# Tag Info

## New answers tagged ideal-gas

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This came to my mind after reading some introduction to maximum entropy probability distributions. Independence can be derived from the following four assumptions: (1) average momentum of particles inside the box is fixed at 0 (2) average kinetic energy of particles inside the box is fixed (3) the gas velocity distribution must be maximum entropy ...

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temperature is the measure of speed of molecules. when you compress the gas molecules start moving faster, which is the same as saying the temperature increases. why do molecules start moving faster? there are many ways of explaining this. here's one. when molecules are squeezed into a smaller volume their location is now more certain, it's locked in a ...

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you have an open system, i.e. the heat is removed from gas through its walls. when the gas cools down, it shrinks, so the piston will squeeze into the chamber. since you still apply the force, gas will heat up, and heat will be removed through walls again. while gas shrinks more it'll be more and more difficult to remove heat from walls, because of finite ...

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To increase the energy and temperature of the gas, you need a force and a displacement of the agent of the force, in this case the piston ($\mathrm{d}E = - p\,\mathrm{d}V$) A constant force with no displacement will not increase the energy and temperature of the gas. Any force, constant or not, with a displacement will increase the energy. The increase ...

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Here's an semi-formal explanation. Define $$f(N,V,T)=\frac{p(N,V,T)}{kT}.$$ While $f(N,V,T)$ is a function of $N,T$ and $V$, the variables $N$ and $V$ are partially redundant, and only the ratio $\rho=\frac{N}{V}$ is needed, since pressure is an intensive quantity. Thus we can write $$f(\rho,T)=\frac{p(\rho,T)}{kT}.$$ Every smooth multivariate function ...

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Compressibility factor comes from the virial expansion, any (monoatomic) gas can be study as an ideal gas with Z=1 but it's obviously just an approximation. The problem is that for the ideal gas law you assume that the particles (atoms) are punctiform without a proper volume. In the real gas model we have to correct volume and pressure because of the finite ...

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H = U + PV dH = dU +PdV + VdP In other words, equation 6 is missing the VdP term.

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Perfect vacuum can't be created. Even if you somehow get rid of all material particles, there still will be blackbody photons from the container, not to mention virtual gravitons. Generally, you can't be 100% sure that some part of space is perfect vacuum - to know that you should measure precisely the energy of that region, but it's forbidden by ...

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Given an ideal piston/cylinder, starting with the piston completely inserted and zero volume, the work to make a perfect vacuum is simply: (distance the piston moves) X (force) = (distance) X (area of the piston) X (exterior pressure). So the work to make a vacuum of volume V, is V X P, where P is the exterior pressure, such as atmospheric pressure. ...

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In quantum scale, particles are appearing and disappearing out of random everywhere all the time, meaning that if you are actually able to create a perfect vacuum at a macro scale, it would be instantly denied by the quantum scale. Also, in the quantum world, there's a small chance of random "teleportation" of any particle or atom from your container to any ...

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The answer is no, or at least it is in the classical vacuum sense. I also don't see a rationale for why creating a vacuum would require infinite energy. An explicit construction is to use a solid-phase reactive chemical "getter" to eliminate (nearly) all gas molecules present; in experimental practice, virtually all man-made materials still outgas ...

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Normally when compressing a gas the temperature increases. If you assume adiabatic compression, the law is $PV^\gamma=k$, where $\gamma=\frac {C_P}{C_V}$ is the ratio of specific heats and is usually about $1.4$ for air. Then, as shown here $\frac {T_2}{T_1}=\left(\frac {P_2}{P_1}\right)^{\gamma-\frac 1\gamma}$ This assumes you don't leak heat to the ...

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There are many problems : $1.$ As pointed out by Olin, gas cannot exist as a gas at $0 K$. $2.$ In ideal gases, interaction between molecules are absent. Hence, there is no potential energy. Remember that Potential energy always has an additive arbitrary constant. $3.$ As pointed by Wojciech, you would need (to take}energy to cool that ...

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Assuming the walls of the container are perfect insulators the final steady states must be indentical, as they are determined only by the gas's volume, internal energy and particle number, all of which are the same in both cases. (Internal energy is the same as no energy is transfered to the gas from outside.) I think the confusion arises because one part ...

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According to the equation of kinetic theory of gases, 'm' is the mass per a single molecule. NA*m = M(molar mass of molecule)

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You have to realize first that Charles' law is the change in volume with respect to temperature for constant pressure while Boyle's law is the change in volume with respect to pressure for constant temperature. So when you combine them, you need to account for these If I take a gas of volume $V_1$, pressure $P_1$ and temperature $T_1$ and let it change have ...

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Here is the definition of an ideal gas: An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly eleastic and in which there are no intermolecular attractive forces. One can visualize it as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other. In such a gas, all ...

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That means the gas molecules are spread out in a 2.4L volume. There's nothing between the air molecules themselves.

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After some thinking, I got the answer. When you decrease the ambient pressure from $p_i$ to $p_{atm}$ , the system goes out of equilibrium, and so the expansion of the gas is not quasi-static. So, all the state variables like pressure and volume aren't defined. So we can't use them for calculating the work done in definition 2.

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