Final answer:
The change in entropy for scenario (a) is calculated using the formula for isothermal expansion, while for scenario (b), an adiabatic reversal expansion, the entropy change is zero.
Step-by-step explanation:
The question deals with the change in entropy of a nitrogen gas system under two specific scenarios. In part (a), the system is expanded isothermally and reversibly, which means the temperature remains constant, and we can use the formula \( \Delta S = nR\ln(\frac{V2}{V1}) \) where \(n\) is the number of moles of gas, \(R\) is the universal gas constant, and \(V2\) and \(V1\) are the final and initial volumes, respectively.
In part (b), the system is expanded adiabatically and reversibly, meaning there is no heat exchange with the surroundings. For an ideal gas undergoing a reversible adiabatic process, the entropy change is zero because the process is carried out without any heat transfer (\(\Delta Q = 0\)), thus using \(\Delta S = \frac{\Delta Q}{T}\), we find \(\Delta S = 0\).