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In two containers we have Xenon. The volume of the first container is 2780 L, temperature is 300 degree Celsius, and n = 200 mol. Volume of the second container is 4000 L, temperature is 1300 degrees Celsius and n = 100 mol. Temperature in both of the containers is constant. How much Xe do we have to transfer from one container to another so that the pressure in both containers is the same?

I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.

After substracting this value from p1 I get the new pressure in container 1 which is 334898.5738 Pa which is the same as the pressure in container 2. From the new p1 I have calculated new n1 = 195.3798 mol, so I have to transfer 4.620 moles, or 606.593 grams to container 2.

My question is: When I take into the consideration the new n2 = 100 mol + 4.620 mol = 104.620 moles, and then I try to check if the p2 new is the same as p1 new I don'5 get the same result. p1 new = 334898.5738 Pa, and p2 new = 342086.14 Pa. Is my calculation wrong?

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    $\begingroup$ Using 10 valid digits is definitely wrong, if not backed up by the accuracy of this value. Surely is there pressure 334898.5738 Pa ? A methodical advice: Stay with symbolic algebra until you have the equation providing the final result. After that, plug in the numbers. $\endgroup$ – Poutnik Jan 31 at 9:57
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    $\begingroup$ For eventual writing and formatting of chemical formulas or equations, see how to use MathJax with mhchem extension $\endgroup$ – Poutnik Jan 31 at 10:19
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    $\begingroup$ Why the votes to close as homework?? The poster gave the problem and explained how the wrong solution was obtained. $\endgroup$ – MaxW Jan 31 at 18:13
  • $\begingroup$ user33683 - I think the tone about significant figures was bad. If you put all the calculations into a formula, then the calculator obviously carries a ridiculous number of significant digits throughout the calculation. My take is that you do want to avoid rounding errors in intermediate calculations. So figure out how many significant figures the answer should have, then carry two extra digits in the intermediate calculations, if you use them. Any more is just wasted effort. I think 3 digits for this problem, so your answer should have been 4.62 moles, or 607 grams. $\endgroup$ – MaxW Feb 1 at 0:18
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I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.

You cannot generally just average both pressures. As $pV=nRT$, then ( for $\mathrm{T}$ and $\mathrm{V}$ constant ) :

$V \cdot \mathrm{d}p=\mathrm{d}n \cdot RT$

$\frac{\mathrm{d}p}{\mathrm{d}n}=\frac{RT}V$.

So pressure averaging would work only in the special case if $\frac{T_1}{V_1}=\frac{T_2}{V_2}$. Therefore, a weighted average must be used.


I suppose the scenario is vaguely defined. It can be the polytropic expansion, or it may be the "locally isothermal" expansion. But for the former, it has like infinite number of possible solutions, depending on the rate of the transfer and of heat exchange.

I consider below the latter, i.e. isothermal conditions in each of containers.


We can consider the ideal behaviour, where

$$p=\frac{nRT}V$$

For equality of pressures, if $\Delta n$ is the moved molar amount of xenon from the container 2 to the container 1, the following equality must be true:

$$p=\frac{(n_1 + \Delta n)RT_1}{V_1}=\frac{(n_2 - \Delta n)RT_2}{V_2}$$

From that, we can easily calculate $\Delta n$:

$$(n_1 + \Delta n)RT_1V_2=(n_2 - \Delta n)RT_2V_1$$

$$\Delta n \cdot R(T_1V_2 + T_2V_1)=n_2RT_2V_1 - n_1RT_1V_2$$

$$\Delta n=\frac{n_2T_2V_1 - n_1T_1V_2}{T_1V_2 + T_2V_1}$$

Now you plug in the numbers here in the final formula, instead of working with literal numbers all the way:

In two containers we have Xenon. The volume of the first container is 2780 L, temperature is 300 degree Celsius, and n = 200 mol. Volume of the second container is 4000 L, temperature is 1300 degrees Celsius and n = 100 mol. Temperature in both of the containers is constant. How much Xe do we have to transfer from one container to another so that the pressure in both containers is the same?

$$\Delta n=\frac{(\pu{100 mol})(\pu{1573.15 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573.15 K})(\pu{4.000 m3})}{(\pu{573.15 K})(\pu{4.000 m3}) + (\pu{1573.15 K})(\pu{2.780 m3})}$$

In the sense below, if to be precise, we should round the temperature values for presented formula, because temperature was not known in 2 decimals accuracy ( otherwise there would be $\pu{300.00 ^{\circ}C}$, resp. $\pu{1300.00 ^{\circ}C}$), internally still computing with non rounded values.

$$\Delta n=\frac{(\pu{100 mol})(\pu{1573 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573 K})(\pu{4.000 m3})}{(\pu{573 K})(\pu{4.000 m3}) + (\pu{1573 K})(\pu{2.780 m3})} \approx \pu{-3.18 mol}$$

The negative value means the $\pu{3.18 mol}$ of xenon is to be moved from the first container to the second one, as we have formally assumed the opposite direction.


Internal full calculator precision of computation still makes good sense, to avoid artefacts of rounding errors. But for presentation of final or intermediate values of quantities from the real world, the values have to reflect error propagation rules. As input data and constants have limited accuracy.

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    $\begingroup$ I still have the feeling the end of the point(1) If you want to answer a homework question, we much prefer that you give a full explanation of the method used to reach the answer of hint-answers-revisited does not mean you are expected to enumerate numbers in the formula to get literal result. $\endgroup$ – Poutnik Feb 1 at 11:39
  • $\begingroup$ I don't think the differential equation $\frac{\mathrm{d}p}{\mathrm{d}n}=\frac{RT}V$ is the right one. $n$ is a variable too since gas is being transferred from one container to another. // Overall your point is correct. Don't expect averaging pressures to work. $\endgroup$ – MaxW Feb 1 at 12:32
  • $\begingroup$ @MaxW But that is the point. There is dn because of this transfer and RT/V is the rate of pressure drop dp/dn, because of this transfer. p and n are linearly proportional and RT/V is the proportionality constant. $\endgroup$ – Poutnik Feb 1 at 12:42

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