1. The problem statement, all variables and given/known data Problem: A sample of dry air has an initial pressure p1 = 1000 hPa and temperature T1 = 300K. It undergoes a process that takes it to a new pressure p2 = 500 hPa with unchanged temperature T2 = T1. Compute the mechanical work per unit mass perfromed by the sample under the following scenarios. a) Isochoric pressure reduction to p2 followed by isobaric expansion to final state. b) Isobaric expansion to final specific volume α2 followed by isochoric pressure reduction to final state. c) Isothermal expansion to final state. A pressure by specific volume graph may be helpful. In a nutshell the given data is as follows T1 = T2 = 300 K p1 = 1000 hPa p2 = 500 hPa -Gas constant Rd = 287 JK-1kg-1 2. Relevant equations pressure volume work equation: w = p*dα equation of state: pα=RdT 3. The attempt at a solution Part a) I started by breaking down the process listed in part a. I drew a pressure by specific volume graph starting first with moving the parcel under constant volume conditions to the new pressure and then moving it to the new specific volume under constant pressure. After this I looked at each piece as a leg in the journey of this air parcel and started figuring out the work involved in the first leg. ∫ dw = ∫ pdα w = 0 I arrived at this solution for the first leg because the parcel doesn't change in specific volume (due to it being isochoric) meaning that dα = 0 . . . so w = 0. The second leg involved changing the parcels specific volume under constant pressure. I applied the same idea a second time. ∫ dw = ∫ pdα The limits of integration are from α1 up to α2. w = p(α2 - α1) From this point I rearranged the equation of state and solved for α α = (RdT)/ p So. . . w = p2 ( ((RdT)/ p2)) - ((RdT)/ p1) ) plugging in the values w works out to be 43050 Jkg-1. Speaking with my professor today, he says that the work done should be negative? Why should it be negative? Since the parcel is expanding it is doing work on the environment, therefore the work is positive right? What am I missing? I calculated part B in almost exactly the same fashion but arrived at a -43050 Jkg-1. As for part C, the process I used was slightly different. ∫ dw = ∫ pdα Using the equation of state I substituted for p. ∫ dw = ∫ (RdT / α) dα ∫ dw = RdT ∫ 1/α dα Limits being α1 up to α2 w = RdT ( ln(α2 / α1)) I again used the equation of state to substitute for each of the α values w = RdT ( ln( (RdT/ p2) * (p1 / RdT) ) ) This simplifies down to . . . w = RdT ( ln(p1/p2) ) Plugging in the corresponding values w = 287 * 300 * (.693147181) w = 59679.97 Jkg-1 Why is this answer different than the first two? Is it because of the process involved or because of a calculation mistake on my part? Thank you for any help you can provide, please let me know if I wasn't clear enough on any of the above parts and I'll try my best to clarify.