., you).
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    Quote Originally Posted by Harold14370 View Post
    Quote Originally Posted by Marwyn View Post
    Thank you for your reply. So that website is wrong? It has the system doing positive work on the book...the person's hand that is.
    The website is not wrong. They said that it was negative work if the motion is opposite the force, which is different than what I said. I have no problem calling it negative work.
    I still don't get the explanation of why you can't use the internal pressure to calculate work done on the surroundings.
    Think of it like this. If you have a cylinder pressurized to 1000 pounds per square inch, and it weighs 100 pounds, would it be any more work for you to lift than an unpressurized cylinder that weighs 100 pounds? No, because the internal pressure does not affect the surroundings (i.e., you).
    but it does affect the surroundings when it expands though doesn't it?
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    It affects its surrounding when it expands against air pressure. In an adiabatic free expansion Free expansion - Wikipedia, the free encyclopedia where the gas expands into an evacuated chamber, no work is done.
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