# Isothermal compression of a gas and the associated entropy change.

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.