BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY 第三章热力学第二定神美文日题及参考答兼 1.Calculate the increase in entropy when 3.50 mol of a nonatomic perfect gas with C 5R/2,is heated from 250K to 700K and simultaneously expanded from20.Lto60.L Solutions:However the change occurred,AS has the same value as if the change happened by reversible heating at constant volume(step 1)followed by reversible isothermal expansion(step 2) s0AS=△S1+△S2 For the first step A=nG.h2=30x83145xh7=48-K and for the second P s0△S=△S1+△S2=76.9JK 2 Consider a system consisting of 15 mol co(g)initially at 15C and 90 atm and confined toa eylinder of cr section 100.0cmThe sample is allowed to carbon dioxide may be considered a perfect gas with Cy=28.8 J-K.mol',and calculate Q,W. AU,ATand△s. Solutions:In adiabatic process,so Q=0 W=-D△V=.15×101×103×100×15×106=-227.3j 4U=Q+W=.227.3J nC15x288-53 AT= -227.3 P2 T1=15+273.15=288.15K T2=T1+△T=282.85K 5=mRT_15x008206x2815-39421 D 9.0 V2=V1+△V=3.942+100×15×103=5.442L S=C,nlh2+mRn 7 7=L5×288xn2828 28.15+1.5x8314m5442 3942=32K TEL010-64434903
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 1 第三章 热力学第二定律英文习题及参考答案 1.Calculate the increase in entropy when 3.50 mol of a nonatomic perfect gas with Cp,m = 5R/2, is heated from 250K to 700K and simultaneously expanded from 20.0 L to 60.0 L. Solutions: However the change occurred, DS has the same value as if the change happened by reversible heating at constant volume(step 1) followed by reversible isothermal expansion(step 2) so DS = DS1 + DS2 For the first step 1 1 2 1 , 44.9 250 700 8.3145 ln 2 3 ln 3.50 - D = = ´ ´ ´ = J × K T T S nCV m and for the second 1 1 2 2 1 2 32.0 20.0 60.0 ln ln 3.50 8.3145 ln - D = = = ´ ´ = J × K V V nR p p S nR so DS = DS1 + DS2 = 76.9 J×K -1 . 2. Consider a system consisting of 1.5 mol CO2(g), initially at 15°C and 9.0 atm and confined to a cylinder of cross-section 100.0 cm 2 . The sample is allowed to expand adiabatically against an external pressure of 1.5 atm until the piston has moved outwards through 15cm. Assume that carbon dioxide may be considered a perfect gas with CV,m = 28.8 J×K -1 ×mol-1 , and calculate Q, W, DU, DT and DS. Solutions: In adiabatic process, so Q = 0 W = -pex DV = - 1.5 ´ 1.01 ´ 105 ´ 100 ´ 15 ´ 10-6 = -227.3 J DU = Q + W = - 227.3 J K nC U T V m 5.3 1.5 28.8 227.3 , = - ´ - = D D = 1 2 1 2 , 2 1 1 2 , ln ln ln ln V V nR T T nC p p nR T T DS = nCV m + = V m + T1 = 15+ 273.15 = 288.15K T2 = T1 + DT = 282.85 K L p nRT V 3.942 9.0 1.5 0.08206 288.15 1 1 = ´ ´ = = V2 = V1 + DV= 3.942 + 100 ´ 15 ´ 10-3 = 5.442 L so 1 1 2 1 2 , 3.2 3.942 5.442 1.5 8.314ln 288.15 282.85 ln ln 1.5 28.8 ln - D = + = ´ ´ + ´ = J × K V V nR T T S nCv m
