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Lets consider I have two vessels of the same volume $V$ each of which contains ideal gas at different pressures and temperatures $(p_1,T_1)$, $(p_2,T_2)$. I pump certain amount of the gas $ n_\Delta$ from one vessel to the other. What is the work required for the pumping (assuming adiabatic process, without friction or other losses).

Is there some elegant solution or simple equation using only pressure, temperatue and molar amount?


Not elegant solution:

We can use well known formula for adiabiatic process

$p V^\gamma = const = K$

We calculate volume fraction of each gas before and after from

$pV=nRT$

so we calculate molar amounts in each vessel before pumping

$n_1 = p_1 V /( R T_1 )$

$n_2 = p_2 V /( R T_2 )$

and attribute volume fraction to each molar amount ($V \propto n$)

$V_1 = V(n_1-n_\Delta)/n_1 $ ... gas left in vessel 1

$V_2 = V$ ... gas already present in vessel 2

$\Delta V = V n_\Delta/n_1 $ ... gas pumped between vessels

Than we can calculate adiabatic compression/expansion for each volume independently $[V_1 \rightarrow V_1'; V_2 \rightarrow V_2'; V_\Delta \rightarrow V'_\Delta]$. To do so we have to determined final volume and pressure.

After the compression we have

$V_1' = V$

$V_2' + V'_\Delta = V $

Therefore

$V'_\Delta = V n_\Delta /(n_2+n_\Delta) $

$V_2' = V n_2/(n_2+n_\Delta) $

Now we know volume of each fraction of gas, so we can use $p V^\gamma = const$ and

$W = (p'V' - pV)/(1-\gamma) $

... but this starts to be rather complicated ... so I wonder if there is more elegant way

NOTE : In addition I'm not sure about entropy when the temperature of pumped gas after compression is different than the gas originally in vessel ( => mixing gas of different temperatures)

EDIT : To make is a sketch of the pumping translated to compression by imaginary piston

enter image description here

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  • $\begingroup$ It looks like you are allowing the volumes to change also. Did you really want to do that? It seems to me, the amount of work required is going to depend on the details of the pump geometry and operating characteristics. $\endgroup$ – Chet Miller Jan 5 at 13:28
  • $\begingroup$ 1) No the change of volume is only imaginary - I translate the problem of pumping into problem of moving wall (piston) of the vessel, Imagine you cut the vessel 1 somewere in the middle so that it forms compartment containing $\delta n$ moles of the gas, than you remove the original wall and move the new wall so that the volumes of the two vessels are again equal (prehaps picture is needed?) 2) This is theoretical question, with ideal gas etc. it should not depend on technical details $\endgroup$ – Prokop Hapala Jan 5 at 13:43
  • $\begingroup$ When you move the new wall, do you allow it to move on its own, or do you move it gradually. Is the new wall adiabatic, so that no heat can be exchanged between the two chambers? When you remove the original wall, does it simply disappear suddenly? $\endgroup$ – Chet Miller Jan 5 at 16:41
  • $\begingroup$ Yes, Move it adiabatically, it is basically piston of the pump. Yes, every wall is perfectly insulating (no heat transfer). ad disappear of original wall - that is the point of the second question (entropy due to mixing gas of different temperatures). But for simplicity let's just assume it does not disappear, or it disappears some time later when all work is already done. What I realized only now - the imprtaint condition is $p'_\Delta' = p_2'$ (that is after compression there is no pressure difference on blue wall). $\endgroup$ – Prokop Hapala Jan 5 at 18:16
  • $\begingroup$ I'm sorry. I'm not really able to understand your specification of this problem. It keeps changing. If you did it by the blue wall first disappearing and then the red wall being forced to move until the two volumes are equal again, I could provide an answer. $\endgroup$ – Chet Miller Jan 5 at 20:22
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OK. This is a two-step process. In the first step A the red wall is inserted (materializes) and the blue wall is removed (de-materializes). In the 2nd step B, the gas in the left compartment is caused to expand adiabatically and reversibly to the initial volume V while the gas in the then right compartment is caused to be compressed adiabatically and reversibly to the initial volume V.

Let the subscript A represent the conditions after step A and the subscript B represent the conditions after step B, and let the subscripts 1 and 2 represent the parameters for compartments 1 and 2.

STEP A

At the end of Step A, the number of moles in each of the two compartments is: $$n_{1A}=n_1-\Delta n$$ $$n_{2A}=n_2+\Delta n$$

The change in internal energy for the system is zero during step A. So, $$n_{1A}C_v(T_{1A}-T_{ref})+n_{2A}C_v(T_{2A}-T_{ref})=n_1C_v(T_1-T_{ref})+n_2C_v(T_2-T_{ref})$$where $T_{ref}$ is an arbitrary reference temperature. Since there is no change in the temperature in the left compartment during Step 1, $$T_{1A}=T_1$$ From these equations, it follows that $$T_{2A}=\frac{(\Delta n) T_1+n_2T_2}{\Delta n+n_2}$$ The final volumes and pressures in the two chambers after this step are: $$V_{1A}=V\left(1-\frac{\Delta n}{n_1}\right)$$$$V_{2A}=V\left(1+\frac{\Delta n}{n_1}\right)$$ $$P_{1A}=P_1=\frac{n_1RT_1}{V}$$ $$P_{2A}=\frac{n_{2A}RT_{2A}}{V_{2A}}$$

STEP B For the adiabatic reversible compression and expansion in Step B, $$V_{1B}=V_{2B}=V$$$$P_{1B}=B\left(\frac{V_{1A}}{V}\right)^{\gamma}=P\left(1-\frac{\Delta n}{n_1}\right)^{\gamma}$$ $$P_{2B}=P_{2A}\left(\frac{V_{1B}}{V}\right)^{\gamma}$$ I leave it up to you to get the work done on each gas and the net work done by the pump in this step.

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  • $\begingroup$ Thanks. This is what i would call not elegant solution of the problem. It is very similar (similarly complex) as my solution, just for modified problem where the wall is removed first. ... I'm still hoping there will be some simple formula at the end. $\endgroup$ – Prokop Hapala Jan 6 at 14:46
  • $\begingroup$ but, you are right that mixing the temperature first makes it easier, because you compress only two different chunks of gas, while in my cas I have to consider 3 chunks of gas separately. $\endgroup$ – Prokop Hapala Jan 6 at 14:51
  • $\begingroup$ Just to explain, in my original motivation (application) the ideal gas is not really a gas, but just approximation for dense hot compressible material in something like hypervelocity impact or implosion (incidently degenerate electron gas compress as ideal gas (physics.stackexchange.com/questions/522349/…)). => The compression is fast (therefore adiabatic) and heat transfer (temperature mixing) is perhaps much slower. $\endgroup$ – Prokop Hapala Jan 6 at 14:58
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    $\begingroup$ I think you are looking at an irreversible deformation in which transport processes and turbulence are important. Also, you should be focusing on the equation of state for your particular material. With irreversible transport processes involved, it is not going to be simple, and maybe you should be considering Computational Fluid Dynamics. $\endgroup$ – Chet Miller Jan 6 at 16:05
  • $\begingroup$ Sure, but I need some simple model which I can apply for each volume element (voxel) in finite-element simulation (that is my vessel - a cubic voxel ). There will be turbulence on large scale, but I assume I can ignore it within one voxel, if the voxel is small enough. Ad Equation of state - yes, possibly later, but for the moment ideal gas seems to be quite good model for hot dense plasma. Basically I wonto to make very simple and fast compressible CDF solver for a game. $\endgroup$ – Prokop Hapala Jan 6 at 17:46

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