[orthoformate]

In this case, yes. More generally ΔS_surr = -Q_sys/T_surr
But I still disagree with your value of ΔV. How did you get it?

I made a calculation error. you are right. ΔV= 0.0156 m³

so W=1,013,250 Pa*0.0156 m³=15806.7 J

so ΔS= ∫PdV/T

so ΔS= 15806.7 J/T

how can we determine the temperature of the surroundings?

[mjc123]

I guess you're meant to assume it's the same as that of the gas, unless you are told otherwise. As the change is isothermal, and heat can flow between the gas and surroundings, we should assume they are in thermal equilibrium before and after the change (if not necessarily during).

[orthoformate]

I guess you're meant to assume it's the same as that of the gas, unless you are told otherwise. As the change is isothermal, and heat can flow between the gas and surroundings, we should assume they are in thermal equilibrium before and after the change (if not necessarily during).

Ah ok.

ΔS_surr=62.48 J/K
ΔS_sys=-4.3 J/K

ΔS_universe=58.18 J/K

Thank you for your help. This has been very enlightening.