BM102 Physical Chemistry (General Chemistry II) PDF

BM102 Physical Chemistry (General Chemistry II) First Chapter State of Matter Introduction All matter exists in three states: gas, liquid and solid. A molecular level representation of gaseous, liquid and solid states is shown in Figure (1). A gas consis...

BM102 Physical Chemistry (General Chemistry II) First Chapter State of Matter Introduction All matter exists in three states: gas, liquid and solid. A molecular level representation of gaseous, liquid and solid states is shown in Figure (1). A gas consists of molecules separated wide apart in empty space. The molecules are free to move about throughout the container. A liquid has molecules touching each other. However, the intermolecular space, permit the movement of molecules throughout the liquid. A solid has molecules, atoms or ions arranged in a certain order in fixed positions in the crystal lattice. The particles in a solid are not free to move about but vibrate in their fixed positions. Figure (1) illustrates the changes in material from one state to another. Figure (1) The material changes from state to another state 1. Gaseous State 1.1. General Characteristics of Gases 1. Expansibility: Gases have limitless expansibility. They expand to fill the entire vessel they are placed in. 2. Compressibility: Gases are easily compressed by application of pressure to a movable piston fitted in the container. 3. Diffusibility: Gases can diffuse rapidly through each other to form a homogeneous mixture. 1 4. Pressure: Gases exert pressure on the walls of the container in all directions. 5. Effect of Heat: When a gas, confined in a vessel is heated, its pressure increases. Upon heating in a vessel fitted with a piston, volume of the gas increases. The above properties of gases can be easily explained by the Kinetic Molecular Theory, which will be considered later in the chapter. The material is not discontinuance for specific state but may whatever one of three states according to the surrounding circumstances. Therefore it must be define the circumstances and conditions, as example the standard or normal condition (STP), i.e., standard pressure and temperature, which means that the pressure is equal to one atmosphere (1.0 atm) and temperature is equal to zero centigrade (0.0℃). In addition, it was an agreement that the 25℃ is taken as room temperature in the entire world regardless of the real temperature. 1.2. General Physical Properties of Gases The gases are characterized by some properties tan the other material states. The gas molecules exist at separate distances and according to that, the attractive forces between gas molecules are so weak. These weak attractive forces permit to continuous, fast, eternal and random movement of gas particles. All gases are capable for mixing to each other's with any ratio to form a solution, which is argued to the intermediate distances between the gas molecules where the gas molecules interpenetrate between that intermediate distances for another gas. It follows this gas can be compressed to less volume and lowering the intermediate distances and adjacent the gas molecules to each other's. The low densities of gases give the gases the ability of diffusion and occupy the total vessel volume. The gas density can be reduced or increased by changing the gas quantity or the occupied volume by controlling the pressure or temperature of both of them. It is clear that gas properties closely related to four variables: Pressure (P), Volume (V), Temperature (T) and Quantity (n) The Pressure (P): is defined as the effective force to the unit area. The gas pressure is a result of the collision strength of its particle with to the area walls of the containing vessel 2  Pressures unite = P (unit) = = kg m-1 s-2 = Pascal (Pa) The Pascal is unit of the international world units. It is defined as the strength which equivalent to one Newton (1.0 N = 1.0 kg m s-2) and effective to one square meter area (1.0 m2). The gas pressure can be measured using a mercury manometer or barometer as in Figure (2). Figure (2) a Mercury Manometer and a Mercury Barometer Patm = ρHg. g. h Patm = (13595 kg/cm3) (9.807 m/s2) (0.760 m) Patm = 101325 kg m-1 s-2 = 101325 Pa = 17696 lb/in2 It was found that the mercury length at sea level and temperature 0.0℃ is equal to 76 cm (760 mm), which leads to take this as atmosphere pressure unit. 1 atm = 76 cm Hg = 760 mm Hg = 760 torr = 14.7 psi 1 atm = 101325 Pa (N m-2) = 101.325 kPa = 1.01325 bar 1.3. General gas equation 1.3.1. Boyle’s Law In 1660, Robert Boyle found out experimentally the change in volume of a given sample of gas with pressure at room temperature. From his observations, he formulated a generalization known as Boyle’s Law. It states that: "At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure". If the pressure is doubled, the volume is halved P∝ ⟾ P= ⟾ PV = K1 ∴ P1V1 = P2V2 = P3V3 = K1 & K1 = constant 3  Figure (7) Graphical representation of Boyle’s law (a) a plot of versus for a gas sample is (isotherm) hyperbola; (b) a plot of versus 1/P is a straight line The Boyle’s law can be demonstrated by adding liquid mercury to the open end of a J-tube. As the pressure is increased by addition of mercury, the volume of the sample of trapped gas decreases. Gas pressure and volume are inversely related; one increases when the other decreases. Figure (8) Demonstration of Boyle’s law 1.3.2. Charles and Gay-Lussac's Law In 1787, Jacques Charles investigated the effect of change of temperature on the volume of a fixed amount of gas at constant pressure. He established a generalization, which is called the Charles’ Law. It states that: at constant pressure, the volume of a fixed mass of gas is directly proportional to the Kelvin temperature of absolute temperature. If the absolute temperature is doubled, the volume is doubled. The Scientist Gay-Lussac's has put the following law, which has been also obtained before by Charles in 1787, which illustrates the relation between the gas volume and its temperature at constant pressure and quantity. 4 “At constant pressure the volume of a quantity of gas increases (or decreases) by of its volume at the absolute temperature if the gas temperature increases (or decreases) by one centigrade”. = K2 ⟾ V = K2 T = = …………….. = K2 = constant Therefore, the Charles and Gay-Lussac's law can be simply paraphrased as follows: “At constant pressure, the volume of a fixed mass of gas is directly proportional to the Kelvin temperature of absolute temperature”. 1.3.3. Amonton,s Law Amonton could deduce relation between the gas pressure and its temperature as following: “At constant volume the pressure of quantity of gas is directly proportional to its absolute temperature”. P∝T ⟾ P = K3 T = = & K3 = constant 1.3.4. Avogadro's Law “At the same condition of pressure and temperature, the equal volume of gases contains the same number of molecules”. V∝n ⟾ V = K4 n = K4 ⟾ = = & K4 = constant In case of equality of the volumes at the same conditions of pressure and temperature: ∵ V1 = V 2 ⟾ ∴ n1 = n2 1.3.5. The Molar Volume At constant T and P, whereas K4 = , i.e., the volume of a certain quantity (n) from any gas is constant at constant pressure and temperature. When this quantity is one mole only, its volume is called the molar volume and symboled as Vm, i.e. the molar volume of any gas at standard conditions (STP) is equal to 22.4 liter. 1.3.6. P – n Relationship “At constant volume and temperature, the gas pressure is directly proportional to its quantity”. P∝n ⟾ P = K5 n = K5 ⟾ = = = K5 = constant 1.3.7. n – T Relationship 5 The text relation: “At constant pressure and volume, the gas quantity is inversely proportional to its absolute temperature”. n∝ ⟾ n= n1T1 = n2T2 …………. = K6 = constant 1.3.8. State Equation of Gases: = 1.3.9. The Values of the Constant (R): 1.0 mol of gas occupies 22.4 L at STP, i.e., pressure is equal to 1.0 atm and temperature 273 K. R= = = 0.0821 L atm mol-1 K-1 K 1.0 Pa = 1.0 kg m-1s-2 R = 8.314 kg m2 s-2 K-1 1 atm = 1.013 x 106 dyne cm-2 R= = 1.988 cal mol-1 K-1 K R= = 4.184 J mol-1 K-1 K Table (1) summarize the values and units of gas constant (R) 0.0821 atm L mol-1 K-1 8.314 Pa m3 mol-1 K-1 8.314 kg m s K 2 -2 -1 8.314 dyne cm mol-1 K-1 8.314 g cm2 s-2 mol-1 K-1 8.314 erg mol-1 K-1 8.314 J mol-1 K-1 1.997 cal mol-1 K-1 1.3.10. Student Exercises 1. The value of pressure of quantity of a gas in container vessel its volume is "4L" is 2.0 atm at temperature 0.0℃. How much will be the pressure of the same quantity of gas at the same temperature if the volume is changed to "2L"? (a) 4 atm (b) 1 atm (c) 3 atm (d) 2 atm View Answer Answer: [a] Hint and Solution: [a] ∵ The gas quantity and temperature are constant. ∴ According to Boyle’s law: P1V2 = P2V2 P1 = P 2 x = 2 atm x = 4 atm 2. If 100 cm3 of a gas under pressure 100 kPa is changed to 125 kPa at the same temperature, what is the final volume? (a) 90 cm3 (b) 80 cm3 (c) 70 cm3 (d) 60 cm3 View Answer Answer: [b] 6 Hint and Solution: [b] ∵ The gas quantity and temperature are constant. ∴ According to Boyle’s law: P1V2 = P2V2 ∴ V2 = V 1 x = 100 cm3 x = 80 cm3 3. A balloon is filled with hydrogen gas with volume 1250 mL under pressure 760 mmHg. This balloon is pushed to mountaintop, the pressure there is 720 mmHg. How much will be the balloon volume assuming temperature constancy? (a) 1768.5 mL (b) 1543.3 mL (c) 1426.6 mL (d) 1319.4 mL View Answer Answer: [d] Hint and Solution: [d] ∵ The gas quantity and temperature are constant. ∴ According to Boyle’s law: P1V2 = P2V2 ∴ V2 = V 1 x = 1250 mL x =1319.4 mL 4. A gas occupies a volume 200 cm3 at temperature 0.0℃ and pressure 760 mmHg. What is the new volume for this gas if the temperature is changed to 100℃ and pressure 760 mmHg? (a) 273.26 cm3 (b) 265.26 cm3 (c) 284.35 cm3 (d) 303.35 cm3 View Answer Answer: [a] Hint and Solution: [a] ∵ The gas quantity and pressure are constant. ∴ According to Charles and Gay-Lussac's Law: = T1 = 0.0 + 273 = 273K & T2 = 100 + 273 = 373 K V1 = 200 cm3 & V2 = ? V2 = V1 x = 200 cm3 x = 273.26 cm3 5. A sample of gas occupies a volume 250 cm3 at temperature 27℃. What is the temperature will be if the gas volume becomes 258.3 cm3 assuming that there is no temperature change. (a) 37℃ (b) 73℃ (c) 80℃ (d) 34℃ View Answer Answer: [a] Hint and Solution: [a] ∵ The gas quantity and pressure are constant. ∴ According to Charles and Gay-Lussac's Law: = T1 = 27 + 273 = 300 K & T2 = ? V1 = 250 cm3 & V2 = 258.3 cm3 7  T2 = T 1 x = 300 K x = 309.96 K ∴ t ℃ = T K – 273 = 310 – 273 = 37℃ 6. A gas cylinder contains 20 kg of nitrogen under constant pressure at certain temperature. How much the temperature will be after its half quantity is consumed? (a) 4 T1 (b) 3 T1 (c) 2 T1 (d) 6T1 View Answer Answer: [c] Hint and Solution: [c] ∵ The gas volume is constant (cylinder volume) and pressure is constant also (V & P are constants). ∴ According to the relation between the gas quantity and its temperature: n1T1 = n2T2 =½ ⟾ =2 n1T1 = n2T2 ⟾ ∴ T2 = T1 x = T1 x = 2 T1 7. Which temperature must be heated a gas in closed vessel at temperature 0.0℃ the duplicate its volume? (a) 243℃ (b) 273℃ (c) 280℃ (d) 434℃ View Answer Answer: [b] Hint and Solution: [b] ∵ The gas quantity and volume are constant. ∴ According to Amonton Law: = T1 = 0.0 + 273 = 273 K & T2 =? P1 = P1 atm & P2 = 2 P1 atm T2 = T1 x = 273 K x = 546 K ∴ t ℃ = T K – 273 = 546 – 273 = 273℃ 8. The volume of one mole of a gas at definite pressure and temperature is 10 L. What is the volume of 5 moles from that gas at the same conditions? (a) 50 L (b) 90 L (c) 120 L (d) 150 L View Answer Answer: [a] Hint and Solution: [a] ∵ The gas pressure and temperature are constant. ∴ According to Avogadro's Law: = V1 = 10 L & V2 =? n1 = 1 mol & n2 = 5 mol 8  ∴ V2 = V 1 x = 10 L x = 50 L 9. Calculate the molecular weight of the gas if you know that 0.824 g of it occupies a volume 260 mL at STP. The molecular weight (M) of the gas is one mole and the volume is 22.4 L at STP. (a) 78 g /mol (b) 71 g/mol (c) 63 g/mol (d) 74 g/mol View Answer Answer: [b] Hint and Solution: [b] M.wt. ⟾ 22.4 L 0.824 g ⟾ 0.26 L M.wt. = = 71 g mol-1 10- Shall increasing the temperature of quantity of gas from 100℃ to 200℃ at constant pressure the volume will be duplicated? Why? (a) Duplicated (True) (False) View Answer Answer: [False] Hint and Solution: [False] ∵ The gas quantity and pressure are constant. ∴ According to Charles and Gay-Lussac's Law: = = = = 1.268 ⟾ V2 = 1.268 V ∴ The gas volume is not duplicate according to Charles and Gay- Lussac's Law: “The gas volume is proportional to the absolute temperature not with Celsius temperature at constant pressure and amount”. 11- The temperature of quantity of gas at constant pressure is changed from 100℃ to 200 K. Shall the gas volume is increased and why? (True) (False) View Answer Answer: [False] Hint and Solution: [False] ∵ The gas quantity and pressure are constant. ∴ According to Charles and Charles and Gay-Lussac's Law: = = = = 0.536 ⟾ V2 = 0.536 V1 ∴ The gas volume is decreased according to Charles and Gay- Lussac's Law: “The gas volume is proportional to the absolute 9 temperature not with Celsius temperature at constant pressure and amount”. 12- Shall the increasing the temperature of quantity of gas from 100℃ to 200℃ at constant volume the pressure will be duplicated? Why? (a) Duplicated (True) (False) View Answer Answer: [False] Hint and Solution: [False] ∵ The gas quantity and volume are constant. ∴ According to Amonton,s Law: = = = =1.268 ⟾ P2 = 1.268 P1 ∴ The gas volume is not duplicate according to Amonton,s Law “The gas pressure is proportional to the absolute temperature not with Celsius temperature at constant volume and amount”. 13- The temperature of quantity of gas at constant volume is changed from 100℃ to 200 K. Shall the gas pressure is increased and why? (True) (False) View Answer Answer: [False] Hint and Solution: [False] ∵ The gas quantity and volume are constant. ∴ According to Amonton,s Law: = = = = 0.536 ⟾ P2 = 0.536 P1 ∴ The gas volume is not duplicate according to Amonton,s Law "The gas pressure is proportional to the absolute temperature not with Celsius temperature at constant volume and amount". 14- What is meant by the molar volume for a gas? What is its value at standard conditions (STP)? The molar gas volume is the volume of one mole of the gas; it was found it is equal to 22.4 liter at standard conditions of temperature and pressure (STP). 10 15. What is the volume of gas at STP, if it occupies a volume 250 cm3 at room temperature under pressure 85 kPa? (a) 393 cm3 (b) 164 cm3 (c) 205 cm3 (d) 192 cm3 View Answer Answer: [d] Hint and Solution: [d] STP means that: Pressure = 1 atm & temperature = 0.0℃ ∵ The quantity (n) is constant. T1= 0.0 + 273 = 273 K & T2 = 25 + 273 = 298 K P1= 101.325 kPa & P2 = 85 kPa V1 =? & V2 = 250 cm3 ∴ According to state equation: = V1 = V 2 x = 250 cm3 x = 192 cm3 16. A sample of gas under pressure 0.814 atm in container 300 cm3 at temperature 20℃. What is the pressure for the same quantity of gas in container 500 cm3 at temperature 50℃? (a) 0.838 atm (b) 0.538 atm (c) 0.735 atm (d) 0.433 atm View Answer Answer: [b] Hint and Solution: [b] ∵ The quantity (n) is constant. T1= 20 + 273 = 293 K & T2 = 50 + 273 = 323 K P1= 0.814 atm & P2 =? V1 = 300 cm3 & V2 = 500 cm3 ∴ According to state equation: = P2 = P 1 x = 0.814 atm x = 0.538 atm 11 17. It was found that the pressure of certain quantity of gas in container its volume is 4 L under pressure 3 atm and temperature 30℃. The gas is compressed to 1.5 atm in container its volume 7.5 L. What will be the temperature in this case? (a) 11℃ (b) 31℃ (c) 14℃ (d) 53℃ View Answer Answer: [a] Hint and Solution: [a] ∵ The quantity (n) is constant. T1= 30 + 273 = 303 K & T2 =? P1 = 3 atm & P2 = 1.5 atm V1 = 4 L & V2 = 7.5 L ∴ According to state equation: = T2 = T1 x = 303 K x = 284 K T K = t ℃ + 273 ⟾ t ℃ = T K – 273 = 284 – 273 = 11℃ 18. Calculate the volume of ammonia gas (NH3) its weight is 6.85 g at temperature 24℃ under pressure 750 mm Hg. (a) 7.844 L (b) 9.944 L (c) 43.655 L (d) 17.343 L View Answer Answer: [b] Hint and Solution: [b] The molecular weight of ammonia gas = 17 g mol-1 PV = nRT ⟾ PV = RT K K V= = = 9.944 L 19. Find the number of moles for carbon monoxide gas CO(g) in 500 ml container at temperature 50℃ under pressure 1.5 atm. Calculate the gas mass? (a) 0.038 mol (b) 0.028 mol (b) 0.045 mol (b) 0.022 mol View Answer Answer: [b] Hint and Solution: [b] The molecular weight of carbon monoxide (CO) = 28 g mol-1 PV = nRT n= = = 0.028 mol K K 20. What is the volume of 10 g of carbon dioxide gas CO2(g) at temperature 27℃ and pressure 2.0 atm. (a) 2.792 L (b) 2.891 L (c) 2.393 L (d) 2.446 L View Answer 12 Answer: [a] Hint and Solution: [a] The molecular weight of carbon dioxide (CO2) = 44 g mol-1 PV = RT K K V= = = 2.792 L 21. A mixture of two gases consists of 2 g of hydrogen gas (H2) and 8 g of oxygen gas (O2). Compare between their volumes at the same conditions of pressure and temperature (or at STP). (a) 2 (b) 4 (c) 6 (d) 8 View Answer Answer: [b] Hint and Solution: [b] At the same conditions of pressure and temperature, Avogadro law can be applied: = ⟾ = = = = =4 ∴ =4 22. A mixture of two gases consists of 2 g of helium gas (He) and 8 g of oxygen gas (O2). Compare between their volumes at the same conditions of pressure and temperature (or at STP). (a) 2 (b) 4 (c) 6 (d) 8 View Answer Answer: [a] Hint and Solution: [a] At the same conditions of pressure and temperature, Avogadro law can be applied: = ⟾ = = = = =2 ∴ =2 23- A mixture of two gases, consists of 2 g of helium gas (He) and 8 g of sulpher dioxide gas (SO2). Compare between their volumes at the same conditions of pressure and temperature (at STP). (a) 2 (b) 4 (c) 6 (d) 8 View Answer 13 Answer: [b] Hint and Solution: [b] At the same conditions of pressure and temperature, Avogadro law can be applied: = ⟾ = = = = =4 ∴ VHe = 4 24. The pressure of quantity of nitrogen gas in a container its volume is 1.0 L is 2.0 atm at temperature 20℃ while the pressure of another quantity of oxygen gas in container its volume is 5.0 L under pressure 1.5 atm and temperature 23℃. The two quantities are collected in a container its volume is 10 L. How much will be the total pressure of the mixture in this container at temperature 25℃? (a) 0.89 atm (b) 0.1 atm (c) 0.93 atm (d) 0.95 atm View Answer Answer: [d] Hint and Solution: [d] Ptotal V = ntotal RT = = = 0.08 mol K K = = = 0.309 mol K K ntotal = + = (0.08 + 0.309) mol = 0.389 mol Ptotal V = ntotal RT ⟾ Ptotal = K K Ptotal = = 0.9506 atm 25. Flask has a capacity of 2000 cm3 contains n mole of nitrogen gas (N2) under pressure 38 cmHg and temperature T K. At addition of 0.32 g of oxygen gas (O2), it was necessary to lower the temperature to 10℃, in order to get the same last pressure. Calculate both of temperature (T). (a) ~ 98.77℃ (b) ~ 97.34℃ (c) ~ 96.55℃ (d) ~ 95.52℃ View Answer Answer: [c] Hint and Solution: [c] V= = 2.0 L & P= atm = 0.5 atm T = 10 + 273 = 283 K & n* = = 0.01 mol PV = (n + n*) RT 14  0.5 atm x 2 L = (n + 0.01 mol) x 0.082 L atm mol-1 K-1 x 283 K 1 = (n + 0.01 mol) x 23.206 mol-1 n + 0.01 mol = mol = 0.0431 mol n = (0.431 – 0.01) mol = 0.033 mol PV = nRT 0.5 atm x 2 L = 0.033 mol x 0.082 L atm mol-1 K-1 x T 1 = 0.002706 K-1 x T T= K = 369.549 K Temperature = 369.549 – 273 = 96.549℃ 26. Flask has a capacity of 2.0 L contains (n) mole of nitrogen gas (N2) under pressure 0.5 atm and temperature T K. At addition of 0.01 mole of nitrogen gas (N2), it was necessary to lower the temperature to 10℃, in order to get the same last pressure. Calculate both of temperature. (a) ~ 98.77℃ (b) ~ 97.34℃ (c) ~ 96.55℃ (d) ~ 95.52℃ View Answer Answer: [c] Hint and Solution: [c] PV = (n + n*) RT 0.5 atm x 2 L = (n + 0.01 mol) x 0.082 L atm mol-1 K-1 x 283 K 1 = (n + 0.01 mol) x 23.206 mol-1 n + 0.01 mol = mol = 0.0431 mol n = (0.431 – 0.01) mol = 0.033 mol PV = nRT 0.5 atm x 2 L = 0.033 mol x 0.082 L atm mol-1 K-1 x T 1 = 0.002706 K-1 x T T= K = 369.549 K Temperature = 369.549 – 273 = 96.549℃ 27. A 10 liter vessel contained 14 g of nitrogen gas (N2), 0.4 g hydrogen gas (H2) and 8 g of oxygen gas (O2) and at temperature 27℃. What is the total pressure of the vessel? (a) ~ 2.14 atm (b) ~ 2.24 atm (c) ~ 2.34 atm (d) ~ 2.44 atm View Answer Answer: [c] Hint and Solution: [c] = = = 0.5 mol = = = 0.2 mol = = = 0.25 mol ntotal = + + = 0.5 + 0.2 + 0.25 = 0.95 mol 15  PV = nRT ⟾ P= P= = 2.337 atm = = 1.23 atm = = 0.492 atm = = 0.615 atm Ptotal = + + Ptotal = 1.23 + 0.492 + 0.615 = 2.337 atm 21. If we take equal masses of oxygen gas O2, hydrogen gas H2 and methane gas CH4 at the same conditions of temperature and pressure. Calculate the ratio between their volumes. ∵ Gas volume is directly proportional with its number of moles at constant temperature and pressure. ∴ To find the ratio between the gases volume, we can apply the universal gas equation as following: PV = RT = = = 32 g mol-1 & = 16 g mol-1 & = 2 g mol-1 =. =. (1) =. =. (2) =. =. (3) By dividing equation (1) to (2) taking in our consideration the equality of the mass of gases, we get: = = = : = 16 : 1 By dividing equation (3) to equation (2): = = = : = 8 : 1 ∴ The volume ratio between the three gases is as following: H2 : CH4 : O2 16 : 8 : 1 16  2. Liquid State We have seen that the molecules in a gas are in constant random motion. The spaces between them are large and the intermolecular attractions negligible. However, in a liquid the molecules are in contact with each other. The forces of attraction between the molecules are strong enough to hold them together. All the same, the molecules are able to move past one another through available intermolecular spaces. The molecules in a liquid move in a random fashion. At any instant, the molecules may form clusters, leaving vacant space or "hole" here and there. A molecule may be defined as: a collection of molecules held close to each other and executing random motion through intervening spaces. Most of the physical properties of liquids are actually controlled by the strengths of intermolecular attractive forces. Therefore, before discussing the properties of liquids, we must have a look at the nature of intermolecular forces. 2.1. PROBERTIES OF LIQUIDS 2.1.1. VAPOUR PRESSURE When a liquid is placed in an open vessel, it evaporates. The molecules in the liquid are moving with different kinetic energies. The molecules that possess above-average kinetic energies can overcome the intermolecular forces that hold them in the liquid. These energetic molecules escape from the liquid surface as vapour. The process by which molecules of a liquid go into the gaseous state (vapours) is called Vaporization or Evaporation. The reverse process whereby gas molecules become liquid molecules is called Condensation. If the liquid is placed in a closed vessel Figure (2.1), the molecules with high kinetic energies escape into space above the liquid. As the number of molecules in the gas phase increases, some of them strike the liquid surface and are recaptured (condensation). A stage comes when the number of molecules escaping from the liquid is equal to the number of molecules returning to the liquid. In other words, the rate of evaporation exactly equals the rate of condensation. Thus, a dynamic equilibrium is established between the liquid and the vapour at the given temperature. Liquid ⇌ Vapour 17 Now the concentration of the vapour in the space above the liquid will remain unchanged with lapse of time. Hence, the vapour will exert a definite pressure at the equilibrium. The vapour pressure of a liquid is defined as: the pressure exerted by the vapour in equilibrium with the liquid at a fixed temperature. Figure (2.1) Illustration of vapour pressure The vapour pressures of various liquids differ considerably, depending upon the identity of the liquid with its particular intermolecular forces. Thus, ethanol having weaker hydrogen bonding than water evaporates faster than water. Hence, we expect that ethanol will have higher vapour pressure than water at a given temperature. As shown by the actual plot vapour pressure versus temperature, the vapour pressures of ethanol and water at 60℃ are about 350 torr and 150 torr respectively. 2.1.1.1. Determination of Vapour Pressure The vapour pressure of a given liquid can be measured by Static method or Dynamic method. 2.1.1.1.1. Barometric Method 18 Vapour pressure may be measured by inserting some of the liquid over mercury in the barometer tube using convolutedly pipette. The difference in the mercury levels before and after insertion of the liquid is a measure of its vapour pressure. Figure (2.3) illustrate the determination of the vapour pressure of water at 20℃. As indicated the vapour pressure of water at this temperature is equal to 17.4 mm Hg. Enough water must be added so that a few drops remain in the liquid state. This is necessary to ensure a condition of equilibrium between the liquid and its vapour. Figure (2.3) The barometric method for measuring the liquid vapour pressure Experimental determination of vapour pressure of the liquid 1- Fill some evacuate closed glass tubes from one side and open from another side with length 100 cm with mercury. Then reverse the tubes into a basin filled with mercury. Consequently the mercury becomes lower for 76 cm Hg and keeps a vacuum over the mercury surface. Therefore, there is no practitioner pressure on mercury column, whereas the mercury pressure is 76 cm Hg at room temperature 25℃. 2- Insert the different liquids to the different mercury columns using convolutely pipette or convolutedly dropper with a certain constant volume. According to the difference in density of the 19 liquids and the mercury density, the all liquids becomes over the mercury surface. The liquid will be vaporized and condensed again till dynamic equilibrium is happened between the liquid state and vapour state for each liquid. 3- Measure the depression of mercury level for the three liquids than the standard level, which is without liquid. 4- The depression than the standard level of mercury is equal to the vapour pressure of the liquid. Figure () Determination of vapour pressures of liquids at temperature 25℃, by interdicting the liquids into the barometer using convolutedly pipette 2.1.1.1.2. The Direct (Static) Method A simplified apparatus used for the static method is shown in Figure (2.4). A sufficient amount of the liquid whose vapour pressure is to be determined is placed in the bulb connected to a mercury manometer and a vacuum pump. All the air from the bulb is removed by working the vacuum pump and the stopcock closed. A part of the liquid evaporates. The system is then maintained at a fixed temperature for enough time so that the equilibrium is established. The difference in the levels of mercury in the manometer is equal to the vapour pressure of the liquid. By adjusting the thermostat at a different temperature, the vapour pressure of the liquid at another temperature can be determined. This method is used for liquids having vapour pressures up to one atmosphere. 20  Figure (2.4) Determination of vapour pressure by Static method 2.1.1.1.3. The Dynamic Method The apparatus used for the dynamic method is illustrated in Figure (2.5). An inert gas is passed through the given liquid at a constant temperature (T). The gas saturated with the vapour of the liquid leaves the flask at the exit tube. Figure (2.5) Determination of vapour pressure by Dynamic method If V be the volume of the gas passed and the loss in weight of the liquid, the vapour pressure is given by the expression Vapour pressure = x (2.1)  Notice that: This method is so suitable for liquids of very low vapour pressure. 21 Example 2.1: At passing 30 L of dry argon gas (Ar) through 50 g of tetra bromide carbon (CBr4) at temperature 27℃, the remaining weight of the liquid at equilibrium is 48.175 g. Calculate the vapour pressure of tetra bromide carbon. (a) 2.5 x 10-3 atm (b) 3.5 x 10-3 atm (c) 4.5 x 10-3 atm (d) 5.5 x 10-3 atm View answer Answer: [c] Hint and Solution: [c] ( ) = 12 + (4 x 80) = 332 g/mol ( ( )) = 50 – 48.175 = 1.825 g Vapour pressure = x ( ) = = 4.5 x 10-3 atm Example 2.2: Calculate saturated vapour pressure of water if you know that when 100 liter of dry helium gas is passed through 65.44 g water and left behind of it 63.13 g of water. (a) 0.038 atm (b) 0.036 atm (c) 0.034 atm (d) 0.032 atm View answer Answer: [d] Hint and Explanation: [d] ( ) = (2 x 1) + 16 = 18 g/mol ( ( )) = 65.44 – 63.13 = 2.31 g Vapour pressure = x ( )= = 0.032 atm 2.1.1.3. Effect of Vapour Pressure on Boiling Points When a liquid is heated, tiny bubbles are formed in it. These rise to the liquid surface and burst. The temperature at which it happens is the boiling point of the liquid. Let us consider an individual bubble. The liquid vaporizes into it and the vapour pressure in the bubble keeps it in form. However, the pressure of the atmosphere exerted on the liquid top tends to collapse the bubble. As the bubble goes to the surface, the vapour pressure in the bubble equals the atmospheric pressure. Thus, the bubble collapses. The boiling point of the liquid may, therefore, be defined as the temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure. 22  Figure (2.6) A liquid boils when the pressure of the vapour within the bubble equals the atmospheric pressure exerted on the bubble at the liquid surface Because the atmospheric pressure varies with altitude and other conditions, the boiling points are reported at 760 torr (1 atm). Therefore, the normal boiling point of a liquid is the temperature at which the vapour pressure of the liquid is 760 torr or 1 atm. As evident from Fig. 2.6, the boiling point of ethanol is 78℃ and of water, 100℃. The boiling point of a liquid can be lowered by reducing the external pressure by vacuum pump. Then the vapour pressure of the liquid is equal to the external pressure at a lower temperature. The boiling point of a liquid can be increased by raising the external pressure. Thus, the vapour pressure of the liquid is equal to the external pressure at a higher temperature. A domestic pressure cooker works on this principle. The pressure inside the cooker is maintained above one atmosphere and the liquid contained in it would boil at a higher temperature than 100℃. Thus, the food is cooked in a shorter time. Boiling point of the liquid (Tb): The temperature which the saturated vapour pressure is equal to the pressure above the liquid surface, if it is equal to 1 atm, then the boiling point is called normal boiling point. The specific vaporization heat (ΔℓV): it is the quantity of heat by calories which is necessary to vaporize one mole of the liquid at normal boiling point and normal atmospheric pressure and its unit is cal mol-1. It is clear that the boiling point of the liquid is dependent on the external pressure above the liquid. The boiling point of water is 100℃ under pressure 1.0 atm, while it is decrease so 23 much in the places higher than the sea level (for example at the top of mountain) according to lowering of atmospheric pressure at mountain top. Example 3: arrange the following liquids according to their boiling degree at pressure 1.0 atm. Liquid water Acetone Ether Acetic acid Ethyl alcohol Vapour pressure atm 0.023 0.243 0.582 0.016 0.058 Solution: Acetic acid > water > Ethyl alcohol > Acetone > Ether Clausius-Clapeyron Equation From thermodynamic relation, it is known that: = (1) ( ) where, ΔH is the molar vaporization heat, Vℓ is the volume of one mole of the liquid, Vvap is the volume of one mole of vapours of the liquid at boiling point and T is the absolute temperature. ∵ (Vℓ) is so small in comparison with the volume which is occupied with one mole of the vapours (Vvap). ∴ (Vℓ) can be neglected and equation (1) becomes as following: = (2) From ideal gas equation: PV = nRT For one mole of vapour (gas): V = ∴ by compensation of the value of V in equation (2), we get: = (3) By separation of variables and integration of both sides of equation. ∫ = ∫ (4) ln P = constant – (5) Assuming the saturated vapour pressure of the liquid are P1 and P2 at absolute temperature T1 and T2 respectively. ln P1 = constant – (6) ln P2 = constant – (7) Subtracting equation (7) from equation (6), we get ln P2 – ln P1 = – + = ( ) 24  ln = ( ) (8) log = ( ) (9) Equation (8) is called Clausius-Clapeyron Equation which illustrates the influence of the absolute temperature on the vapour pressure, where P2 and P1 are the saturated vapour pressure of the liquid at absolute temperatures T2 and T1 respectively. (where P2 > P1 and T2 > T1). ΔHvap is the molar vaporization heat. ΔHVap. = ΔℓV x M (1) Where M is the molecular weight and R is the gas constant. Example 1: What is temperature degree the bromine (Br2) is heated to start boiling at pressure 0.75 atm, knowing that the boiling point of bromine is 58℃ and the specific heat of vaporization is 46.4 cal g-1 and the atomic weight of Br is 80 g/mol. (a) ~ 46℃ (b) ~ 48℃ (c) ~ 50℃ (d) ~ 52℃ View answer Answer: [c] Hint and Explanation: [c] The molecular weight of bromine (Br2) = 2 x 80 = 160 g/mol T1 ? P1 0.75 atm T2 58℃ P2 = 1 atm ΔHV = ΔℓV x M = 160 g mol x 46.4 cal. g-1 = 7424 cal mol-1 -1 log = ( ) T2 = 28 + 273 = 331 oK T1 = ? & T2 = 58 + 273 = 331 K P1 = 0.75 & P2 = 1 atm T2 > T1 & P2 > P1 ( ) K log = K (( )K ) ( ) 0.124938 = 1620.56 ( ) ( ) 0.0255 T1 = 331 – T1 ⇛ 0.0255 T1 + 1T1 = 331 1.0255 T1 = 331 ⇛ T1 = 322.8 K ∴ t1 ℃ = 322.8 – 273 = 49.8℃ The temperature of boiling of bromine is 49.8℃ Student Exercises 1. Calculate the pressure of methanol (CH3OH) at temperature 50℃, if you know that its boiling point is 64.7℃, its specific vaporization heat is 263 cal g-1 and its molecular weight is 32 g/mol. 25 (a) 0.565 atm (b) 0.576 atm (c) 0.587 atm (d) 0.598 atm View answer Answer: [a] Hint and Explanation: [a] ΔHV = ΔℓV x M = 263 cal g-1 x 32 g mol-1 = 8416 cal mol-1 log = ( ) T1 = 50 + 273 = 323 K & T2 = 28 + 273 = 331 K P1 = ? & P2 = 1 atm T2 > T1 & P 2 > P1 ( )K log = K (( )K ) log = ( ) ⇛ ∴ log = SHFIT log 0.248 = 1.77 ∴ = 1.77 ⇛ ∴ P1 = = 0.565 atm 2. The vapour pressure of mercury at 290℃ is 197.3 mmHg, while it is 305.6 mmHg at 310℃. Calculate its molar vaporization heat. (a) ~ 14500 cal mol-1 (b) ~ 14000 cal mol-1 (c) ~ 15000 cal mol-1 (d) ~ 15500 cal mol-1 View answer Answer: [b] Hint and Explanation: [b] log = ( ) T1 = 290 + 273 = 563 K & T2 = 310 + 273 = 583 K P1 = 197.3 mm Hg & P2 = 305.6 mm Hg T2 > T1 & P 2 > P1 ( )K log = K (( )K ) log 1.5489 = ( ) 0.1900 = ( ) ΔHV = =14126.1 cal mol-1 26 3. The boiling point of normal butane C4H10 is 0.6℃ and its specific vaporization heat is 96.6 cal g-1. What is its vapour pressure at 20℃? (a) 1.958 atm (b) 1.858 atm (c) 1.758 atm (d) 1.658 atm View answer Answer: [a] Hint and Explanation: [a] The molecular weight of butane = 58 g mol-1 ΔHV = ΔℓV x M = 96.6 cal g-1 x 58 g mol-1 = 5602.8 cal mol-1 log = ( ) T1 = 0.6 + 273 = 273.6 K & T2 = 20 + 273 = 293 K P1 = 1 atm & P2 = ? T2 > T1 & P2 > P1 ( )K log = K (( )K ) log P2 = 0.2958 SHIFT log 0.2958 = 1.958 ⟾ ∴ P2 = 100.2958 = 1.958 atm 4. Calculate the boiling point of water at the top of mountain, if you know that the vapour pressure of water at the top of the mountain is 550 mmHg and its specific vapour pressure is 540 cal/mol. (a) 97.06℃ (b) 94.06℃ (c) 93.06℃ (d) 91.06℃ View answer Answer: [d] Hint and Explanation: [d] The molecular weight of water = 18 g mol-1 ΔHV = ΔℓV x M = 540 cal g-1 x 18 g mol-1 = 9780 cal mol-1 log = ( ) T1 = ? & T2 = 100 + 273 = 373 K P1 = 550 mm Hg & P2 = 760 mm Hg T2 > T1 & P2 > P 1 ( )K log = K (( ) K ) log 1.3818 = 2133.987 ( ) 0.140445 = 2133.987 ( ) 6.58 x 10-5 = ( ) 0.024548 T1 = 373 - T1 0.024548 T1 + T1 = 373 1.024548 T1 = 373 ∴ T1 = 364.06 K ⟾ t = 91.06℃ 27 5. If you know that the vapour pressure of benzene is 75 mmHg at temperature 20℃ and 118 mmHg at 30℃. What is the molar vaporization heat of benzene in unites of (cal mol-1)? (a) ~ 8639 (b) ~ 8457 (c) ~ 8220 (d) ~ 8007 View answer Answer: [d] Hint and Explanation: [d] log = ( ) T1 = 20 + 273 = 293 K & T2 = 30 + 273 = 303 K P1 = 75 mm Hg & P2 = 118 mm Hg T2 > T1 & P2 > P1 ( )K log = K (( )K ) 0.1968 = ( ) ∴ ΔHV = 8007.23 cal mol-1 2.1.2. SURFACE TENSION This property of liquids arises from the intermolecular forces of attraction. A molecule in the interior of a liquid is attracted equally in all directions by the molecules around it. A molecule in the surface of a liquid is attracted only sideways and toward the interior. The forces on the sides being counterbalanced the surface molecule is pulled only inward the liquid. Thus, there is a tendency on the part of the surface molecules to go into the bulk of the liquid. The liquid surface is, therefore, under tension and tends to contract to the smallest possible area in order to have the minimum number of molecules at the surface. It is for this reason that in air, drops of a liquid assume spherical shapes because for a given volume a sphere has the minimum surface area. The surface tension (γ) is defined as: the force in dynes acting along the surface of a liquid at right angle to any line 1 cm in length. 28  Figure 2.7 Figure 2.8 Surface tension is caused by the net The inward forces on inward pull on the surface molecules the surface molecules minimize the surface area and form a drop. Units of Surface Tension As included in the above definition the unit of surface tension in CGS system is dynes per centimeter (dyne cm–1). In SI system, the unit is Newton per meter (Nm–1). Both these units are related as: 1 dyne cm–1 = 1 m Nm–1 2.1.2.1. Relation between contact angle and surface tension The topic of wetting has received tremendous interest from both fundamental and applied points of view. It plays an important role in many industrial processes, such as oil recovery, lubrication, liquid coating, printing, and spray quenching. In recent years, there has been an increasing interest in the study of superhydrophobic surfaces, due to their potential applications in, for example, self-cleaning, nanofluides, and electrowetting. Wettability studies usually involve the measurement of contact angles as the primary data, which indicates the degree of wetting when a solid and liquid interact. Small contact angles (> 90°) correspond to low wettability. This chapter will begin with an introduction of the fundamental science behind wetting and contact angle phenomena, followed by a comprehensive description of the various techniques used to measure contact angles, as well as their applications and limitations in terms of the geometric forms of solid samples. Most of the techniques can be classified into two main groups: the direct optical method and the indirect force method. 29 Calculations based on measured contact angle values yield an important parameter—the solid surface tension, which quantifies the wetting characteristics of a solid material. The criteria of calculating solid surface tension based on experimental contact angle values will be discussed. Finally, the most up-to-date contact angle measurement techniques will be presented and discussed. Fig. 2.9. Illustration of contact angles formed by sessile liquid drops on a smooth homogeneous solid surface Fig. 2.10. Surface tension is caused by the unbalanced forces of liquid molecules at the surface 2.1.2.2. Settlement of liquids on solid surface materials: The liquids staying on the solid material with definite angle "θ" which is measured internal the liquid and is called contact angle. When the liquids tend to be spread at the solid surface, the contact angle "" is < 90o, i.e.; (0 – 90o). In this case, the liquid wets the solid material, for example water at glass surface as illustrated in Figure 2.11. 30  Figure 2.11. (a) Illustrates the contact angle "θ" in case of spreading water at glass surface (b) Illustrates the contact angle "θ" in case of spreading water in a capillary tube The water in the capillary tube, the liquid surface take concave upwards shape. Over there, contrariwise liquids which does not wet the solid surfaces, consequently not tend to be spread at surface of solid materials and the contact angle "θ" > 90o, i.e.; between (90o – 180o). Example of this the mercury on glass surface and in a capillary tube, the mercury surface take Convex upwards shape as illustrated in Figure (2.12). Figure 2.12 (a) Illustrates the contact angle "θ" in case of spreading mercury at glass surface (b) Illustrates the contact angle "θ" in case of spreading mercury in a capillary tube 2.1.2.3. Determination of Surface Tension The methods commonly employed for the determination of surface tension are: 2.1.2. 3.1. Capillary-rise Method A capillary tube of radius r is vertically inserted into a liquid. If the liquid has a tendency to dump the glass like water and glass, the liquid rises to a height h and forms a concave meniscus. This rise of the liquid in the capillary tube is argued to difference in applied pressure inside (x) the capillary tube and outside (Y) of it as shown in figure (2.13). As we know that 31 the pressure on the flat surface is more than the concave surface and the curved of the half radius at the point (Y) is too much higher (a = so high). Therefore the liquid rises in the capillary tube to a certain value equal to. The surface tension (γ) acting along the inner circumference of the tube exactly supports the weight of the liquid column. Figure 2.13. Illustrate the difference in applied potential in the flat and concave surface It is clear from figure (A) that, at point such as "x" under the liquid surface directly in the capillary tube in concave face from the surface, the pressure is less than the vapour pressure over the liquid surface with a value of. In the other, hand the pressure at point "y" which the liquid surface is uniform and the half radius of curvature "a" is so large. Consequently, the difference of the pressure for both liquid sides is very low, i.e.; "∆P ≈ 0.0". When the vapour pressure in the space over ―x‖ the same over "y" as in figure (B). The vapour pressure in space over "x" is the same over "y" as illustrated in figure (B). ∴ The pressure at point "y" in more than the pressure at point "x" with a value. Py – Px = (2.2) This difference in pressure will push the liquid surface inside the capillary tube to highest level, The increment of this level will stop when it is equal to the weight of liquid column in the tube and equal to the pressure difference "∆P" , i.e.; Weight of liquid column = pressure difference 32  h g (ρ – ρ-) = (2.3) where: h is the difference height of the liquid in the capillary and liquid surface, (ρ – ρ-) the density of liquid and vapour density and G acceleration of gravity As the vapour density "ρ-" is so small in comparison with the liquid density (ρ-