Solution Cont. 2)Properties at each state nRT 500mol×8.3145J/(mol.K)×273+25)K P= 5 V 2m3 3 =6.19393×103Pa=6.11atm Reversible adiabatic ideal gas n==k=1.67(monotomic gas) K-1 1.67-1 1MPa 1.67 =144.26K =-2）产 11.67 5.906m3 T=T3=144.26KV3=V=2m3 500mol×8.3145J/(mol·K×144.26K P3= =299855Pa 2m3 上游充通大 March 23,2018 14 SHANGHAI JIAO TONG UNIVERSITYMarch 23, 2018 14 1 1 1 nRT p V  3 3 2 3 1 T T V V     144.26 K 2 m 1 1.67 1 1.67 2 2 1 1 1MPa 298K 144.26K 6.11MPa p T T p                     1 1 1.67 1 3 3 2 1 2 6.11MPa 2m 5.906m 1MPa p V V p                    3 5 500 mol 8.314 5 J/(mol K) 273 25 K 2 m 6.193 93 10 Pa 6.11 atm         3 3 500 mol 8.314 5 J/(mol K) 144.26 K 299 855 Pa 2 m p      Reversible + adiabatic + ideal gas n k     1.67(monotomic gas) 2) Properties at each state Solution Cont. 5 3