To determine the final temperature (T2) and change in entropy (∆S) for the given process of oxygen gas, we can use the first law of thermodynamics and the ideal gas equation. By calculating the number of moles of the gas, we can solve for T2 using the equation ∆U = Q - W = nCv(T2 - T1), and then determine ∆S using the equation ∆S = Cv ln(T2/T1) - R ln(p2/p1).
In this question, we are given information about the initial and final states of the oxygen gas and the values of work (W) and heat (Q) for the process. We are asked to determine the final temperature (T2) in oR and the change in entropy (∆S) in Btu/oR.
To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy (∆U) of a system is equal to the heat added to the system minus the work done by the system. Since the oxygen gas is behaving as an ideal gas, we can use the equation ∆U = Q - W = nCv(T2 - T1) to solve for T2, where n is the number of moles of the gas and Cv is the molar specific heat at constant volume. We can also determine ∆S by using the equation ∆S = Cv ln(T2/T1) - R ln(p2/p1), where R is the gas constant.
Given that one-quarter lbmol of oxygen gas is present, we can convert this to moles by multiplying by the conversion factor 1 lbmol/16 lb, resulting in 0.015625 moles. We can then calculate the value of T2 using the equation ∆U = Q - W = nCv(T2 - T1). Finally, we can calculate ∆S using the equation ∆S = Cv ln(T2/T1) - R ln(p2/p1).