# Tag Info

## New answers tagged equilibrium

0

I agree with you that most books do not follow a logical path when defining thermodynamics terms. Even great books such as Fermi's and Pauli's. The first thing you need to define is the concept of thermodynamic variables. Thermodynamic variables are macroscopic quantities whose values depend only on the current state of thermodynamic equilibrium of ...

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As far as I understand, the answer is: not only ergodicity but Poincare reccurence theorem "contradicts a bit" with the second law of thermodynamics. The point is that actually time that every ergodic system (for instance, Boltzmann billiard, as Sinai proved) is in some measurable part of full phase space of system is proportional to the phase volume of ...

10

You have to be careful to distinguish between microstates and macrostates. Thermodynamic equilibrium is a macrostate which consists of a mixture of all possible microstates of energy $E$ weighted by a Boltzmann weight $e^{- \beta E} / Z$. A state in macroscopic thermal equilibrium can be thought of as "moving through phase space" ergodically (i.e. the ...

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Think of an inflated balloon. Pressure inside the balloon is greater than atmospheric pressure. Entire system is in mechanical equilibrium because of tension provided by balloon surface. If you keep inflating the balloon, pressure inside the balloon keeps increasing, and so does tension in balloon surface, until it can take no more and bursts, thus ...

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A deodorant can contains a liquid hydrocarbon, typically a propane/butane mixture, and the pressure inside the can is due to the vapour pressure of this hydrocarbon. The pressure can be set to any desired value by varying the composition of the propellant - more propane makes a higher pressure while more butane makes a lower pressure. For a deodorant the ...

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Take gas in a container with a piston connecting to its surrounding as an example. When, at the moment you release the piston, the pressure difference between the surrounding and the container is very small, the piston will move slowly because there is not a lot force to accelerate it. In this situation, the system's response (temperature, pressure change) ...

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A thermodynamic process is called reversible if an infinitesimal change of the external condition reverses the process. Consider a gas enclosed by a freely moving piston in a cylinder. Let us say it is in mechanical equilibrium with the atmosphere, that is, the pressures on the piston match. If you increase the external pressure infinitesimally the piston ...

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In a water pipe you might have high pressure in one end and low pressure in the other. In other words this means that a force is pressing from one end, trying to make the water move to the other end - while another but smaller force is pushing from the other end. Like a strong man and a weaker man pushing on each other. Who will win? Will the water move ...

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Your substitution is almost correct. You can have multiple forces each supplying their own acceleration vector, but it doesn't make a lot of sense to include the mass in the sum because it would mean you have to distribute the mass among all the constituent forces just to add them again. $$\sum _{i}{\vec F_i}=\vec F_{total}$$ $$\vec F_{total}=m\vec a_{total}=... 0 Say \Sigma F=F_1+F_2+...=0. Then your substitution, which is mathematically correct, would physically imply the following: if F_1 alone had acted on the body then it would accelerate to a_1, if F_2 alone had acted on the body then it would accelerate to a_2, and so on. So F_1+F_2+...=0 implies a_1+a_2+...=0, while individual components ... 0 You cannot substitute the force \vec F with m\vec a. This is second law of Newton. The law talks about equivalency not sameness. Indeed we have \Sigma \vec F\equiv m\vec a. For substitution of the force, you should consider to the first law of Newton that defines the force. 2 You can't meaningfully make that substitution. Newton's second law (for constant mass) is actually:$$ \sum F = ma  In other words, the acceleration is proportional to the total force.

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I believe the issue is that the sketch doesn't show all forces acting on the bar. (My guess is that this is a question concerning only rotation about the y-axis) There must be forces at point A as well. The bar seems fixed at this point so that a torque will appear here to balance the others in order to keep no rotation. The torque 159 Nm comes from force \$...

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