In the process of taking a gas from state a to state c along the curved path sho

Question

In the process of taking a gas from state a to state c along the curved path shown in Fig. 15-29, 88 J of heat leave the system and 57 J of work are done on the system. Here is a summary of what is Q_a c= -88 J U_b - U_a = -10 J W_a c= -57 J P_a = 3.5 times P_d Determine the change in internal energy, U_c - U_a. When the gas is taken along the path abc, the work done by the gas is W = -125 J. How much heat Q is added to the gas in the process abc? If U_b - U_a = -10 J, what is Q for the process bc? If P_a = 3.5P_d, how much work is done by the gas in the process cda? What is Q for path cda?

Explanation / Answer

c) delta U= -10J W= -57J

Q=delta U- W = -10 J+ 57J = 47 J

d) W= Wad+ Wdc + Wac =2.5Pd V + Pd x deltaV +-57J

e) for any cyclic process there is no change in temperature since it return to original state.Thus no change in temperature indicate no change in internal inergy.

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