北京化工大学：《物理化学》课程教学资源（双语练习题）第一章 气体英文习题及参考答案 gas problems

BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY 第一章气体英文司题及参考答素 1.A gas mixture consists of 320 mg of methane,175 mg of argon,and 225 mg of neon.The partial pressure of neon at 300 K is 66.5 Torr.Calculate (a)the volume and(b)the total pressure of the mixture. Solutions:(a)The volume occupied by each gas is the same,since each completely fills the container.Thus V=mNe)RT=0.225/20.18)x62.36x300 3.14L p(Ne) 66.5 (b)The total pressure is determined from the total amount of gas, n=n(CH4)+n(Ar)+n(Ne) nCH,)=0320 1604=1,95x102m0l 0.175 (Ar)= =4.38×10-3mol 39.95 nWe)=0225 =1.115×10-2mol 20.18 so n (total)=n(CH4)+n(Ar)+n(Ne)=3.548 x 102mol oal)=toaR7_3548x10-x6236x300-21270r V 3.14 2.A vessel of volume 22.4L contains 1.5 mol H2 and 2.5 mol N2 at 273.15 K.Calculate (a) the mole fractions of each component,(b)their partial pressures,and(c)their total pressure. Solutions:(a)Mole fractions are W2)=W)=2.5 =0.63 n(total)1.5+2.5 H2)=1-yN2)=1-0.63=0.37 (c)According to the perfect gas law p(total)V=n(total)RT s0poa0=Woa0RT-L-5+25)x8314x27315-4053Pa V 22.4×10-3 (b)The partial pressures are pN2)=p(total)yN2)=405.53×0.63=255.48kPa pH)=p(total)yH)=405.53×0.37=150.05kPa TEL:010-64434903

BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL：010-64434903 1 第一章 气体英文习题及参考答案 1.A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300 K is 66.5 Torr. Calculate (a) the volume and (b) the total pressure of the mixture. Solutions: (a) The volume occupied by each gas is the same,since each completely fills the container. Thus: L p Ne n Ne RT V 3.14 66.5 (0.225 20.18) 62.36 300 ( ) ( ) = ´ ´ = = (b) The total pressure is determined from the total amount of gas, n = n(CH4) + n(Ar) + n(Ne) n CH mol 2 4 1.995 10 16.04 0.320 ( ) - = = ´ n Ar mol 3 4.38 10 39.95 0.175 ( ) - = = ´ n Ne mol 2 1.115 10 20.18 0.225 ( ) - = = ´ so n (total) = n(CH4) + n(Ar) + n(Ne) = 3.548 ´ 10-2 mol Torr V n total RT p total 212 3.14 ( ) 3.548 10 62.36 300 ( ) 2 = ´ ´ ´ = = - 2. A vessel of volume 22.4L contains 1.5 mol H2 and 2.5 mol N2 at 273.15 K. Calculate (a) the mole fractions of each component, (b) their partial pressures, and (c) their total pressure. Solutions: (a) Mole fractions are 0.63 1.5 2.5 2.5 ( ) ( ) ( ) 2 2 = + = = n total n N y N y(H2) = 1 – y(N2) = 1 – 0.63 = 0.37 (c) According to the perfect gas law p(total) V = n(total) R T so kPa V n total RT p total 405.53 22.4 10 ( ) (1.5 2.5) 8.314 273.15 ( ) 3 = ´ + ´ ´ = = - (b) The partial pressures are p(N2) = p(total) y(N2) = 405.53 ´ 0.63 = 255.48 kPa p(H2) = p(total) y(H2) = 405.53 ´ 0.37 = 150.05 kPa