- #1
oramao
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Consider a thermodynamic system of two independent variables, like a gas inside a recipient,
at equilibrium.
We are now going to carry out two different transformations at constant volume:
Transformation 1) A reversible process where an amount of heat Q is "given" to the system
Transformation 2) An irreversible process where the same amount of heat Q is "given" to the system.
In both transformations the change of Internal energy (E) is the same, once volume is constant for both (Q=change of Internal energy)
Therefore if we choose as independent variables, Internal Energy and volume, we obtain:
Transformation 1) Initial state E1 and V and final state E2 and V
Transformation 2) Initial state E1 and V and final state E2 and V
As there are only two independent variables, the final state of the system is the same for transfomations 1 and 2 and therefore the final entropy is also the same in both cases. This is not what the 2nd law tells us
at equilibrium.
We are now going to carry out two different transformations at constant volume:
Transformation 1) A reversible process where an amount of heat Q is "given" to the system
Transformation 2) An irreversible process where the same amount of heat Q is "given" to the system.
In both transformations the change of Internal energy (E) is the same, once volume is constant for both (Q=change of Internal energy)
Therefore if we choose as independent variables, Internal Energy and volume, we obtain:
Transformation 1) Initial state E1 and V and final state E2 and V
Transformation 2) Initial state E1 and V and final state E2 and V
As there are only two independent variables, the final state of the system is the same for transfomations 1 and 2 and therefore the final entropy is also the same in both cases. This is not what the 2nd law tells us