Entropy Change of Surroundings during Isothermal Expansion

To determine the change in entropy of the surroundings ($\Delta S_{\text{surr}}$), we apply the First Law of Thermodynamics and the principles of statistical thermodynamics to the given process.

1. Work done during irreversible isothermal expansion: For an ideal gas expanding against a constant external pressure ($P_{\text{ext}}$), the work done on the system is given by: $w = -P_{\text{ext}} (V_2 - V_1)$

Given:

• $P_{\text{ext}} = 3.0 \text{ atm}$
• $V_1 = 1.0 \text{ L}$
• $V_2 = 2.0 \text{ L}$

Substituting these values: $w = -3.0 \text{ atm} \times (2.0 \text{ L} - 1.0 \text{ L}) = -3.0 \text{ L atm}$

Using the conversion factor $1 \text{ L atm} = 101.3 \text{ J}$: $w = -3.0 \times 101.3 \text{ J} = -303.9 \text{ J}$

2. Heat transferred to the system: For an isothermal process of an ideal gas, the change in internal energy ($\Delta U$) is zero: $\Delta U = 0$

According to the First Law of Thermodynamics ($\Delta U = q_{\text{sys}} + w$): $q_{\text{sys}} = -w = -(-303.9 \text{ J}) = 303.9 \text{ J}$

3. Heat exchanged with the surroundings: Since heat absorbed by the system is supplied by the surroundings: $q_{\text{surr}} = -q_{\text{sys}} = -303.9 \text{ J}$

4. Entropy change of the surroundings: The surroundings remain at a constant temperature of $T = 300 \text{ K}$ throughout the process. Thus, the entropy change of the surroundings is calculated as: $\Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T} = \frac{-303.9 \text{ J}}{300 \text{ K}} = -1.013 \text{ J K}^{-1}$

Therefore, the correct option is (C).