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For commercial uses, usually these gases are in liquefied form .You can find more details, for LPG lets say , here: https://en.m.wikipedia.org/wiki/Liquefied_petroleum_gas In general you are right.


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The Ideal Gas Law PV=nRT explains what happens. It can be written as PV/T=nR. For our purposes nR can be considered a constant, PV&T are all expressed in absolute units. Thus if the container volume is constant, P/T is a constant.


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Apart from volume and temperature, there is one more parameter affecting pressure, that is number of molecules. At filling up the container you added more molecules in, consequntly you get higher pressure.


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This can be concluded by reviewing Gibbs equation for upstream and downstream stagnation conditions. $$T_0ds_0=dh_0-\frac 1{\rho_0}dP_0$$ Because across the shock wave is an adiabatic process, $dh_0=0$ Then Gibbs equation becomes $$ds_0=-\frac 1{\rho_0T_0}dP_0=-\frac {R}{P_0}dP_0$$ We know entropy increases. This leads to conclusion that the stagnant ...


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For a bit more empirical approach, try the compressibility factor. With this factor, the ideal gas equation becomes PV = znRT, where z is described in very great detail by https://en.wikipedia.org/wiki/Compressibility_factor. For places on the "z plot" that differ substantially from a value of z=1, you will find that argon starts behaving more and more ...


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The result is problematic that the compressibility factor is greater than 1. The well-known Van der Waals equation is, $$(P+\frac {n^2a}{V^2})(V-nb)=nRT$$ And ideal gas EOS is, $$P_{ideal}V=nRT$$ where n is molar number and R is universal gas constant. You can use particle number if you prefer that way. From VDW equation, we get $$ P+\frac ...


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The state of equilibrium is characterised by a minimum in free energy $F=U-TS$ (Helmholtz for simplicity), not a minimum in potential energy. What this means is that while the system is indeed attempting to minimise the potential energy $U$, it is simultaneously trying to maximise the entropy $S$. The balance, i.e. which term dominates, is determined by the ...


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First, the geometric extent of the (quantum mechanical) wave packet doesn't mean that the particle (atom in your case) has become diluted all over the volume like if it were fog. Instead, the right interpretation of the wave packet is in terms of probabilities. For example, assume that there is 1 atom in a box. The initial wave packet is nearly localized ...


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1) Your question states that, until the compressor is switched off, a constant pressure of $5\; atm$ is maintained in the box. This answers the 1st part of your question. (However, perhaps you mean that the compressor delivers a certain amount of gas per second while the 2nd hole is open, and you wish to know the pressure of the gas in the box when ...


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The derivation of ideal gas equation from Hamilton's equations will take the same procedure as what you have seen in Wikipedia. Since you said you haven't understood the way in which the equation is derived I will give you a step by step explanation on it. So we have a system of perfect gas molecules. Of course they are non-interacting. We are going to ...


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The piston moves to left if $P_2\gt P_1$ and to right if $P_2\lt P_1$. In above answers, Lucas's beginning part is fine. $$P_1V_1=m_1RT_1$$ $$P_2V_2=m_2RT_2$$ with $P_1=P_2$,$V_1=V_2$,$T_2=T_1-ΔT$ We then have, $$\frac{m_2}{m_1}=\frac{T_1}{T1-ΔT}$$ When temperature on both side is raised by $\delta T$ $$P′_1V′_1=m_1RT′_1$$ $$P′_2V′_2=m_2RT′_2$$ with ...


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Rigorous calculation The first hypothesis is hand-wavy Aristotelian logic. Whatever does happen has to be explainable by the Ideal Gas Law. At all times, $p_1 = p_2$ must hold, otherwise, the piston would move to equalize the pressures. So, $$\begin{eqnarray} p_1 =& p_2 \\ \frac{N_1 k_b T_1}{V_1} =& \frac{N_2 k_b T_2}{V_2} \\ \frac{V_2}{V_1} ...


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Let $T_{20}$ be the initial temperature of tank 2 and $T_{10}=T_{20}+\Delta T$ be the initial temperature in tank 1. Let $\delta T$ be the equal rise in the temperature of both thanks. Assuming that the piston does not move, we would have $$p_{2f}=p_2\frac{T_{20}+\delta T}{T_{20}}$$and$$p_{1f}=p_1\frac{T_{20}+\Delta T+\delta T}{T_{20}+\Delta T}$$ Since ...


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$$P_1V_1=m_1RT_1$$ $$P_2V_2=m_2RT_2$$ $P_1=P_2$,$V_1=V_2$,$T_1=T_2+\Delta T$ Then, we have:$$\frac{m_2}{m_1}=\frac{T_1}{T_1-\Delta T}$$ After heating: $$P'_1V'_1=m_1RT'_1$$ $$P'_2V'_2=m_2RT'_2$$ $P'_1=P'_2$,$T'_1=T'_2+\Delta T$ Then, we have:$$V'_2=V'_1\left(\frac{m_2(T'_1-\Delta T)}{m_1T'_1}\right)=V'_1\left(\frac{T_1}{T_1-\Delta ...


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Your second thesis is right. At least the result. The explanation not, as the example in the answer of Diracology might illustrate. Assume, that the piston does not move (we fix it). So $V$ didn't change on either side. Obviuosly, $n$ didn't change on either side. Therefore, $p$ is proportional to $T$ (in each side separately!). So if $T$ rises by the same ...


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I actually did some of the algebra work and using the equations given, I managed to show that the pressure from 1 is less than 2. The volume being held constant is important here. Just to note I'm purely using algebra here, so please check through my working and see if it makes sense to you. For notation sake, I've set anything with b as before the tube is ...


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The piston cannot move if the pressures are maintained equal (assuming same cross section are at both extremes). Otherwise you would violate the second Newton's law. The fact that the second container has more energy does not imply the piston should move. Imagine two containers filled with the same gas linked by a piston. Assume same pressure and same ...


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I think the idea is basic. In the experiment you say that "The Volume of the containers also stays constant." Which the tubes in the diagram are part of the containers, the piston can not move in order to keep the volume constant. Now if the piston could move, then the volume would change, considering the Ideal Gas Law, the piston would move to the left and ...



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