Two moles of monatomic ideal gas is taken through a cyclic process shown on `P - T` diagram in Fig. Process `CA` is represented as `PT` = constant. If efficiency of given cyclic process is
`1 - (x)/(12 1n 2 + 15)`
then find `x`,

Explanation

For process `AB`,
`T_(A) = 300 K, T_(B) = 600 K`
`W = nR Delta T = nR (T_(B) - T_(A)) = 300 nT = 600 R`
`Q = nC_(P) Delta T = 2 xx (5)/(2) R (300) = 1500 R`
For process `BC`,
`W = nRT 1n (V_(f))/(V_(i)) = nRT 1n (P_(i))/(P_(f))`
`= nRT 1n 2 = 1200 1n 2`
For process `CA`,
`W = int PdV = int_(600)^(300) (K)/(T) (2 nRT)/(K) dT`
`= - 2 nR (300) = - 1200 R`
`Q = n C_(V) Delta T + W`
`= 2 xx (3)/(2) R (-300) - 1200 R`
`= - 900 R - 1200 R = - 2100 R`
`eta = (600 R + 1200 R 1n - 1200 R)/(1500 R + 1200 R 1n 2)`
`eta = 1 - (21)/(12 1n 2 + 15)`
Comparing with given equation, `x = 21`.