Chapter 1 State of SubstancesGas1-1TheIdeal Gas Equationof State1-2Dalton's Law ofPartialPressures1-3 TheKinetic-MolecularTheory1-4Real Gases2Liquid2-1 Evaporation of Liquids2-2VaporPressure,C-C equation2-3BoilingPoint of Liquid3Solid
1 Chapter 1 State of Substances 1 Gas 1-1 The Ideal Gas Equation of State 1-2 Dalton’s Law of Partial Pressures 1-3 The Kinetic-Molecular Theory 1-4 Real Gases 2 Liquid 2-1 Evaporation of Liquids 2-2 Vapor Pressure ,C-C equation 2-3 Boiling Point of Liquid 3 Solid
4Plasma(（等离子体）等离子体即指其中离子所带的正、负电量是相等的[地球上只能在实验条件下产生氛气，D,→D,++e的物质是自然界中大量存在，宇宙中绝大多数（大于99%）以等离子状态存在的，地球电离层、极光、闪电
2 4 Plasma (等离子体) 等离子体即指其中离子所带的正、负电量是相等的 [地球上只能在实验条件下产生]氘气，D2 → D2 + + e－ 自然界中大量存在，宇宙中绝大多数（大于 99% ）的物质是 以等离子状态存在的，地球电离层、极光、闪电
5Superhidensity（超高密度态）Atveryhighpressures(greaterthan 2o0atm),somegaseousmixtures consist of more than one phase.Eg.helium and xenon在高压下物质的结构会产生很大的变化例1：绝缘→导体，半导→导→超导例2：104Pa下，95℃的会凝固成冰（热冰）例3：106Pa时，H,(1.)可导电，并具有稳定的电导率
3 5 Superhi density (超高密度态) At very high pressures(greater than 200 atm),some gaseous mixtures consist of more than one phase. Eg. helium and xenon 在高压下物质的结构会产生很大的变化 例1： 绝缘 → 导体，半导 → 导 →超导 例2 ：104 Pa下，95℃的会凝固成冰（热冰） 例3 ：106 Pa时，H2 (l.)可导电，并具有稳定的电导率
1 GasI-l The Ideal Gas Eguation of State(1)Ideal GasIt is an ideal model: A gas which has quantity but no volume,without interactions,without losing kinetic energy when collideswitheachother.具有质量的几何点，只有位置而不占有体积，分子间无相互引力，发生碰撞没有动能损失Under high temperature and low pressure(T> oC; Pseveralatmosphere pressures, P→ 0)The distance among molecules are much bigger than the moleculesize,which means that compared with the gas volume,the moleculevolume is very tiny.Interactions among molecules can be neglected
4 1 Gas 1-1 The Ideal Gas Equation of State (1) Ideal Gas It is an ideal model : A gas which has quantity but no volume, without interactions,without losing kinetic energy when collides with each other. 具有质量的几何点，只有位置而不占有体积，分子间无相互引力， 发生碰撞没有动能损失. Under high temperature and low pressure( T > 0℃；P ~several atmosphere pressures，P → 0 ) The distance among molecules are much bigger than the molecule size, which means that compared with the gas volume, the molecule volume is very tiny . Interactions among molecules can be neglected
2TheIdealGasEquationof StateBoyle's law V= constantp/P (T,n constant)Charles's law V = constantc X T (P, n constant)Avogadro's law V = constant Xn (T, P constant)V = a constant (T n/P)PV=nRT
5 (2 )The Ideal Gas Equation of State Boyle's law V = constantB /P (T ,n constant) Charles's law V = constantC ×T (P, n constant) Avogadro's law V = constantA ×n (T, P constant) V = a constant (T n/P) PV = n RT
PV=nRTP--- PaV .--m3n --- molT --- K (Kelvin)R--gas constantR=PVInT= (1atm)(22.4141L/(1mol)(273.15K)=0.082atm·L/mol·K=8.314 J/mol·K
6 PV = n RT P- Pa V -m3 n - mol T - K (Kelvin) R-gas constant R = PV/nT = (1atm)(22.4141L/(1mol)(273.15K) = 0.082 atm·L/mol·K = 8.314 J / mol • K
Calculatepand M:PV=nRT= mRT / Mp= m / V = PVM /RTV = PM /RT:. M=pRT/P
7 Calculateρand M： PV = nRT = m RT / M ρ= m / V = PVM / RTV = PM / RT M = ρRT/ P
Example:It's found that 0.896g of a pure gaseouscompound containing only N and O occupies 0.524L atapressure of 0.973 X 105Pa and a temperature of 28.0℃.What are the molecular weight and molecular formula ofthegas?Solution : n =m / M=PV / RT.M= mRT / PV= 0.896X 8.314 X301 / (0.973 X105X 0.524 X10-3)= 43.98g/mol.. The molecule is N,O
8 Example: It’s found that 0.896g of a pure gaseous compound containing only N and O occupies 0.524L at a pressure of 0.973×105Pa and a temperature of 28.0℃. What are the molecular weight and molecular formula of the gas? Solution ：n = m / M = PV / RT ∴M = mRT / PV = 0.896×8.314×301 / (0.973×105×0.524×10-3 ) = 43.98g/mol ∴ The molecule is N2O
1-2Dalton's Lawof Partial Pressurespartial pressure the pressure of each gas in a mixturePtotal =Pi + P2 +... = ZP;:PV= n,RTP,V = n,RT ...P,V = n,RTPtotal V=(P, + P, + ... + P) V= (ni + n2 + ... + n) RT = n total RT.. P / P total = n; / n total = XiP; = Ptotan XiXi:molarfraction（摩尔分数）
9 1-2 Dalton’s Law of Partial Pressures partial pressure :the pressure of each gas in a mixture . Ptotal = P1 + P2 + . = ∑Pi ∵P1V = n1RT P2V = n2RT . PiV = niRT Ptotal V = ( P1 + P2 + . + Pi ) V = (n1 + n2 + . + ni ) RT = n total RT ∴Pi / P total = ni / n total = Xi Pi = Ptotal·Xi Xi :molar fraction（摩尔分数）
Example:When a mixture of potassium chlorate and manganesedioxide was heated to decompose, its quantity decreased O0.480g andan oxygen of 0.377 dm3was obtained by draining water method (排水集气法)。Thetemperatureis294Kandpressureis9.96X104PaPlease calculate the molecular weight of oxygen.(And the saturatedvaporpressureis2.48X103Pa.)SolutionsMnO2,△2 KCI03(s.) → 2 KCI(s.)+ 3021Mo2 = mo2 / no2Ptotal =PH20 + Po2:: Po2= Ptotal -Ph20 = 9.96 X 104—2.48 X 103 = 9.71 X 104(Pa)no2 = Po02:Vtotal / RT= 9.71× 104X0.377X10-3/ (8.314X294)=0.0150 (m0l)Mo2 = mo2 / no2 = 0.480 / 0.0150 = 32.0 (g / mol)
10 Example : When a mixture of potassium chlorate and manganese dioxide was heated to decompose, its quantity decreased 0.480g and an oxygen of 0.377 dm3 was obtained by draining water method (排 水集气法)。The temperature is 294 K and pressure is 9.96×104 Pa. Please calculate the molecular weight of oxygen. (And the saturated vapor pressure is 2.48×103 Pa.) Solutions： MnO2 ,△ 2 KClO3 (s.) → 2 KCl(s.) + 3O2↑ MO2 = mO2 / nO2 Ptotal = PH2O + PO2 ∴PO2 = Ptotal－PH2O = 9.96×104－2.48×103 = 9.71×104 (Pa) nO2 = PO2·Vtotal / RT = 9.71×104×0.377×10－3 / (8.314×294) = 0.0150 (mol) MO2 = mO2 / nO2 = 0.480 / 0.0150 = 32.0 (g / mol)