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:

E

_{1} = (5/2)*NkT

_{1}after decomposition we have a double number of molecules and translational levels only:

E

_{2} = (3/2)*(2N)kT

_{2}Since, for hypotesis, the internal energy of the gas doesn't change in the decomposition, we must have:

E

_{1} = E

_{2}and so:

(5/2)T

_{1} = (3/2)*2 T

_{2}-->

**T**_{2} = (5/6)T_{1}and the final T is

*less* than the initial one.

What about pressure?

P

_{1}V = n

_{1}RT

_{1}P

_{2}V = n

_{2}RT

_{2}since n

_{2} = 2n

_{1} and T

_{2} = (5/6)T

_{1}, we have:

**P**_{2} = (5/3)P_{1}the pressure increases.

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

E

_{1} = (7/2)*NkT

_{1}E

_{2} = (3/2)*(2N)kT

_{2}-->

**T**_{2} = (7/6)T_{1}and the final T is now

*greater *than the initial one.

The pressure:

P

_{1}V = n

_{1}RT

_{1}P

_{2}V = n

_{2}RT

_{2}since n

_{2} = 2n

_{1} and T

_{2} = (7/6)T

_{1}, we have:

**P**_{2} = (7/3)P_{1}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.