- #1
jaumzaum
- 434
- 33
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)NA) = R ln2
ΔS (oxygen) = k ln ((2V)/V)NA) = R ln2
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 ln2
ΔS (oxygen) = k ln ((2V)/V)NA) = R ln2
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?