复旦大学：《普通化学 General Chemistry》课程教学资源（课堂讲义）Chap 8 - Changes in States, Colligative Properties and Phase Properties, and Phase Diagram

PART 8-Changes in States, Colligative Properties, and Phase Diagram Reference: Chapter 5, 12 in textbook

PART 8 – Ch i St t Changes in States, Colligative Properties and Phase Properties, and Phase Diagram Reference: Chap , ter 5 12 in textbook 1

States of materials Solid Liquid Gas ,● 器∴

States of Materials Solid Liquid Gas 2

Gas and ideal gas Properties of a Gas Low density, No fixed shape Can be significantly expanded and compressed Can be mixed by any ratios o Ideal gas- Assumption Gas molecules have no volume No intermolecular interaction

Gas and Ideal Gas z Prope es o a Gas rties of a Gas „ Low density, No fixed shape „ Can be significantly expanded and compressed „ Can be mixed by any ratios Can be mixed by any ratios z Ideal Gas Ideal Gas – Assumption Assumption „ Gas molecules have no volume; „ No intermolecular interaction 3

Ideal gaS: PV. t P VS. V →P*∨= constant or 。∨vs.T →∨/T= constant or V1/T1=V2/T2 。Pvs.T →P/T= constant or P1/T1=P2/T2 Note: the units of Pv. t 4

Ideal Gas: P, V, T z P sV v s. V „ Æ P * V = constant or P1 * V1 = P 2 1 1 * V 2 z V vs. T „ Æ V / T = constant or V1 / T1 = V 2 / T 2 z P vs. T „ Æ P/T t t P / T = cons tan t o r P1 / T1 = P 2 / T 2 z Note: the units of P V T Note: the units of P, V, T 4

Ideal gas law Ideal Gas equation P*V=n*R*t P: 101.3 kPa =1 atm=760 mmHg =760 torr V:1m3=1×103dm3(L)=1×106cm3(mL) T:K=C°+273.15 ,n: of moles of all gases R=8.31J·mo1K1=0.082atm·L·mo1·K1 Avogadros Law Under same P, T, 1 mole of gas occupies the same V At the Standard Temperature Pressure(STP),1 mole of any ideal gas is 22. 4 liter 5

Ideal Gas Law z Ideal Gas Equation „ P * V = n * R * T ¾ P： 101 3. kPa = 1 atm = 760 mmHg = 760 = 760 mmHg = 760 torr ¾ V： 1 m3 = 1 × 103 dm3 ( L ) = 1 × 106 cm3 (mL) ¾ T： K = C° + 273.15 ¾ n ： # of moles of all gases ¾ R = 8.31 J • mol-1•K-1 = 0.082 atm • L • mol-1 • K-1 z Avogadro’s Law „ Under same P T 1 mole of gas occupies the same V Under same P, T, 1 mole of gas occupies the same V. „ At the Standard Temperature & Pressure (STP), 1 mole of any ideal gas is 22.4 liter. 5

Practice Q1: N2H4 is a fuel, prepared by the following reaction 2NH3(9)+ Naoci(aq)>N2H4(aq)+ Nacl(aq)+H2o( To synthesize 15.0 kg n2H4, how much volume of 10C, 3.63 atm of N3(g is needed? Solution: n(N2H4)=m(N204)/MW(N204)=1.50 X 1049/32 g/mol 469 mol (NH3)-n(NH3) RT/ P 2n(N2H4) RT/P 2X469×0.082X(273+10)/363 =599X103L 6

Practice z Q1：N H is a fuel prepared by the following reaction: 2H4 is a fuel, prepared by the following reaction: 2NH3 (g) + NaOCl (aq) → N2H4 (aq) + NaCl (aq) + H2O (l) To synthesize 15.0 kg N2H4, how much volume of 10oC, 3.63 atm of NH3 (g) is needed? z Solution: n = m / MW = 1 50 x 104 n g / 32 g/mol = 469 mol (N2H4) = m (N2O4) / MW (N2O4) = 1.50 x 104 g / 32 g/mol = 469 mol V(NH3) = n (NH3) RT/ P = 2n (N2H4) RT/ P = 2 x 469 x 0.082 x (273 + 10) / 3.63 = 5.99 x 103 L 6

Partial Pressure Daltons Law Under constant temperature& volume, the total pressure of a gas mixture equals the partial pressure of each ga as component Poal=PA+PB+Pc+……+P Gas Partial Pressure(Pi) P of a gas component solely occupying whole volume

Partial Pressure z Dalton’s Law „ Under constant temperature & volume, the total pressure of i t l th ti l f a gas mix ture equals the parti al pressure of each gas component. Ptotal = PA + P B + P C + …… + P i z Gas Partial Pressure (Pi) „ P of a gas component solely occupying whole volume 7

Molar fraction Each gas component is valid for the Ideal Gas Law PA= nART/V, PB=nBRT/V P=nRT/V Ptat=nRT/∨+ngRT/V+…+nRTV (nA+n+……+n)RT/∨ ntotal RT/V total (n RT/V)/(nota RT/v)=n, I ntotal n; /ntotal =Xi(molar fraction) 2 Pi=Ptotal XX 8

Molar Fraction Each gas component is valid for the Ideal Gas Law: PA = n ART/ V, P B = n BRT/ V …… ， Pi= niRT/ V Ptotal = n ART/ V + n RT/ V + n BRT/ V + RT/ V + …… + niRT/ V = ( n A + n B + … … + ni ( A B i )RT/ V = ntotal RT/ V Pi / Ptotal = (niRT/ V) / (ntotalRT/ V) = ni / ntotal Æ ni / ntotal = Xi (molar fraction) 8 Æ Pi = Ptotal × Xi

Partia|∨o|ume Partial volume of a gas component Volume of a gas component under T and ptotal Vtoa=VA+Vg+Vc+…+V Molar Fraction(Xi) Xi=n; /ntotal=P; /Ptotal=Vi/Vtotal P1×Vtoa= Ptotal×V

Partial Volume z Partial Vol me of a gas component Partial Volume of a gas component „ Volume of a gas component under T and Ptotal „ Vtotal = VA + VB + VC + …… + Vi z Molar Fraction (Xi) Molar Fraction (Xi) „ Xi = ni / ntotal = Pi / Ptotal = Vi / Vtotal „ Pi × Vtotal= Ptotal × Vi 9

Practice Q2: At 25C. 1 atm. we collected a mixture of h and saturated water vapor 152 mL If the saturated water vapor is 23. 76 mmHg, calculate: (a) Partial pressure of H2 (g); (b) Amount of substance of H2(g); (c)Volume of this H2 (g) if it is dry Solution (1)PH2= Ptotal- PH20=760-2376=736.2 mmHg (2) PH2 Vtotal =nH2RT H2= PH2 total/RT =7362x0.152/624X298.15=62X103mo 3)PH2= VH2 Ptotal VH2= PH2Vtotal /Ptotal=736.2 X 152/760=147 ml 10

Practice z Q2: At 25°C, , 1 atm we collected a mixture of H2 and saturated water vapor 152 mL. If the saturated water vapor is 23.76 mmHg, calculate: (a) Partial pressure of H2 ( ) (b) A t f b t f H H2 (g); (b) Amount of substance of H ( ) ( )V l f 2 (g); (c) Volume of this H2 (g) if it is dry. z Solution: (1) PH2 = Ptotal ( ) H2 total- PH2O = 760 - 23.76 = 736.2 mmHg (2) PH2Vtotal = nH2RT nH2 = PH2Vtotal / RT = 736.2 x 0.152 / 62.4 x 298.15 = 6.2 x 10-3 mol (3) PH2Vtotal= VH2 Ptotal VH2 = PH2Vtotal / Ptotal = 736.2 x 152 / 760 = 147 ml 10