網(wǎng)絡有哪些適合做營銷的產(chǎn)品

1、Gases葉姍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具有擴散性和可壓縮性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. 理想氣體分子之間沒有相互吸引和排斥，分子本身的體積相對于氣體所占有體積完全可以忽略。 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:為了行車的安全，可在汽車中裝備上空氣袋，防止碰撞時司機受到傷害。這種空氣袋是用氮氣充脹起來的，所用的氮氣是由疊氮化鈉與

5、三氧化二鐵在火花的引發(fā)下反應生成的?？偡磻牵?NaN3+Fe2O3(s) 3Na2O(s)+2Fe(s)+9N2(g)在25。748mmHg下，要產(chǎn)生75.0L的N2，計算需要疊氮化鈉的質(zhì)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的摩爾分數(shù)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: 可以用亞硝酸銨受熱分解的方法制取純氮氣。反應如下： NH4NO2(s) 2H2O(g) + N2(g)If T:19、P:97.8kPa，以排水集氣法在水面上收集到的

9、氮氣體積為4.16L，計算消耗掉的亞硝酸銨的質(zhì)量。solution： T=（273+19）K=292K p=97.8kPa V=4.16L292K 時，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例題 氧是人類維持生命的必需氣體，缺氧生命就會死亡，過剩的氧會使人致病，只有在氧氣的分壓維持21kP

10、a才能使人體維持正常代謝過程。在潛水員自身攜帶的水下呼吸器中充有氧氣和氦氣（He在血液中溶解度很小，N2的溶解度大，可使人得氣栓?。?某潛水員潛至海水30m處作業(yè)，海水的密度為1.03gcm-3，溫度為20。在這種條件下，若維持O2、He混合氣中p(O2)=21kPa氧氣的體積分數(shù)為多少？以1.000L混合氣體為基準，計算氧氣的分體積和氦的質(zhì)量。（重力加速度取9.807m/s2）解：T=（273+20）K=293K海水深30m處的壓力是由30m高的海水和海面的 大 氣 共 同 產(chǎn) 生 。 海 面 上 的 空 氣 壓 力 為760mmHg，則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若混合氣體體積為1.000L時，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 理想氣體狀態(tài)方程式僅在足夠低壓力下適合于真實氣體。 產(chǎn)生偏差的主要原因是:氣體分子本身的體積的影響分子間力的影響5.4.2 van der Waa

13、ls EquationnRTnbVVnap)(22a,b分別稱為van der waals常量。(V-nb)=Videal等于氣體分子運動的自由空間b:Correction for volume of molecules為 分子間吸引力正比于(n/V)2 內(nèi)壓力 p=a(n/V)2 pideal=preal+a(n/V)2氣體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: 分別按理想氣體狀態(tài)方程式和van der waals方程式計算1.50mol SO2在30攝氏度占有20.0L體積時的壓力，并比較兩者的相對誤差。如果體積減少為2.00L，其相對誤差又如何？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