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Author Topic: Does the temperature increase when a gas decomposes?  (Read 5995 times)

Offline Freak

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One mole of certain gas is confined in a container. Suppose every molecule has decomposed and given rise to- two new molecules. Let's assume (for this purpose ) it requires negligible amount of energy for the molecule to decompose (or not any energy at all). So what will we observe?
The amount of matter has been doubled by the process, so pressure should increase- that's obvious. What happens to the temperature?
Does it stay the same OR increase?
Once I thought it should increase but the ideal gas formula seems to suggest the other.
Here
R/V=P/nT=constant (from PV=nRT)

when n increase P tends to increase and T tends to decrease to keep R/V a constant.
It is also possible that Temperature increase and so pressure increase -..But this time pressure should increase comparatively higher for it is pushed up by both increase in Pressure and Temperature.
Please help me
« Last Edit: 12/08/2008 22:59:40 by chris »


 

lyner

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Re: Does the temperature increase when a gas decomposes?
« Reply #1 on: 10/08/2008 18:09:11 »
Temperature corresponds to mean Kinetic Energy of the particles.
If each big particle splits into two, without significant energy transfer, the total KE of the two smaller particles will be the same - so the temperature will not change.
 

Offline Freak

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Re: Does the temperature increase when a gas decomposes?
« Reply #2 on: 11/08/2008 05:50:20 »
Paul B from yahoo answers community gave me the following solution:

At first, I thought that pressure should double and temperature should stay the same. WRONG! [But see added comments at end]

Back to first principles.

Delta E = q + w

You haven't changed the volume of the container, and you haven't supplied any heat energy. So q and w are 0 and DeltaE is 0.

So you now have twice as many molecules, but the same total energy. So the energy per molecule is halved.

Say for simplicity that all this energy is kinetic energy. You have now halved the kinetic energy per molecule, so you have halved the temperature.

P = nRT/V

You have doubled n but halved T, so P is unchanged.

Does this make sense? YES! Each molecule of A is moving just as fast as each molecule of A2 was the start with. But you have twice as many of them. So there will be twice as many collisions with the walls of the vessel, but each collision will only transfer half as much momentum, cancelling out.

Try this one out on your colleagues and teachers.

[Second thoughts: I erred in neglecting internal modes. Once these are included, there is no change in number of degrees of freedom. There will still be some reduction in temperature, for the kinds of reasons stated above, because splitting the molecule will replace one or more vibrational modes (which carry energy < 1/2 RT because of quantum effects) by translational or rotational modes. Only exception is if we imagine splitting the atoms of a monatomic gas without supplying energy, when my original analysis will still be valid. Wow!]
« Last Edit: 11/08/2008 05:53:03 by Freak »
 

Offline lightarrow

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Does the temperature increase when a gas decomposes?
« Reply #3 on: 13/08/2008 11:22:17 »
One mole of certain gas is confined in a container. Suppose every molecule has decomposed and given rise to- two new molecules. Let's assume (for this purpose ) it requires negligible amount of energy for the molecule to decompose (or not any energy at all). So what will we observe?
The amount of matter has been doubled by the process, so pressure should increase- that's obvious. What happens to the temperature?
Does it stay the same OR increase?
Once I thought it should increase but the ideal gas formula seems to suggest the other.
Here
R/V=P/nT=constant (from PV=nRT)

when n increase P tends to increase and T tends to decrease to keep R/V a constant.
It is also possible that Temperature increase and so pressure increase -..But this time pressure should increase comparatively higher for it is pushed up by both increase in Pressure and Temperature.
Please help me
PV =nRT
T doesn't vary, assuming negligible amount of energy for the molecule to decompose. What varies is n and P varies according to it.
 

Offline Freak

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Does the temperature increase when a gas decomposes?
« Reply #4 on: 13/08/2008 14:55:47 »
One mole of certain gas is confined in a container. Suppose every molecule has decomposed and given rise to- two new molecules. Let's assume (for this purpose ) it requires negligible amount of energy for the molecule to decompose (or not any energy at all). So what will we observe?
The amount of matter has been doubled by the process, so pressure should increase- that's obvious. What happens to the temperature?
Does it stay the same OR increase?
Once I thought it should increase but the ideal gas formula seems to suggest the other.
Here
R/V=P/nT=constant (from PV=nRT)

when n increase P tends to increase and T tends to decrease to keep R/V a constant.
It is also possible that Temperature increase and so pressure increase -..But this time pressure should increase comparatively higher for it is pushed up by both increase in Pressure and Temperature.
Please help me
PV =nRT
T doesn't vary, assuming negligible amount of energy for the molecule to decompose. What varies is n and P varies according to it.
Say specie A-A have energy 1/2(Ma+Ma)v^2= Ma*v^2 when the specie decompose it will have give rise to two A each having Energy 1/2Ma*v^2 which is half of the original species. As far as I know temperature is proportional to average kinetic energy per molecule in the system, so it is kinda straight forward to conclude that temperature will drop with the drop in energy.
I am expecting responses. Thx..
 

Offline lightarrow

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Does the temperature increase when a gas decomposes?
« Reply #5 on: 14/08/2008 19:16:06 »
One mole of certain gas is confined in a container. Suppose every molecule has decomposed and given rise to- two new molecules. Let's assume (for this purpose ) it requires negligible amount of energy for the molecule to decompose (or not any energy at all). So what will we observe?
The amount of matter has been doubled by the process, so pressure should increase- that's obvious. What happens to the temperature?
Does it stay the same OR increase?
Once I thought it should increase but the ideal gas formula seems to suggest the other.
Here
R/V=P/nT=constant (from PV=nRT)

when n increase P tends to increase and T tends to decrease to keep R/V a constant.
It is also possible that Temperature increase and so pressure increase -..But this time pressure should increase comparatively higher for it is pushed up by both increase in Pressure and Temperature.
Please help me
PV =nRT
T doesn't vary, assuming negligible amount of energy for the molecule to decompose. What varies is n and P varies according to it.
Say specie A-A have energy 1/2(Ma+Ma)v^2= Ma*v^2 when the specie decompose it will have give rise to two A each having Energy 1/2Ma*v^2 which is half of the original species. As far as I know temperature is proportional to average kinetic energy per molecule in the system, so it is kinda straight forward to conclude that temperature will drop with the drop in energy.
I am expecting responses. Thx..

I had a rethinking. The subject is less simple than I thought before, thanks to have pointed to it.

Let's assume that only rotational levels are present, in addition to the translational ones, in the undissociated molecule. Then the total energy of the molecules is:

E1 = (5/2)*NkT1

after decomposition we have a double number of molecules and translational levels only:

E2 = (3/2)*(2N)kT2

Since, for hypotesis, the internal energy of the gas doesn't change in the decomposition, we must have:

E1 = E2

and so:

(5/2)T1 = (3/2)*2 T2

--> T2 = (5/6)T1

and the final T is less than the initial one.

What about pressure?

P1V = n1RT1

P2V = n2RT2

since n2 = 2n1 and T2 = (5/6)T1, we have:

P2 = (5/3)P1

the pressure increases.


Now let's see what happens if, instead, we have at the beginning vibrational levels also:

E1 = (7/2)*NkT1

E2 = (3/2)*(2N)kT2

--> T2 = (7/6)T1

and the final T is now greater than the initial one.

The pressure:

P1V = n1RT1

P2V = n2RT2

since n2 = 2n1 and T2 = (7/6)T1, we have:

P2 = (7/3)P1

the pressure increases even more than in the previous case.


So it's not possible to establish a general rule: at low temperatures, the decomposition seems to decrease the temperature; the opposite at high temperatures. (Weird!)

The pressure, instead, always increases.
« Last Edit: 14/08/2008 19:36:04 by lightarrow »
 

Offline Freak

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Does the temperature increase when a gas decomposes?
« Reply #6 on: 16/08/2008 07:29:03 »
Thx Lightarrow .
I think it is this very reason for having different reaction enthalpy at different temperature.
Even if chemical bonds had no real strength, to cleave them there would always be some change in temperature. And it would be hard to determine whether change in temperature is due to energy redistribution in chemical bonds or in thermal energy in new set of species (this species have different s.p heat content so they exhibit different temperature having the same energy).
To get rid of this ambiguity I believe Bond dissociation energy is defined as the standard enthalpy change when a bond is cleaved by homolysis,with reactants and products of the homolysis reaction at 0 K (absolute zero) newbielink:http://en.wikipedia.org/wiki/Bond_dissociation_energy [nonactive].
« Last Edit: 18/08/2008 05:04:41 by Freak »
 

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Does the temperature increase when a gas decomposes?
« Reply #6 on: 16/08/2008 07:29:03 »

 

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