- #1

dRic2

Gold Member

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Consider an irreversible isothermal expansion of an ideal gas from state ##A## to state ##B## and suppose I know the amount of heat given to the system to perform the expansion - I'll simply call it ##Q##.

First of all I need to evaluate the change in entropy between the two states. In order do this I forget about the irreversibility and I evaluate the change in entropy like this:

$$ T dS = de + pdV $$

I suppose the process is a reversible thermal expansion so, at constant temperature the change in internal energy ##de## is zero:

$$ dS = \frac p T dV = \frac {nR} V dV $$

$$ \Delta S_{A→B} = nR \ln \frac {V_{B}} {V_{A}}$$

This result should be correct because the entropy is a function of state so it does not care about the path taken (reversible or irreversible) to go from ##A## to ##B## as long as the initial and final states are the same.

I also know that, for a closed system, the following is true:

$$dS = dS_e + dS_i$$

where ##dS_e## is the entropy received by the system from the surroundings and it is equal to ##Q/T## where ##T##, in our case, is a constant. ##dS_i## is the entropy generated inside the system due to irreversibility.

Integrating and re-arranging the above equation:

$$\Delta S_i = \Delta S_{A→B} - \Delta S_e = \frac {Q_{rev}} T - \frac Q T = nR \ln \frac {V_{B}} {V_{A}} - \frac Q T$$

Since the generation of entropy should be greater than zero then ##Q_{rev} - Q > 0 ## or ##Q_{rev} > Q## which is very strange to me.

Did I miss something ? Is It wrong ?