9.47 B

[Michelle Dong 1F]

Initially a sample of ideal gas at 323 K occupies 1.67 L at 4.94 atm. The gas is allowed to expand to 7.33 L by 2 pathways:
(b) isothermal, irreversible free expansion. Calculate deltaS total, deltaS system, and delta S surroundings for each pathway.

Why is delta S total not 0 J/K anymore? What are the concepts behind this?

[Deap Bhandal L1 S1J]

For a reversible path, work is done, so there is a q(surr) this means entropy change for the surroundings is not 0. In fact it is the exact opposite of the entropy change in the system so they cancel out and make total entropy change equal to 0. for the irreversible process no work is being done, q(surr) is 0, entropy change for the surroundings is 0, so the total entropy change is non-zero.

[Sara Varadharajulu]

delta s total will only be zero when the system is at equilibrium---for reversible isothermal expansion, it is like the system is at equilibrium, so you can assume delta s total is 0.
since part b talks about irreversible expansion, the system is not at equilibrium. this means that delta s total is a non-zero value. since delta s surroundings is 0 (bc no work is being done and change in internal energy is 0 since its isothermal expansion which means q=0 and q/T=0), delta s total = delta s system.