1. Given data

• One mole of ideal gas
• Temperature: $T = 300\,\text{K}$
• Isothermal expansion from $V_1 = 1.0\,\text{L}$ to $V_2 = 2.0\,\text{L}$
• Constant external pressure: $P_{\text{ext}} = 3.0\,\text{atm}$
• Conversion: $1\,\text{L atm} = 101.3\,\text{J}$

We need to find the entropy change of surroundings, $\Delta S_{\text{surr}}$.

2. Find work done by the gas

For expansion against constant external pressure,

$w = -P_{\text{ext}}\Delta V$

Here,

$\Delta V = V_2 - V_1 = 2.0 - 1.0 = 1.0\,\text{L}$

So,

$w = -(3.0)(1.0) = -3.0\,\text{L atm}$

Converting into joules:

$w = -3.0 \times 101.3 = -303.9\,\text{J}$

3. Use first law of thermodynamics

For an ideal gas in an isothermal process,

$\Delta U = 0$

And,

$\Delta U = q + w$

Therefore,

$0 = q + (-303.9)$

$q = +303.9\,\text{J}$

This is the heat absorbed by the system.

4. Heat exchanged by surroundings

The surroundings lose the same amount of heat:

$q_{\text{surr}} = -303.9\,\text{J}$

Since the surroundings are at temperature $300\,\text{K}$,

$\Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T}$

Thus,

$\Delta S_{\text{surr}} = \frac{-303.9}{300} = -1.013\,\text{J K}^{-1}$

5. Final answer

$\boxed{\Delta S_{\text{surr}} = -1.013\,\text{J K}^{-1}}$

So the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.