Final answer:
The change in entropy for the compression and heating of a monoatomic perfect gas involves calculations using the formula for entropy change, considering both volume and temperature variations.
Step-by-step explanation:
The question asks about the change in entropy when two moles of a monoatomic perfect gas are compressed to half its initial volume and simultaneously heated to twice its initial temperature. In thermodynamics, entropy is a measure of the disorder or randomness of a system. When a gas is compressed, its volume decreases, which tends to decrease its entropy. However, when the temperature of the gas is simultaneously doubled, the increase in temperature leads to an increase in entropy.
Entropy change for an ideal gas can be calculated using the formula ΔS = nRln(V2/V1) + nCvln(T2/T1), where n is the number of moles, R is the universal gas constant, V1 and V2 are the initial and final volumes, T1 and T2 are the initial and final temperatures, and Cv is the molar heat capacity at constant volume for a monoatomic ideal gas. For a monoatomic ideal gas, Cv is given by (3/2)R. Thus, we need to calculate the entropy change considering both the compression and heating processes.