One mole of an ideal gas at 300K in thermal contact with surroundings expands isothermally from 1.0L to 2.0L against a constant pressure of 3.0atm. In this process, the change in entropy of surroundings (ΔSsurr) in JK⁻¹ is (1Latm=10³J):

ΔSsurr = Δq/T
For isothermal irreversible expansion against constant external pressure:
Δq = -w = PΔV
ΔV = 2.0L - 1.0L = 1.0L
P = 3.0 atm
Δq = 3.0 atm × 1.0 L = 3.0 Latm
Since 1 Latm = 10³ J
Δq = 3.0 × 10³ J
T = 300 K
ΔSsurr = (3.0 × 10³ J) / 300 K = 10 J/K = +10 JK⁻¹
However, this is not among the options. Let's check the calculation again.
The work done by the system is given by:
w = -PΔV = -3.0 atm × (2.0 L - 1.0 L) = -3.0 Latm = -3000 J
Since the process is isothermal, the heat absorbed by the surroundings is equal to the work done by the system:
q_surr = -w = 3000 J
The change in entropy of the surroundings is given by:
ΔS_surr = q_surr / T = 3000 J / 300 K = 10 J/K
There seems to be an issue with the provided options. The correct answer based on the provided data and standard thermodynamic principles is 10 J/K. None of the provided options match this result.