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1、Gases叶姗5 Gase 5.1 properties of Gases 5.2 The Ideal Gas Equation 5.3 Daltons Law of Partial Pressure 5.4 The kinetic-Molecular Theory5.1 properties of Gases5.1.1 about P,V,T of gases5.1.2 Ideal gasesuDiffuse and effuse u具有扩散性和可压缩性The basic characteristics of gasPressureuPascal -PauatmummHg Volume an

2、d TemperatureStandard Temperture and PressureuStandard Temperture: 0 or 273.15KuStandard Pressure: 760mmHg or 1atm (1.01325 105Pa)The Ideal Gas物理模型 A gas that behaves exactly as described by the ideal gas equation is called an ideal gas. 理想气体分子之间没有相互吸引和排斥，分子本身的体积相对于气体所占有体积完全可以忽略。 5.2 The Ideal Gas E

3、quationpV = nRT R- a proportionality constantSTP，p =101.325kPa, T=273.15Kn=1.0 mol, Vm=22.414L=22.41410-3m31133KmolJ314.8K15.2731.0molm1022.414Pa101325nTpVRR=8.314 kPaLK-1mol-15.2.1 Using the Ideal Gas Equation to Solve Problems1. Calculating one of the four parameters (p, V, T, n ) Scope of Applica

4、tion： a real gas that not low temperature and pressure is not too high.pV = nRT2. calculating molar mass of gasM = Mr gmol-1pVmRTMRTMmpVnRTpVMmn3. The gas density calculation =RTpMpVmRTM = m / VpRTM5.2.2The calculation of the volume of gas例:为了行车的安全，可在汽车中装备上空气袋，防止碰撞时司机受到伤害。这种空气袋是用氮气充胀起来的，所用的氮气是由叠氮化钠与

5、三氧化二铁在火花的引发下反应生成的。总反应是：6NaN3+Fe2O3(s) 3Na2O(s)+2Fe(s)+9N2(g)在25。748mmHg下，要产生75.0L的N2，计算需要叠氮化钠的质量。 6NaN3+Fe2O3(s) 3Na2O(s)+2Fe(s)+9N2(g)6mol 9molMr(NaN3)=65.01 P=748mmHg=99.73kPa T=298Km(NaN3)=390.06g V(N2)=223.6Lm(NaN3)=？ V(N2)=75.0Lm(NaN3)= =131gL6223L075g06390.5.3 Daltons Law of Partial Pressure 3

6、.1 Daltons Law of Partial Pressure 3.2 vapor pressure3.3 Application of Partial pressurecomponent gas ： The ideal gas mixture for each gas called component gas Partial pressure: The pressure of each gas in a mixture is called the partial pressure.VRTnpBB3.1 Daltons Law of Partial PressureDaltons Law

7、 of Partial Pressures The total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases in the mixture : p = p1 + p2 + or p = pB VnRTp ,2211VRTnpVRTnpVRTnnVRTnVRTnp2121 n =n1+ n2+ x B B的摩尔分数VRTnpBBpxpnnpxnnppBBBBBBVnRTp 3.2 vapor pressurePtotal=Pgas + Pwat

8、er vaporThe pressure of water vapor formed by evaporation of water in aclosed container depends only on the temperature of the water and is called the vapor pressure3.3 Application of Partial pressure Sample problem: 可以用亚硝酸铵受热分解的方法制取纯氮气。反应如下： NH4NO2(s) 2H2O(g) + N2(g)If T:19、P:97.8kPa，以排水集气法在水面上收集到的

9、氮气体积为4.16L，计算消耗掉的亚硝酸铵的质量。solution： T=（273+19）K=292K p=97.8kPa V=4.16L292K 时，p（H2O）=2.20kPaMr (NH4NO2)=64.04n(N2)=0.164molNH4NO2(s) 2H2O(g) + N2(g)64.04g 1molm(NH4NO2)=? 0.164mol m(NH4NO2)= =10.5gK292molKJ31484.16L)kPa202897(1 -1 -.mol1mol164. 0g04.64例题 氧是人类维持生命的必需气体，缺氧生命就会死亡，过剩的氧会使人致病，只有在氧气的分压维持21kP

10、a才能使人体维持正常代谢过程。在潜水员自身携带的水下呼吸器中充有氧气和氦气（He在血液中溶解度很小，N2的溶解度大，可使人得气栓病） 某潜水员潜至海水30m处作业，海水的密度为1.03gcm-3，温度为20。在这种条件下，若维持O2、He混合气中p(O2)=21kPa氧气的体积分数为多少？以1.000L混合气体为基准，计算氧气的分体积和氦的质量。（重力加速度取9.807m/s2）解：T=（273+20）K=293K海水深30m处的压力是由30m高的海水和海面的 大 气 共 同 产 生 。 海 面 上 的 空 气 压 力 为760mmHg，则p=g hw+ kPa101mmg760mmHg760

11、=9.807 m/s2 1.03103kgcm-330m+101kPa=3.03103 kgcm-1 s-2 +101kPa =303kPa+101kPa =404kPa%2.5%10040421)O()O()O(kPa21)O(2222iixVVxVVpppg63. 0K293molKJ314. 8L)052. 0000. 1 (kPa404molg0026. 4)He()He(0026. 4)He(mL52L52. 0)O(111r2TRVPMmMV若混合气体体积为1.000L时，5.4 The kinetic-Molecular TheoryGrahams Law of Effusion

12、uRate of effusion of gas1 (MM gas 2 )1/2 = r Rate of effusion of gas1 (MM gas 1)1/2real gasesreal gasesThe behavior of real and ideal gases Van der Waals EquationVan der Waals Equation5.4.1 The behavior of real and ideal gases 理想气体状态方程式仅在足够低压力下适合于真实气体。 产生偏差的主要原因是:气体分子本身的体积的影响分子间力的影响5.4.2 van der Waa

13、ls EquationnRTnbVVnap)(22a,b分别称为van der waals常量。(V-nb)=Videal等于气体分子运动的自由空间b:Correction for volume of molecules为 分子间吸引力正比于(n/V)2 内压力 p=a(n/V)2 pideal=preal+a(n/V)2气体10a(Pam6mol-2)104b(m3mol-1)He0.034570.2370H20.24760.2661Ar1.3630.3219O21.3780.3183N21.4080.3913CH42.2830.4278CO23.6400.4267HCl3.7160.408

14、1NH34.2250.3707NO25.3540.4424H2O5.5360.3049C2H65.5620.6380SO26.8030.5636C2H5OH12.180.8407Table5-1 Some van der Waals ConstantsSample problem: 分别按理想气体状态方程式和van der waals方程式计算1.50mol SO2在30摄氏度占有20.0L体积时的压力，并比较两者的相对误差。如果体积减少为2.00L，其相对误差又如何？解：已知T =303K，V=20.0L，n=1.50mol，a=0.6803Pa m6 mol-2，=0.563610-4m3 mol-1189kPa20.0L303KmolK8.314J1.5mol