(3)/b (M/m)R=914K Wh=(murThhn(/h=1.05x10J 净功W=W+Wn P(atm) =5.47×103J (4)Q=Qa+qb 6 △Eb+Wbe2 b =4.09×104J O V() 25 50 7=W/q1=13%W = Wbc +Wda 净功 M R PV T b a b ( / ) (3)  = 1.05 10 J 4 ( ln( / ) =  bc b Vc Vb W = M/)RT 5.47 10 J 3 =  6 2 o 25 50 P(atm) V(l) a b c d Q1 = Qab +Qbc (4) = DEab +Wbc 4.09 10 J 4 =  η = W/Q1 =13% = 914K