Final answer:
The change in entropy (ΔS) for an ideal gas during a reversible adiabatic process is zero because no heat is exchanged with the surroundings. Hence, the total change in entropy (ΔS_total) is also zero for such a process.
Step-by-step explanation:
To calculate the change in entropy (ΔS) of 123 g of carbon monoxide (CO) during an adiabatic reversible expansion, one should use the thermodynamic definition of entropy. However, since no specific heat is provided for CO, and the problem implies an ideal gas behavior, usually entropy change for an ideal gas in an adiabatic process is zero because no heat is transferred into or out of the system (ΔQ = 0). In adiabatic processes, all expansion work is done at the expense of the internal energy of the gas, and since entropy is a function of heat exchange at a given temperature, the net change in entropy for the gas in such a process is zero.
For the second part of the question, computing the total entropy change ΔS_total during the expansion would typically require considering the surroundings as well. In a truly adiabatic and reversible expansion, there is no heat exchange with the surroundings, thus no change in the entropy of the surroundings, and the total entropy change would likewise be zero. In the real world or in irreversible processes, ΔS_total might be different, but these require additional information.