A thermodynamic sys-tem is taken from state ?a to state ?c in ?Fig. P19.38 along either path ?abc or path ?adc?. Along path ?abc?, the work W done by the system is 450 J. Along path ?adc?, ?W is 120 J. The internal energies of each of the four states shown in the figure are Ua = 150 J, Ub = 240 J, Uc = 680 J, and Ud = 330 J. Calculate the heat flow Q for each of the four processes ?ab?, ?bc?, ?ad?, and ?dc?. In each process, does the system absorb or liberate heat?

Solution 42P This question is based on the first law of thermodynamics, which is expressed mathematically as Q = U + W , where Q is the amount of heat added to a system, U is the change in internal energy of the system and W is the work done by the system. Along path abc, W) abc = 450 J From path a to b, the volume remains constant, so W) = 0…..(1) ab So, W) = 450 J …..(2) bc Along the path adc, W) = 120 J adc From path d to c, the volume remains constant, so W) dc= 0…..(3) So, W) ad = 120 J …..(4) Now, change in internal energy from a to b U ab= U bU a= 240 J 150 J = 90 J …..(5) Similarly. U bc= U cU b = 680 J 240 J = 440 J …..(6) U dc= 680 J 330 J = 350 J …..(7) U = 330 J 150 J = 180 J …..(8) ad Heat flow calculations: Heat flow for ab, From equations 1 and 5, Q) ab= U ab+ W ab = 90 J + 0 J = 90 J Heat flow for bc, From equations 2 and 6, Q) = U + W = 440 J + 450 J = 890 J bc bc bc Heat flow for dc, From equations 3 and 7, Q) dc= U dc+ W) dc= 350 J + 0 = 350 J Heat flow for ad, From equations (4) and (8), Q) = U + W) = 180 J + 120 J = 300 J ad ad ad Q value is positive for each path. Therefore, the system will absorb heat.