Final answer:
The change in entropy of the surroundings during the isothermal reversible expansion of one mole of an ideal gas is calculated using the heat exchanged and the temperature of the surroundings. The correct answer, determined by the formula for entropy change, is -5 J/K.
Step-by-step explanation:
The student is asking about the change in entropy of the surroundings when one mole of an ideal gas undergoes an isothermal reversible expansion from 2.0 L to 3.0 L against a constant external pressure of 4 atm. Since the process is isothermal (temperature remains constant), we can use the formula:
\(\Delta S = \frac{q_{rev}}{T}\)
Where \(\Delta S\) is the change in entropy, \(q_{rev}\) is the heat exchanged during a reversible process, and \(T\) is the temperature in Kelvin. The heat exchanged by the surroundings is opposite in sign to the heat absorbed by the system. To find \(q_{rev}\), we calculate the work done by the gas during expansion, which is given by:
\(w = -P_{ext} \times \Delta V\)
Where \(w\) is work, \(P_{ext}\) is the external pressure in Pa (1 atm = 101.325 kPa), and \(\Delta V\) is the change in volume in cubic meters. Since this is an ideal gas, the relationship between pressure, volume, and moles is given by\(PV = nRT\), allowing us to simplify the expression to:
\(\Delta S = -\frac{nRT \ln(\frac{V_2}{V_1})}{T}\)
After converting the units and substituting the values, we can solve for \(\Delta S\) of the surroundings, which is equal to the negative value of the entropy change of the system since it's an isolated system and entropy is a state function.
Given this, the change in entropy of the surroundings (\(\Delta S\)) will be a negative value showing the decrease in disorder due to the heat released to the surroundings. The correct answer to this question is -5 J/K.