Hi all. Trying to find work done for a gas... Here is the question: A quantity of air is taken from state a to state b along a path that is a straight line in the pv-diagram (figure is attached). If Va = 0.07, Vb = .11, Pa = 100,000 , Pb = 140,000. What is the work W done by the gas in this process? Assume that the gas may be treated as ideal. Ok i know the area under the graph is work. But instead of doing it that way can i use this way????: W = (integral of) p dV and since p is changing linearly: p = mV + b where b = 0 and m = (Pb - Pa)/(Vb - Va) = k so p = k* V substituting that back into the Work equation i get: W = (integral of) k * V dV => W = k * (integral of ) V dV => W = k/2 (Vb2 - Va2) ok so after all this i end up with 3600 J which is wrong (actual answer is 4800). Can any1 tell me what i'm doing wrong? or why this technique doesn't work? once again i know u can just find the area of the graph, but i want to know why this way is wrong.. Thanks all.