one mole of an ideal gas at 300k in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant presses of 3.0 atm. In this process. The change in entropy of surrroundings `(DeltaS)` in `J^(-1)` is
(1 L atm = 101.3 J)

Explanation

To solve the problem of finding the change in entropy of the surroundings (\( \Delta S \)) when one mole of an ideal gas expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm at a temperature of 300 K, we can follow these steps:

### Step 1: Understand the relationship between heat and work in an isothermal process
In an isothermal process, the change in internal energy (\( \Delta U \)) is zero for an ideal gas. Therefore, the heat added to the system (\( \Delta Q \)) is equal to the work done by the system (\( \Delta W \)):
\[ \Delta Q = \Delta W \]

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