Why are moles (quantity of a gas) and temperature inversely related? [closed]








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I'm learning about the combined gas law, and while doing calculations, I found that a doubling in the quantity of moles causes the temperature to be cut in half (all other things being equal).


I don't understand the mechanics of the change. Why does twice as much gas in the same amount of space cause the temperature to drop?


I would assume that there would be no change, since the total heat mass of the gas is unchanged. The particles in the gas are moved closer together, true, and more influenced by intermolecular bonds; but that seems to cut both ways, as a more crowded cluster of particles might produce more collisions? All in all, I understand the formula, but I don't get how the particles are interacting.


Also, is it true to say, mechanistically, that an increase in volume *produces* an increase in temperature? I don't understand how that would work, but it seems like a parallel to the example above (particles with less space to move have less energy; you can either shrink the container or crowd it, and reduce the temperature accordingly).


I understand how Charles' Law works in the other direction: increasing temperature gives the particles more energy, which causes them to collide harder and more frequently with their container, which will then expand.


But does merely making a container larger cause the gas within to rise in temperature? How would more space increase the total heat mass contained within the system? Where is the extra energy coming from?


A decrease in V causes an increase in P, and an increase in P causes an increase in T, but a decrease in V does *not* increase T?


I would appreciate your help very much, this is driving me a little crazy.


ps. Down voting questions is rather ungenerous. It's a question, not an assertion. I am confessing that I don't understand. If I know too little to ask the question well, forgive me. Your help is appreciated.




You have misunderstood the equation


$$pV = nRT$$


If we double the amount, then to keep in the fixed volume $V$ the initial pressure $p$, you have to set the half of the initial absolute temperature.


$$pV = (2n)R(\frac{T}{2})$$


Because for constant volume and temperature, it would have the double pressure:


$$(2p)V = (2n)RT$$


But that does not mean the doubling the amount decreases temperature.


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