6. Application of Dalton's law of partial pressures 1)求分压: Example: Suppose that hydrogen is collected over water at 25C. How much H2(expressed in moles)atmospheric pressure of 74.3 cmHg? Solution Patm- PH, t PHo Pam=74.2cmHg the vapor pressure of water at 25 C is 23.8mmHg(=2.38cmHg P =74.2-2.4=71.8cmHa p71.876×0.223 8.61×10-3mol RT0.08206×298.15
6. Application of Dalton’s law of partial pressures 1)求分压: Example：Suppose that hydrogen is collected over water at 25c .How much H2 (expressed in moles) atmospheric pressure of 74.3 cmHg?
四.阿玛加分体积定律 Amagat's law of partial volumes 1什么叫分体积:混合气体中某组分单独存在,并 且和混合气体的温度,压强相同时,所具有的体积 称为混合气体中第组分的分体积 2分体积定律:当温度,压力相同时,混合气体的 总体积等于各组分分体积之和 3 Deduction ∑n,RT ∑-=∑
四．阿玛加分体积定律 Amagat’s law of partial volumes 1 什么叫分体积：混合气体中某组分i单独存在，并 且和混合气体的温度，压强相同时，所具有的体积 Vi 称为混合气体中第i组分的分体积 2 分体积定律：当温度，压力相同时，混合气体的 总体积等于各组分分体积之和 3 Deduction:
五.格拉罕姆扩散定律 Graham' s law of effusion 1 Difference between effusion and diffusion The mixture of one gas with another is called diffusion Effusion means gas molecules move from a high-pressure region to a low pressure region through a pinhole 2 Graham's law: Thomas Graham discovered that the effusion rate of a gas is inversely proportional to the square root of its molar mass 在恒压条件下,某一温度下气体的扩散速率与其密 度(或摩尔质量)的平方根成反比
五．格拉罕姆扩散定律 Graham’s law of effusion 1. Difference between effusion and diffusion The mixture of one gas with another is called diffusion. Effusion means gas molecules move from a high-pressure region to a lowpressure region through a pinhole 2. Graham’s law: Thomas Graham discovered that the effusion rate of a gas is inversely proportional to the square root of its molar mass 在恒压条件下，某一温度下气体的扩散速率与其密 度（或摩尔质量）的平方根成反比
L M 3.表达式 PI M Nm 4证明:由pV Nu 3 P= constant→∝ M P∞M 4(p= constant) VP, VM
(p ) M M ρ ρ u u ρ M ρ p u V N m N u ρ m pV M M ρ ρ u u ： constant 1 constant 3 4 3. 1 2 1 2 2 1 2 1 2 1 2 2 1 = = = = = = = = .证 明 ：由 表 达 式
5. Application of Graham's law 1)利用此定律可以测定未知气体的分子量(或原子 量) 2)可以分离同位素 isotope 例如:工业分离铀235 3U02+8HNO3=3U02(NO2+2NO+4H2O UO2 NO2=003+NO+NO2+O2 UO,+H=UO,+HO UO+4HF=UF+2Ho UF4+F2=UF6 M,238.05+61×8.998352.04 1.0043 M,V235.04+61×8.998V349.03 (1:235UF6,2:38UF
5. Application of Graham’s law 1)利用此定律可以测定未知气体的分子量（或原子 量） 2)可以分离同位素 isotope 例如：工业分离铀235 (1 : UF ,2 : UF ) 1.0043 349.03 352.04 235.04 61 8.998 238.05 61 8.998 UF F UF UO 4HF UF 2H O UO H UO H O UO (NO) UO N O N O O 3UO 8HNO 3UO (NO) 2NO 4H O 6 238 6 235 1 2 2 1 4 2 6 2 4 2 3 2 2 2 2 2 3 2 2 2 3 2 2 2 = = + + = = + = + = + + = + = + + + + = + + M M r r