BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY 由压缩因子图查出Z=0.95 PV =ZnRT mo,-六-40gm=43m _202.70×103×40×103） m02)=n02)M(02)=(344.31×31.999g=11k TEL:010-64434903 4BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL：010-64434903 4 4.019 5043 20270 = = = C r p p p 由压缩因子图查出 Z = 0.95 pV = ZnRT mol 344.31mol 0.95 8.314 298.15 202.70 10 40 10 (O ) 3 3 2 = ÷ ÷ ø ö ç ç è æ ´ ´ ´ ´ ´ = = ZRT pV n m(O2 ) = n(O2 )M (O2 ) = (344.31´ 31.999)g = 11kg