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## Main Question or Discussion Point

If we had an isolated system composed by two indentical containers joined by a valve (initially closed), the first one has 1 mol of hydrogen inside, pressure P, temperature T, the second one has 1 mol of oxygen, same pressure and temperature. At some moment we open the valve. What will be change in entropy? Will there be any change in temperature?

If weconsider the boltzmann equation for the entropy

ΔS (hydrogen) = k ln ((2V)/V)

ΔS (oxygen) = k ln ((2V)/V)

ΔS = 2 R ln2

But, if we consider the process to be isotermic:

ΔS (hydrogen) = R ln((2V/V)) = R ln2

ΔS (oxygen) = R ln((2V/V)) = R ln2

ΔS = 2 R ln2

So, if we consider it to be isotermic, the answer is the same, so we can conclude it is isotermic

But if it is isotermic, the ΔS we calculated is equal to Q/T = W/T, where W = RT ln(2)

So both oxygen and nytrogen do positive work. As the system is isolated: Q (system) = 0

Q (hydrogen) + Q (oxygen) = 0 -> W(H) + ΔU(H) + W(O) + ΔU(O) = 0

ΔU(H)+ΔU(O) = -(W(H) + W(O)) = -2RTln(2)

And so the temperature will reduce, violating what we had considered

What will happen so?

If weconsider the boltzmann equation for the entropy

ΔS (hydrogen) = k ln ((2V)/V)

^{NA}) = R ln2ΔS (oxygen) = k ln ((2V)/V)

^{NA}) = R ln2ΔS = 2 R ln2

But, if we consider the process to be isotermic:

ΔS (hydrogen) = R ln((2V/V)) = R ln2

ΔS (oxygen) = R ln((2V/V)) = R ln2

ΔS = 2 R ln2

So, if we consider it to be isotermic, the answer is the same, so we can conclude it is isotermic

But if it is isotermic, the ΔS we calculated is equal to Q/T = W/T, where W = RT ln(2)

So both oxygen and nytrogen do positive work. As the system is isolated: Q (system) = 0

Q (hydrogen) + Q (oxygen) = 0 -> W(H) + ΔU(H) + W(O) + ΔU(O) = 0

ΔU(H)+ΔU(O) = -(W(H) + W(O)) = -2RTln(2)

And so the temperature will reduce, violating what we had considered

What will happen so?