Final answer:
In an isobaric process, the helium gas changes entropy. The final temperature of the gas can be calculated using the equation ΔS = Cp * ln(T-f/Ti). Given the initial temperature of 280 ∘C and a change in entropy of 40 J/K, the final temperature is approximately 290.6 ∘C.
Step-by-step explanation:
An isobaric process is a process that occurs at constant pressure. In this case, the helium gas undergoes an isobaric process where its entropy increases by 40 J/K. Entropy is a measure of the disorder or randomness of a system. In an isobaric process, the change in entropy can be calculated using the equation: ΔS = Cp * ln(T-f/Ti), where Cp is the molar heat capacity at constant pressure, T-f is the final temperature, and Ti is the initial temperature.
Given that the initial temperature is 280 ∘C, which is equivalent to 553 K, and the change in entropy is 40 J/K, we can rearrange the equation to solve for the final temperature: T-f = Ti * e^(ΔS / Cp). Using the molar heat capacity of helium gas at constant pressure, which is approximately 20.79 J/(mol*K), we can calculate the final temperature to be approximately 563.8 K or 290.6 ∘C.