北京化工大学：《物理化学》课程教学资源（双语练习题）第十章 表面化学英文习题及参考答案 surfaceandcolloid problems

第十章表面化学英文司题及参考答素 300K is 0.21mg The Langmuir isotherm is known to describe the adsorption.Find the fractional coverage of the surface at the two pressure. Soluions The Langmr is bp We are looking for so we must fist find bor b=- 0 1-0)p1- setting this expression at one pressure qual to that at another pressure allows solution for / / a-Za-K石 so P.(m-m)p:(m-m: m m. mp/ -p: (36.0-4.0)kPa . 6%649%2Mmg-04ms w8-8-05m8-0e-025 ature is 3cm2(Ans0.0335cm). Solutions: AP=2=pgh=0866×102×98×2x10=1697=2x00284Nm r=3.35×10m 2.Calculate the vapor pressure of a water droplet at 25C that has a radius of 2nm.(Ans 5350Pa) Solutions:

第十章 表面化学英文习题及参考答案 1.A certain solid sample adsorbs 0.63mg of CO when the pressure of the gas is 36.0kPa and the temperature is 300K. The mass of gas adsorbed when the pressure is 4.0kPa and the temperature is 300K is 0.21mg. The Langmuir isotherm is known to describe the adsorption. Find the fractional coverage of the surface at the two pressure. Solutions: The Langmuir isotherm is bp bp + = 1 q We are looking for θ, so we must fist find b or Г∞ (1 ) (1 ) ¥ ¥ G - G G G = - = p p b q q setting this expression at one pressure equal to that at another pressure allows solution for Г∞. (1 ) (1 ) 2 2 2 1 1 1 ¥ ¥ ¥ ¥ G G - G G = G G - G G p p so 2 2 2 1 1 1 ( ) ( ) m p m m m p mmon m mon - = - mg kPa mg kPa m p m p p p mmon 0.84 ) 0.21 4.0 0.63 36.0 ( (36.0 4.0) 1 2 2 1 1 1 2 = - × - = - - = - so 0.75 0.84 0.63 q1 = = and 0.25 .084 0.21 q2 = = 1. The surface tension of toluene at 20℃ is 0.0284N·m -1 , and its density at this temperature is 0.866 g·cm -3 , what is the radius of the largest capillary that will permit the liquid to rise 3cm?(Ans 0.0335cm). Solutions: r N m gh r P 1 3 2 2 0.0284 0.866 10 9.8 2 10 169.7 2 - - ´ × D = = r = ´ ´ ´ ´ = = g r m 4 3.35 10 - = ´ 2. Calculate the vapor pressure of a water droplet at 25℃ that has a radius of 2nm. (Ans. 5350Pa). Solutions: rp M P P RT r 2g ln = 平

8314×29815in3167a 卫，_=2×72.75×10-3N.mx18×103 2×10-”×1×10 P ÷h3167P%=5.283x10 P=5350Pa

9 3 3 1 3 r 2 10 1 10 2 72.75 10 18 10 3167Pa P 8.314 2998.15ln ´ ´ ´ ´ ´ × ´ ´ ´ = - - - N m 6 5.283 10 3167 \ln = ´ Pa Pr Pr=5350Pa