For the cycle in Fig. find the work output and the net heat transfer if the 0.1 kg of air is contained in a piston-cylinder arrangement.
Text Solution
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Sol. The temperature and V 3 are
T 1 = 
= 278.7 K T 2 = T 3 = 
= 2230 K
V 3 = 
= 
= 0.64 m 3 Using the definition of work for each process, we find
W 1–2 = 0 W 2–3 = mRT ln 
= (0.1)(0.287)(2230)ln 
= 133.1 kJ
W 3–1 = P(V 1 – V 3 ) = (100)(0.08 – 0.64) = – 56 kJ
The work output is then W net = 0 + 133.1 – 56.0 = 77.1 kJ. Since this is a complete cycle, the first law for a cycle provides us with
Q net = W net = 77.1 kJ
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