- #1
Marwyn
- 3
- 0
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.
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.