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How would Dalton's law be affected when there are two ideal gases in a container at different temperatures?

Let the gas with higher temperature be gas A and the gas with lower temperature be gas B. Then heat will be transferred from gas A to gas B due to which kinetic energy of the molecules of gas A will decrease and kinetic energy of molecules of gas B will increase. Hence, molecules of gas B should make more collisions to the walls of the container because of the increased kinetic energy and hence the total pressure exerted by gas B should increase and using the same argument, pressure exerted by gas A should decrease.

Now, we can no longer use P(A) + P(B)= P(T)

; P(A) is the partial pressure of gas A,

P(B) is the partial pressure of gas B

and P(T) is the total pressure exerted by the mixture of gas A and gas B

Am I correct with this logic?

A glass bulb of volume 400 ml is connected to another bulb of volume 200 mL by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 293 K and 1.00 atm. The larger bulb is immersed in steam at 373 K ; the smaller, in melting ice at 273 K. Find the final common pressure.

In the above problem, we cannot simply use PV= nRT to calculate final pressure because we don't know the final common temperature. So, we have to individually calculate the partial pressure of both the gases and then sum them up to get the total pressure. But according to me, as stated above, that should not be correct.

Is my reasoning correct or I am missing out on something? Also, how should I solve using my logic and please tell if there are other ways to go about solving this problem.

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  • $\begingroup$ According to the problem description, the two bulbs do not reach a common final temperature. The temperatures in the two bulbs are at 273 and 373 in the final state. Apparently, the bulbs are isolated from one another. So even though the pressures of the gas in the two bulbs are equal, the temperatures are not. $\endgroup$
    – Chet Miller
    Mar 7, 2016 at 16:26
  • $\begingroup$ When the stopcock is opened, moles of one gas will interact with moles of other gas, and because there is temperature difference, heat will flow and hence kinetic energy of the molecules of those gases will be affected. How will I take into account the pressure and temperature change at the same time? $\endgroup$
    – user23923
    Mar 7, 2016 at 16:39
  • $\begingroup$ In the end, when the system equilibrates, all the gas in one bulb will be at 273 and all the gas in the other bulb will be at 373. We are not concerned with what happens in between the initial state and the final state (in this example). Our focus is only on the final state. There will be negligible heat flowing through the tube of "negligible volume" at the end. You know the total number of moles and you know the two volumes and temperatures. This is enough information to find the final pressure and the distribution of gas moles between the two bulbs. $\endgroup$
    – Chet Miller
    Mar 7, 2016 at 17:09
  • $\begingroup$ The final state is when the stopcock is opened and the gases in individual bulbs interact, and we are required to find the final pressure corresponding to that state. According to you, what is the final temperature when the gases interact. $\endgroup$
    – user23923
    Mar 7, 2016 at 17:17
  • $\begingroup$ As I already said, they want you to assume that the final temperatures in the two bulbs are different. I don't know how to say it any other way. The temperatures do not equilibrate. $\endgroup$
    – Chet Miller
    Mar 7, 2016 at 17:26

3 Answers 3

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Determine total number of moles of air in bulbs: $$n=\frac{PV}{RT}=\frac{(1)(0.4+0.2)}{(0.082)(293)}=0.025$$

Consider final state of system. Let $P_F$ be the final equilibrium pressure, $n_1$ be the final number of moles in the larger bulb, and $n_2$ be the final number of moles in the smaller bulb: $$n_1=\frac{(P_F)(0.4)}{(0.082)(373)}=0.0131P_F$$ $$n_2=\frac{(P_F)(0.2)}{(0.082)(273)}=0.0089P_F$$ $$n=n_1+n_2=0.022P_F=0.025$$ $$P_F=1.14\tag{atmospheres}$$

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We can have two "gases" at different tempertures, neutrons can be considered to be gases.

If we consider a light water reactor then the neutrons formed by fission in the fuel pins are fast neutrons with high energies, they can be considered to be a very hot gas.

If they enter a medium where the temperture is much lower then when they strike atoms in the medium (such as water, graphite or even high pressure hydrogen gas) then exchange of energy occurs between the neutrons and the atoms which make up the medium.

If we choose a medium such as deutrium gas where the chance of capture is low (rather than something like boron trifluoride) then the neutrons will bang around from collision to collision and after a short while the translational energy of the neutrons will be lowered to the same level as gas molecules of at the same temperture.

Based on this observation from the operation of thermal nuclear reactors I strongly hold the view that as soon as you mix a hot gas with a cold gas that exchange of energy between the two gases will occur which results in the cold gas warming and the hot gas cooling to the same temperture.

While it is possible in some reactors to observe a change in the neutron energy spectrum as you change position in the reactor I can not think of an experiment which would allow you to do this with "normal" gases.

I hate to have to say it, but when a chemist spends 20 years of their life thinking about radiaoctivity and nuclear issues they start to see the world through a special lens.

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  • $\begingroup$ How does this answer the question? $\endgroup$
    – jimchmst
    Dec 14, 2022 at 4:32
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You cannot have two IDEAL gasses in a homogeneous mixture at different temperatures. As the collisions are stipulated to be non-elastic, all the kinetic energy of the particles (temperature) will be instantly transferred and spread out so that all the gas particles follow the Boltzmann distribution.

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  • $\begingroup$ Maybe not instant exchange of energy but it will be very fast, maybe you should consider the example of a beam of fusion neutrons (14 MeV) in a large tank of high pressure deutrium gas, or maybe C4D10 gas (perdeuterobutane). The neutrons can be considered to be gas (like noble gas atoms) which then hit the other atoms (which are parts of molecules). I am sure that if we have an infinite tank of C4D10 that the neutrons will soon be at thermal energies. $\endgroup$
    – Nuclear Chemist
    Dec 13, 2022 at 14:09

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