# Tag Info

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If a fluid system is experiencing a transient irreversible change, the temperature and pressure within the system will typically be non-uniform spatially. This is the result of two factors not present in (slow) reversible changes: the inertia (mass) of the fluid (which allows non-uniformities in velocity and density) and the viscous nature of the fluid (...

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For quasi-stationary processes you can take the time derivative, no problem. Apart from the "state equation", there is a "process equation", for example, for an adiabatic expansion $PV^{\gamma}=\rm{const}$ or so. It is also quasi-stationary. What you can loose in taking the time derivatives are quick processes like sound or shock wave propagation and their ...

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But for an ideal gas, internal energy is only a function of temperature and so internal energy remains constant here,no change in average kinetic energy of gas particles takes place, so where does the chaos come from to increase entropy of the system. 'Chaos' is not a very well defined term in context of statistical physics. It is not necessary to use it ...

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The chaos comes from by changing of volume or pressure of the system. The average kinetic energy doesn't change, but number of collisions increases (if pressure increase) or length of paths increases (if volume increases).

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I think there may be some confusion as to terminology: compressibility is defined as $$-\frac 1V \frac{\partial V}{\partial p}$$ where something like temperature or entropy is held constant, whereas the compressibility factor is defined as a certain ratio. (ref: https://en.wikipedia.org/wiki/Compressibility) In the ideal gas, particles do not interact ...

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The compressibility factor was originally derived from empirical testing of gases to correct for the observed non-ideal behavior at more extreme pressures and temperatures. Although it cannot be derived from first principles in the kinetic theory of gases, the experimentally derived factors can be reconciled using Van Der Waal's equation that deals with the ...

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If you isolate the balloon from everything else then after filling it with gas the balloon will soon attain the temperature of the gas. Then the balloon and gas will be in thermal equilibrium and no more heating/cooling will take place. Upon colliding with balloon skin the atom is not always lose energy but half the time it will gain energy (if the two ...

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Your balloon is not an isolated system, it can exchange heat with the environment as the "skin" material is not a perfect insulator. The expanding gas inside a balloon will cool as it expands, but the outside air will warm it up again after some time, the speed will depend on the balloon material and volume, but eventually the heat exchange between ...

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Assuming that your preliminary calculations are correct, you now have everything that you need: n, R and T. And unless you have been given a modulus of elasticity for the tyre, you must assume that it is rigid, so you have V as well (as a commentator hinted above). Just plug them into the ideal gas equation. If you know the modulus of elasticity of the tyre,...

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Your approach is almost correct! To see what's wrong, consider the case $d = 0$. Then for a small change in volume, the change in potential energy of the water is zero (since you're just moving water from the left side to water at the same height on the right side). But the gas has definitely done $p \, dV$ work. The mistake is that you've neglected the ...

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Your intuition that the same amount of fluid goes down and then up by the same amount is incomplete, you are forgetting what happens inside the fluid. It is easier to see using solid blocks as in the figure below: Here you can see that the effect of moving block 1 down is to shift block 2 to the right, and moving block 3 back up the same amount that ...

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Enthalpy change of an ideal gas is given by the formula $dH = n c_p dT$ when it undergoes a change in temperature $dT$. This formula is no longer valid when you undergo a phase transition : for a given quantity of water at a given temperature to vaporize requires a huge amount of enthalpy. See for example this wikipedia article. Also, as it was noted in ...

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A few clarifications on this thread in case anyone is reading in the future and is getting confused. I know however, that the temperature did in fact change, hence it's not an adiabatic process. That isn't really how an adiabatic process is defined. The temperature can change within a system (and often does) and it still be adiabatic. An adiabatic ...

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I believe that the answer can be made very simple. If the engine is reversible, its internal processes are, by definition of "reversibility", all quasistatic, regardless of the internal details of the machine. On the other hand, because the machine acts on the gas through a quasistatic process, then by definition this process must be reversible.

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Where am I going wrong? There is no liquid or solid that behaves like an ideal gas. So, there is no phase changing for any ideal gas. Ideal gas isn't a specific gas like oxygen, hydrogen, etc. It is an ideal behavior assumption. Liquid oxygen (for instance) isn't an ideal gas and so we cannot use the formula $\mathrm dh=C_p\mathrm dT$ because this formula ...

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Work is always force times displacement in the direction of the force. The only place where the gas is doing work is at the bottom surface that is moving downward. The force it is exerting there is $PA$, where $P$ is the gas pressure and $A$ is the cross sectional area of the tube. If the lower surface moves downward a differential distance dx, the work ...

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Does the fact that the engine exchanging heat with the gas is reversible imply that the trasformation of the ideal gas (no matter what trasformation is) is a reversible trasformation? Yes. A reversible engine is an engine that only performs reversible transformations. A reversible transformation in which two systems interact (the engine and the gas-...

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A heat engine is reversible if all processes related to working system in that are reversible. So, in this case, the heat exchanging between system and hot source (ideal gas) must be reversible. On the other hand, we know that In a reversible process, the entropy of the world maintains constant I.e. $$\mathrm dS_{\textrm{world}}=0$$ In a reversible ...

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Let me take a stab at this (I am not claiming this is correct... but it may guide your thinking). The "proper" definition of the Ohnesorge number is $$\rm{Oh} = \frac{\mu}{\sqrt{\rho\sigma L}}$$ If we multiply top and bottom by $v$ - a "characteristic velocity" and by $L^2$ - a "characteristic area" and rearrange a bit, we get \rm{Oh} = \frac{\mu L^2 \...

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The reason we talk about heat engines in thermodynamics is that they capture the most general thermodynamic process possible with two bodies. Every transformation can be summed up as "object A put in/took out this much heat, object B put in/took out this much heat, this much work came in/out", which is what happens if you stick a heat engine between them. ...

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Your impulse cannot be converted to pressure because you are considering only one collision. Impulse actually has units of force times time. And there are a few trillion collisions per second on any normal container, so you have to spread the impulses through time to get the proper force. The easiest way is probably to estimate the average time between two ...

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A gas can be approximated as an ideal gas. You then assume that the particles don't feel each other and that they are infinitely small. The potential is zero. The particles can only have kinetic energy. If you would make a simulation of such a system the particles can literally move through each other. If you make this ideal gas approximation, it is ...

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You can model the gas a a collection of hard spheres of some radius $r$, and do the correction relative to the limit as $r\to\infty$ perturbatively. What you find is that for a fixed pressure and temperature amd number of molecules the first order correction to the law replaces $V$ with $V-\frac{4n\pi r^3}{3}$. This tells you that the volume to use is the ...

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Entropy is function of state multiplicity. If you have the same gas, you wouldn't change state multiplicity and thus cannot change its entropy. That's different if two gases are different.

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What is meant by a “change in volume of a system”? "Change in volume of a system" means "change in volume of a system", not anything else. System is a hypothetical concept. There is no specified system before we define it. We ourselves choose and define system. When someone talks about a system defined by himself/herself, he/she talks about that system not ...

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By definition an ideal gas is a gas composed by non-interacting point particles moving randomly. It is not meant to represent a real gas, but a simplification of a gas with more complex interactions, such as vibrations and rotations. For example, most real gases at room temperature can be approximated by an ideal gas, since only translational degrees of ...

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An ideal gas particle is considered to be a non-interacting point mass i.e. a zero-dimensional object. As such it has no internal degrees of freedom - it cannot rotate or vibrate. The only degrees of freedom it has are the three translational ones.

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It's the volume that determines the pressure (for a given amount of gas at a given temperature), so if you consider a rectangular container, increasing the surface area of two opposite sides and/or increasing the distance between those sides will increase the volume and therefor decrease the pressure. But increasing the total area while keeping the volume ...

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The argument in the video seems flawed: it is true that the surface area has increased, but that is irrelevant, because when you calculate pressure you look at the number of molecules hitting a unit of area, which depends on the gas density, and here on the volume, but not on the total area. The derivation is made here: https://en.wikipedia.org/wiki/...

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As you increase the surface area you must bring parts of the surfaces closer together and this will increase the rate at which the surfaces are hit by the molecules which will compensate for the increase in area.

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