![]() ![]() Envisage the external pressure being a load resting on the piston.Imagine I have two pistons, two filled with different amount of gases at same volume and so different pressure.Now if I use this reasoning on expansion, Is diminished by an amount determined by the weight that is lowered,Īnd it is this energy that is transferred into the system.' 'when a gas is compressed, the ability of the surroundings to do work (b) The explanation given by Atkins' Physical Chemistry (same page) is: Based on Atkins' Physical Chemistry Ninth Edition page 50, we use p ext even for compression. (a) In the case of compression, the opposing force is obviously caused by p but not p ext. Using this reasoning, I cannot explain my following thought processes: Many of the texts and related questions on this site I encountered give a same general idea: because work done= opposing force x displacement. They could then use the intermediate measurements to calculate work, and then total those work values up to figure the total work done.For irreversible expansion and compression, why is d w= - p extd V but not d w= - pd V ? Those are measurements of intermediate states. Sometimes they might take measurements while things are happening. The delta values are taken at the beginning and end. A variation would be W = V(delta P), where V is volume, and delta P is the change in pressure. W stands for work, P is the pressure of the system (for gases), and delta V is the change in volume for the system. To figure out the total work done on, or by, a gas system, they use the formula W = P (delta)V. They look at the initial and final states. When scientists measure the work done on, or by, gases, they look at the system at the beginning and the end of the project. If you compress a balloon, you do work, and transfer energy to the air inside the balloon. The air does work, and transfers energy to the balloon. Because the inner surface of the balloon is flexible, that surface moves outward. If you heat a balloon (carefully), the molecules of air in the balloon gain energy and strike the inner walls of the balloon with greater force. The gas would then be doing work and transferring energy to the container. When a gas tries to expand, it exerts an increasing force on the surfaces of a container and may make those surfaces move. What else? Work is also linked to the expansion and compression of gases. We've already talked about moving objects. Work transfers energy from one object to another. If you lifted the brick again after your arm had rested, that would be work. No work was done if no movement happened. Even though you put forth a lot of effort to hold the brick up, did you do any work on the brick? Nope. Slowly, your arm gets tired, the brick feels heavier and heavier, and you finally have to stop to let your arm rest. ![]() Your arm is straight out in front of you and it's pretty tough to hold. Imagine that you are holding a brick above the ground. ![]() You have to exert a force AND move something to qualify as doing work. Driving to your job is not work because you just sit, but the energy your car engine uses to move the car does work. Your fingers are applying a force and moving the keys. Tapping on the keyboard and making the keys move is work. Sitting and looking at a computer screen is not work. A newton-meter is the same thing as a joule, so the units for work are the same as those for energy – joules. A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work. The work is calculated by multiplying the force by the amount of movement of an object ( W = F * d). Work is done when a force that is applied to an object moves that object. Is that work? To a physicist, only parts of it are. You might head off to your job one day, sit at a computer, and type away at the keys. ![]()
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