Isothermal compression of a gas and the associated entropy change.

  • Thread starter Matt15
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1 mole of an ideal gas is compressed isothermally from 1 to 10 bar at 298 K. Calculate the total entropy change associated with this process if it is carried out
1) Reversibly
2) Irreversibly




S (system) =nRln(vf/vi) for an isothermal process.


I think im getting a bit confused here. My first attempt was to use S = qrev/T.

For a reversible isothermal change, W = -nRTln(vf/vi) so w = - 1 * 8.31 * 298 *ln(1/10) = 5704 J. As it's isothermal internal energy must be 0 so q = -w. -> q= -5704

so the entropy change is given by -5704/298 = -19.14 J/k/mol. As this heat gets transferred to the surroundings the entropy change there is just 5704/298 = 19.14j/k/mol -> total entropy change is zero.

My main problem came when trying to do it for an irreversible process. If i use the formula S (system) =nRln(vf/vi) i get -19.14 j/k/mol. Which makes sense as entropy is a state function, but i have no idea how to calculate the entropy change for the surroundings if the process is irreversible.

Thanks in advance
 

Answers and Replies

  • #2
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I've had another go at the question. If i use the formula for the entropy change of a system s = nRln(vf/vi) and use it in the case of irreversible free expansion, w = 0 -> q=0 so entropy change in the surroundings = 0. Therefor total entropy change is just given by nRln(vf/vi) it's self. So that means the total entropy change of the irreversible isothermal compression is nRln(1/10) = -19.14 j/mol/k. This looks wrong though because the entropy is decreasing ?
 

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