During process B→C, the system is neither expanding nor contracting. Thus, (c)W= 0. (d) The sign of ΔEintmust be the same (by the first law of thermodynamics) as that of Qwhich is given as positive. Thus, ΔEint> 0. During process C→A, the system is contracting. The environment is doing work on the system, which implies W<0. Also, ΔEint<0 because ¦ΔEint= 0 (for the whole cycle) and the other values of ΔEint(for the other processes) were positive. Therefore, Q= W+ ΔEintmust also be negative. (e)Q< 0. (f)W< 0. (g)ΔEint< 0. (h) The area of a triangle is 12(base)(height). Applying this to the figure, we find
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