Pressure, Volume and Temp Change...WORK done??? n = 9.95 moles of an ideal gas are slowly heated from initial pressure P1 = .922 atm and initial volume V1 = 679 L to final pressure Pf = 1.16 atm and final volume Vf = 1073 L. Find T1, the initial temperature of the gas. If the plot of this process on a P-V diagram is a straight line, find W, the work done on the system. P1 = .922 atm x 1.013 x 10^5 Pa = 93398.6 Pa V1 = 679 L / 1000 = .679 m^3 PV = nRT (93398.6)(.679) = (9.95)(8.314)(T) T = 767 K For the Work done I would just need to find the area under the curve for the P-V diagram...correct? Well that is what I did and my answer is considered wrong and I cannot understand why... When V = .679 m^3, P = 93398.6 Pa. And when V = 1.073 m^3, P = 117508 Pa. The area of the rectangle under the curve is (1.073 - .679) x 93398.6 = 36799.0484 J. The area of the triangle under the curve is 1/2bh = (.5)(1.073-.679)(117508 - 93398.6) = 4749.5518 J So W SHOULD = 36799.0484 + 4749.5518 = 41548.6002 J = 41.5 kJ...but this is considered WRONG....Can someone please advise what I am doing wrong?? Very frustrated here!!