
#1
Aug1311, 05:30 PM

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There's something I'm not understanding about work done on a system with constant velocity. The total net work is zero right? Say two friends are pushing a block on opposite sides. One friend manages to dominate the other so they go in one direction. The floor is frictionless. The total work done on the block has to be zero right? The two boys are pushing in opposite directions with equal magnitudes of force. What about raising a book with constant velocity against gravity. The total work is zero again but the book is gaining potential energy. Does the energy come from the kinetic energy that the book has while it's being raised up?
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. 



#2
Aug1411, 09:40 AM

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Hi Marwyn, welcome to PF.




#3
Aug1411, 10:19 AM

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In the case of gravity, the total work done by gravity and you is zero. Gravity is a conservative force, so if it were the only force acting on the book, energy would be conserved: if the book had an initial upward velocity, it would lose kinetic energy as it flies up, gaining potential energy equal to the loss in kinetic energy. However, when you raise the book at constant velocity, the book does not lose kinetic energy. The book gained energy because you did work against gravity.
If a process is quasistatic, then the internal pressure is defined at all times, and you can use that to calculate work done. If it is not quasistatic, then the internal pressure isn't welldefined, and you have to calculate work done by some other means. 



#4
Aug1411, 04:19 PM

P: 3

work and expansion of a gasThat's what I was confused about with irreversible expansions. They simply used the external pressure to calculate the work done by the system on the surroundings as it expands. 



#5
Aug1411, 04:20 PM

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#6
Aug1611, 12:12 AM

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AM 



#7
Aug1611, 12:17 AM

P: 976

The problem with the external pressure being zero is that it requires one of two things to be true. Either it requires the internal pressure to also be zero (which means you either have no gas, or it is at absolute zero), which is obviously not very useful, or it requires the problem to be a transient condition (basically free expansion of a gas). Since most of basic thermodynamics is founded on the assumption that processes are effectively reversible and quasisteady state, this situation would basically force you to throw out the standard equations and use a far more complicated analysis.
(My thermo is a bit rusty though, so I'm not promising that this is 100% correct) 



#8
Aug1811, 12:43 PM

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