# Tag Info

## Hot answers tagged ideal-gas

21

If gas A and gas B are of different density, then the situation sketched is not in equilibrium: the water level on the side of the light gas will be higher. There, the containers are moving down, and you have to push your containers through this net difference in level. You do need to put in energy here, which is probably the piece that you are trying to ...

13

The joule is the amount of energy needed to apply one newton of force for a distance of one meter: $$\rm J=N\cdot m=\frac{kg\,m^2}{s^2}\tag{1}$$ Where the 2nd equality comes from the definition of the newton (mass times acceleration): $\rm N=kg\,m/s^2$. The pascal is defined as one newton of force applied to a one-square-meter area: $$\rm ... 11 The fog you are seeing is condensation of atmospheric water, not sublimed CO_2. The water fog is made very near the boiling surface, and then sinks slowly, exactly as it does in rainclouds. Therefore, just because you can see fog gathering on the floor does not mean that the CO_2 is confined there. The CO_2 molecules have a speed, in random ... 10 The internal energy of an ideal gas is independent of volume when considered as a function of volume and temperature. If we choose to consider internal energy as a function of volume and some other thermodynamic variable we will find that the dependence of the energy on volume will change because we are keeping a different variable constant as volume is ... 10 The notation df denotes differential of function f. The differential df is a map $$df:\mathbb{R}\rightarrow \Omega^1(\mathbb{R})$$ where Ω1(ℝ) is the set of linear maps from ℝ to ℝ. The linear map corresponding to point p∈ℝ is often written as dfp $$df(p)=df_p$$ Note that in less formal settings p ... 10 If your an engineer and your looking for a material to spin your turbine, your goal should be to maximize pressure and minimize temperature. Well no. Temperature is generally a constraint of your heat source, and your goal is to maximize profit. That goal does, however, map to physical properties in logical ways, but it's much more complicated that ... 8 Chemical factors The more "localized" the electrons are the higher frequencies they naturally vibrate at (like a shorter guitar string playing a higher note). Gases must be simple, small molecules otherwise they would condense. Small molecules can't have electrons that are delocalized over many atoms. All substances have tightly localized electrons that let ... 7 According to a NASA page, the density in the middle of the Sun is about 150 g/cm3. That's about 9 × 1025 protons in a 1cm3 box, or 450 million to a side, and using that spacing for a voltage calculation reveals a typical interaction energy of 65 eV or so. (If you've never seen this unit before, that is the energy used by a 1V battery to move an electron's ... 7 This follows from the equipartition theorem. The equipartition theorem states that in thermal equilibrium, the average energy of each degree of freedom (each independent way the system can move) is k_B T/2, where T is the temperature and k_B (or just k) is called the Boltzmann constant. There are three independent directions in which a gas particle ... 7 It's a steady state. If there were a pressure gradient, there would be net force on the gas (ignoring gravity). There's no net force here because the air isn't accelerating. Thus the pressure is constant. The number density varies across the box inversely to the temperature so the ideal gas law holds. 7 The ideal gas law is derived from a model (the ideal gas), and like every other model it applies where it's underling assumptions are good approximations to reality. So, important assumptions for the idea gas law: Point particles In the ideal gas, the particles occupy no volume. A real gas in which the atoms of molecules occupy a vanishing fraction of the ... 6 dmckee gives some good qualitative considerations, but we can also develop rules for when the ideal gas law is and isn't appropriate. To start: The law applies perfectly in the case of a gas when P\rightarrow 0. The law does not apply to liquids. Between these two states is a gray area. In that case you should look at the compressibility factor, ... 6 According to the second law, thermal equilibrium between two systems means that they both have the same temperature T. The fact 2 that PV coincide whenever two gases are at thermal equilibrium (and, I assume, for the same n) means that PV is only a function of T. In othor words, there is a function g() such that$$\begin{align}PV&=g(T)& ...

6

The above equation solves for the average kinetic energy of a gaseous particle at a given temperature. k is known as Boltzman's constant, $k_B = 1.3806503 × 10^{-23}\frac{m^2kg}{s^2K}$ and is equal to the ideal gas constant divided by Avagadro's number, $\frac{R}{N_A}$. So where does the equation come from? The short answer: The equation above is ...

6

If a constant pressure of $1\,\rm Pa$ is exerted on a piston, and pushes it back so as to liberate a volume of $1\,\rm m^3$, then the work done by pressure on the piston amounts to $1\,\rm J$.

6

There's actually not one simple answer to your question, which is why you are a bit confused. To specify your problem fully, you must specify exactly how and whether the gas swaps heat with its surroundings and how or even whether it is compressed. You should always refer to the full gas law $P\,V=n\,R\,T$ when reasoning. Common situations that are ...

6

The potential energy for a diatomic molecule is not $$U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} |\vec{q}_1 - \vec{q}_2|^2$$ but is instead $$U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} (|\vec{q}_1 - \vec{q}_2| - r_0)^2,$$ where $r_0$ is the equilibrium bond distance. The important difference here is that in your version, any displacement of the vector ...

5

Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because $E=\frac ... 5 Giving the value simply of$k_B T$is generally more useful, because I can plug that into anything. Sure, I might need to know the ideal gas energy, and multiply by$3/2$. But maybe I need to put it into a partition function, and I just need$k_B T$. Maybe I'm worried about a harmonic oscillator and I just have the two degrees of freedom. The 3/2 is ... 5 As the comments to the question have stated, in real gasses ( contrasted to ideal gasses which just bounce around elastically) there exist both elastic and inelastic scatterings controlled by quantum mechanical interactions. Photons are generated leading to what we call Black Body radiation and an isolated gas volume will lose energy according to the ... 5 The pressure argument This would indeed be a straight-forward application of Gay-Lussac's law. In the case of the Patriots, we are told that the pressure was 2 psi less than the league minimum of 12.5 psi (select the "Ball" option in the first drop down menu, you'll see it clearly in Section 1 of Rule 2 on the first page). Gay-Lussac's law says that $$... 5 Prahar is correct that generally we have an energy contribution of {1 \over 2} kT per degree of freedom in a system - so that atoms in a gas of atoms (e.g. Helium) will have an average energy of {3 \over 2} kT. Often people talk about thermal energy being 'kT' because of the exponential expression in N_i = N_0 {g_i \over g_0} e^{-{E_i \over kT}} ... 5 The Equipartition theorem states that each degree of freedom has an average energy of 1/2KT. This is valid at large enough temperatures where quantum mechanics does not play a role. A = 3/2 KT(3 degrees of freesom) B = 3/2 KT C = 1/2 KT (1 vibrational degree of freedom in a di-atomic molecule) D = KT (2 axis of rotation, the third has very low Moment of ... 5 Photons are radiation so their equation of state is$$ p = \frac{\rho}{3} $$where \rho is the energy density. So we have$$ p = \frac{mc^2}{3V} = \frac{1\times 9\times 10^{16}\,\,{\rm J}}{0.003\,{\rm m}^3} = 3 \times 10^{19}\,\,{\rm Pa}$$It's a huge pressure. Not a surprising fact because the actual mass of photons we can produce is negligible. One ... 5 Pressure and volume have an inverse relationship when$n$and$T$are constant. How do you imagine the pressure in the balloon is increased? Either$n$or$T$must increase, or$V$must decrease. Additionally, balloons are roughly constant-pressure systems. The rubber membrane is a very weak elastic, so the internal pressure of the balloon is at almost ... 5 In addition to Bernhard's answer, just because three gases (Gas A,B and air - which is itself a mixture of nitrogen, oxygen, and other gases) have different densities, it does not mean they will remain seperated when in a container. In fact, as entropy of the system increases over time, Gas A, B and air will make an even (if heterogeneous) mixture. 5 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 ...

4

See the reference Physical, Chemical, and Biologic Properties of Tobacco, Cigarette Smoke, and other Tobacco Products. The reference considers 500mg of cigarette smoke, and says it is 478mg gas phase and 22mg liquid phase. The gas phase contains 350mg air, 50mg water vapor, 50mg carbon dioxide and 20mg carbon monoxide. The liquid phase contains 1.3mg ...

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