理论1、2.doc

第一题： 热力学（24分）Problem 1: Thermodynamics (24 points) 为了自己在二月份的18岁生日晚会，Peter计划将自己父母家花园中的一个小屋变成一个带有人造沙滩的游泳池。为了估计加热水和房间的花费，Peter获得了天然气的组成和它的价格。For his 18th birthday party in February Peter plans to turn a hut in the garden of his parents into a swimming pool with an artificial beach. In order to estimate the costs for heating the water and the house, Peter obtains the data for the natural gas composition and its price. 1.1 请写出天然气中主要成分及其甲烷和乙烷（表1）燃烧的化学方程式，假设氮气在封闭的环境中不参加反应。计算甲烷和乙烷燃烧反应的焓，熵，标准状态下的吉布斯自由能。(1.013105 Pa, 25.0C)，假设反应中的产物均为气态。 热力学参数和天然气的组成如表1所示 1.1 Write down the chemical equations for the complete combustion of the main components of natural gas, methane and ethane, given in Table 1. Assume that nitrogen is inert under the chosen conditions. Calculate the reaction enthalpy, the reaction entropy, and the Gibbs energy under standard conditions (1.013105 Pa, 25.0C) for the combustion of methane and ethane according to the equations above assuming that all products are gaseous. The thermodynamic properties and the composition of natural gas can be found in Table 1. 1.2 天然气的密度是0.740 g L-1 (1.013105 Pa, 25.0C)，由the public utility company. 测定。1.2 The density of natural gas is 0.740 g L-1 (1.013105 Pa, 25.0C) specified by PUC, the public utility company. a) Calculate the amount of methane and ethane (in moles) in 1.00 m3 of natural gas (natural gas, methane, and ethane are not ideal gases!). a）计算1.00m3天然气甲烷和乙烷的量(用mol表示)，天然气和甲烷和乙烷均不是理想状态下。 b) Calculate the combustion energy which is released as thermal energy during the burning of 1.00 m3 of natural gas under standard conditions assuming that all products are gaseous. (If you do not have the amount from 1.2a) assume that 1.00 m3 natural gas corresponds to 40.00 mol natural gas.) b）计算燃烧1.00m3天然气放出的能量（假设所有产物均为气态），如果你从1.2a没有查找到数据，就假定1.00m3天然气相当于40mol天然气。According to the PUC the combustion energy will be 9.981 kWh per m3 of natural gas if all products are gaseous. How large is the deviation (in percent) from the value you obtained in b)? 如果按照PUC的计算，在产物都是气态的情况下，每立方米的天然气燃烧将会放出9.981 kWh能量，按照你在b中的计算，将会有多大的偏差。（用百分比表示）The swimming pool inside the house is 3.00 m wide, 5.00 m long and 1.50 m deep (below the floor). The tap water temperature is 8.00C and the air temperature in the house (dimensions given in the figure below) is 10.0C. Assume a water density of = 1.00 kg L-1 and air behaving like an ideal gas. 屋内的游泳池有3m宽，5m长，1.5m深（低于地表面）。自来水的温度是8.00C，屋内（尺寸见下面的图）空气的温度是10.0C。假设水的密度为1.00kg.L-1,空气被视为理想气体。1.3 Calculate the energy (in MJ) which is required to heat the water in the pool to 22.0C and the energy which is required to heat the initial amount of air (21.0% of O2, 79.0% of N2) to 30.0C at a pressure of 1.013105 Pa. 计算将游泳池的水加热到22C所需要的能量（用千焦表示）以及在1.013105 Pa的压力下将初始的空气(21.0% 的氧气, 79.0%的氮气)加热到30C所需要的能量In February, the outside temperature is about 5C in Northern Germany. Since the concrete walls and the roof of the house are relatively thin (20.0 cm) there will be a loss of energy. This energy is released to the surroundings (heat loss released to water and/or ground should be neglected). The heat conductivity of the wall and roof is 1.00 W K-1 m-1. 二月，德国北部的室外温度大概为5C，因为混凝土的墙壁及屋顶比较薄(20.0 cm)会损失一些能量，这些能量将会被释放到周围的环境中（释放到水中和地面的能量忽略不计），墙壁和屋顶的热传导系数为1.00 W K-1 m-1.1.4 Calculate the energy (in MJ) which is needed to maintain the temperature inside the house at 30.0C during the party (12 hours).1.4 计算在晚会中如要保持室内温度为30.0C需要多少能量。（12小时） 1.00 m3 of natural gas as delivered by PUC costs 0.40 and 1.00 kWh of electricity costs 0.137 . The rent for the equipment for gas heating will cost him about 150.00 while the corresponding electrical heaters will only cost 100.00 .1.00 m3的PUC提供的天然气耗费0.40 ，1.00 kWh的电耗费0.137 .气体加热装置的租金为150.00 ，而电加热的租金为100.00 . 1.5 What is the total energy (in MJ) needed for Peters “winter swimming pool” calculated in 1.3 and 1.4? How much natural gas will he need, if the gas heater has an efficiency of 90.0%? What are the different costs for the use of either natural gas or electricity? Use the values given by PUC for your calculations and assume 100% efficiency for the electric heater. 1.5 在1.3和 1.4中，Peter的冬季游泳池一共需要多少的能量（用千焦表示），如果气体加热的效率为90，需要多少天然气，在用气体和电加热中有什么区别，比较如果按照PUC的价格天然气的开支和假定电加热的效率为100的情况下的电加热的开支。Problem 2: Kinetics at catalyst surfaces (23 points) Apart from other compounds the exhaust gases of an Otto engine are the main pollutants carbon monoxide, nitrogen monoxide and uncombusted hydrocarbons, as, for example, octane. To minimize them they are converted to carbon dioxide, nitrogen and water in a regulated three-way catalytic converter. 第二题：动力学和表面催化如不考虑其他的化合物，CO，NO，以及一些不饱和的碳氢化合物是主要的污染物。为了降低它们的危害，常常是通过三向催化将它们转化成CO2，NO2，及水。2.1 Complete the chemical reaction equations for the reactions of the main pollutants in the catalyst. To remove the main pollutants from the exhaust gas of an Otto engine optimally, the -value is determined by an electro-chemical element, the so called lambda probe. It is located in the exhaust gas stream between engine and the three-way catalytic converter. 2.1完成主要污染物催化反应的化学方程式 为了去除Otto engine 中的主要污染物，的值是被电气化学因素决定的，即所谓的lambda探针。它位于机器和三向催化转换器中间的废气流中。2.2 Decide the questions on the answer sheet concerning the probe. The adsorption of gas molecules on a solid surface can be described in a simple model by using the Langmuir isotherm: where is the fraction of surface sites that are occupied by the gas molecules, p is the gas pressure and K is a constant. The adsorption of a gas at 25 C may be described by using the Langmuir isotherm with K = 0.85 kPa-1. 2.2回答答题纸中有关于探针的问题气体分子在固体表面的吸附可以用Langmuir吸附等温线来简单的考虑。（公式见原试卷）表示的是气体分子占据的表面位点的比率，p表示气压，K是一个常数。气体在25 C的吸附情况可以用Langmuir吸附等温线来描述，这里的K0.85 kPa-1.2.3 a) Determine the surface coverage at a pressure of 0.65 kPa.计算在压力为0.65kPa下的2.3 b) Determine the pressure p at which 15 % of the surface is covered.计算表面覆盖率为15时的压力2.3 c) The rate r of the decomposition of gas molecules at a solid surface depends on the surface coverage (reverse reaction neglected): r = k Give the order of the decomposition reaction at low and at high gas pressures assuming the validity of the Langmuir isotherm given above (products to be neglected). 气体分子的分解效率取决于表面的覆盖率，（逆反应忽略）r = k假设分解的效率很低，以及在高压的情况下以上的Langmuir吸附等温线依然有效（产品的反应忽略）2.3 d) Data for the adsorption of another gas on a metal surface (at 25C) 金属表面吸附其他气体的参数（图见原试卷）If the Langmuir isotherm can be applied, determine the gas volume Va,m