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Chemistry

Chemistry. States of matter – Session 1. Session Opener. Session Objectives. Session Objectives. 1. Definition and differences between solids, liquids and gases. 2. Measurable properties of gas: Mass, volume and temperature 3. Boyle’s law 4. Charles’ law 5. Avogadro’s hypothesis

gareth-cummings
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Chemistry

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  1. Chemistry

  2. States of matter – Session 1

  3. Session Opener

  4. Session Objectives

  5. Session Objectives • 1. Definition and differences between solids, liquids and gases. • 2. Measurable properties of gas: Mass, volume and temperature • 3. Boyle’s law • 4. Charles’ law • 5. Avogadro’s hypothesis • Ideal gas equation • Dalton’s law • 8. Amagat’s law of partial volume • 9. Molecular mass of mixture of gases

  6. Solids, Liquids and Gases

  7. (2) Intermolecular forces negligible. Gaseous state-assumptions (1)The molecules are very loosely packed voids (3) Molecules move very rapidly in all directions in a random manner. (4) The molecules collide with one another and also with walls of the container perfectly elastic collisions. (5) The mass, volume and pressure of gases can be calculated.

  8. Measurable Properties of Gases Mass Volume Volume of a gas is the space occupied by itsmolecules and is equal to the volume of the container. 1m = 10 dm = 100 cm 1m3 = 103 dm3 = 106cm3 1L = 1 dm3 = 103cm3

  9. Measurable Properties of Gases Pressure Pressure is measured in terms ofatmospheric pressure, which is the pressure exerted by the atmospheric gases on the surface of the earth. 1 atm = 76 cm of mercury = 760 mm of mercury 1 atm = 101.325 kPa

  10. Measurable Properties of Gases Centigrade or Celsius scale (°C) Temperature Fahrenheit Scale (°F) Kelvin scale (K) 100° C = 180° F –273.15° C = 0K or 0° C = 273.15K and t° C = (t + 273.15) K STP or NTP Temperature = 0° C = 273.15 K = 273 K Pressure = 1 atm = 76 cm = 760 mm = 101.325 kPa

  11. Question

  12. Illustrative example 1 The temperature which is same on Fahrenheit and Celsiusscale is (a) 40 (b) 90(c) 30 (d) none Solution: Let that temperature be x

  13. Boyle’s Law Boyle’s law : (at constant temperature and fixed mass) P1V1=P2V2=constant [at a constant temperature]

  14. Questions

  15. So, Illustrative example 2 A sample of gas occupies 2 L under apressure of 800 atmosphere. What will be its volume if the pressure is decreased to 500 atmosphere. Assume that the temperature of the gas sample does not change? Solution: V2=? V1=2 L P2=500 atm P1=800 atm P1V1=P2V2 According to Boyle’s law

  16. Illustrative example 3 A gas–filled freely collapsible balloon is pushed from the surface level of a laketo a depth of 100m. Approximately what percent of its original volume will balloonfinally have? Assuming that the gas behaves ideally. Solution:

  17. Solution

  18. Charle’s law: (at constant pressure and fixed mass) Charle’s law

  19. Avogadro’s law (at constant pressure and temperature) “equal volumes of all gases contain equal number of molecules under similar conditions of temperature and pressure”.

  20. Do you know? Loschmidt number. It is the number of molecules present in 1cc of a gas or vapour at STP. Its value is2.617 x 1019 per cc

  21. Ideal gas equation Boyle’s law, Charles’ law, Avogadro’s hypothesis, Combining these, we get

  22. Ideal gas equation The gas constant, RUnits of gas constant (R) R = 0.0821 L atm K–1 mole–1 R = 8.314 107 ergs K–1 mole–1 = 8.314 JK–1 mole–1 R = 1.987 cal K–1 mole–1 = 5.1891019 eVK–1mol–1

  23. Do you know? Gas constant per molecule is known as Boltzmann constant (K)

  24. If Wg of a gas of molecular mass M, occupies a volume V, under pressure P at temperature T, then Density Density and molar mass relation Number of moles of gas :

  25. Vapour density [Since molecular weight of H2 is 2] = vapour density of a gas

  26. Dalton’s Law a moles of He

  27. Dalton’s Law b moles of O2

  28. Dalton’s Law c moles of CO2

  29. Dalton’s Law Total gas pressure, Dalton’s law of partial pressure

  30. Question

  31. Illustrative example 4 At 27° C a cylinder of 20 litres capacity contains three gases He, O2 and N2 0.502 g, 0.250 g and 1.00 g respectively. If all these gases behave ideally, calculate partial pressure of each gas as well as total pressure. Solution: Let number of moles of He, O2 and N2 be n1, n2, n3respectively.

  32. Solution n2 = 0.25 g/32.0 g mole–1 = 0.0078 moles of O2 n3 = 1.00 g/28.0 mole–1 = 0.0357 moles of N2 Total number of moles in the gaseous mixture n = 0.1255 + 0.0078 + 0.0357 = 0.169 moles From gas equation, = 0.208 atm

  33. Solution Partial pressure of helium = 0.1545 atm = 0.0096atm = 0.044 atm

  34. Applications of Dalton’s law of partial pressure a. To determine the pressure of a dry gas Pdry gas = pmoist gas – Aqueous Tension (at t° C) b. To calculate partial pressure In a mixture of non-reacting gases, Partial pressure = mole fraction x total pressure

  35. Question

  36. Illustrative example 5 A certain quantity of a gas occupies100 mL when collected over waterat 150C and 750mm pressure. It occupies 91.9 mL in dry state at NTP. Find the aqueous vapour pressure at 150C. Solution: P= aqueous vapour pressure P1(dry gas)=750-p P2=760 mm V1=100 ml V2=91.9 ml T1=15+273=288 K

  37. Solution

  38. = mole fraction Amagat’s Law of Partial Volume V = VA + VB + VC + . . . . . . (in mixture of gases) According to Amagat’s law,

  39. Molecular mass of the mixture of gases For example, air contains 79% nitrogen and 21% oxygen approximately.Now mass of one mole of air would be (0.79 × 28) + (0.21 × 32) = 28.84 gm/mole. = S Molecular mass of a gas × mole fraction of a gas xi (mole fraction of a gas) =

  40. Class exercise

  41. Class exercise 1 A gas at a pressure of 5 atm is heated from 0° C to 546° C and compressed to one-third of its original volume. Hence final pressure is(a) 10 atm (b) 30 atm(c) 45 atm (d) 5 atm Solution: Hence, the answer is (c).

  42. Class exercise 2 What weight of CO2 at STP could be contained in a vessel that holds 4.8 g of O2 at STP(a) 5.5 g (b) 6.6 g(c) 7.7 g (d) 3.3 g

  43. Solution According to ideal gas equation Comparing both equations Hence, the answer is (b).

  44. Class exercise 3 The volume of helium is 44.8 L at (a) 100° C and 1 atm (b) 0° C and 1 atm (c) 0° C and 0.5 atm d) 100° C and 0.5 atm

  45. Solution At STP, volume of 1 mole of helium = 22.4 L New volume = 2 × 22.4 = 44.8 L (Keeping pressure constant) T2 = 546 K = 273° C (Keeping temperature constant) Hence, the answer is (c). P2 = 0.5 atm

  46. Class exercise 4 The density of O2 at NTP will be (a) 1.43 g/L (b) 1.45 g/L (c) 1.55 g/L (d) 1.59 g/L Solution: PV = nRT Hence, the answer is (a).

  47. Class exercise 5 Which of the following gases has the vapour density 14? (a) O2 (b) CO2(c) CO (d) NO Solution: • If vapour density = 14 Hence, the answer is (c).

  48. Class exercise 6 For a particular gas at NTP, the pressure will be _______ at 100° C. (a) 1.2 atm (b) 1.34 atm(c) 1.4 atm(d) 1.9 atm Solution: Hence, the answer is (b).

  49. Let the vapour pressure of water = p mm Pressure of dry gas, P1 = (750 – p) mm V1 = 100 ml P2 = 760 mm T1 = 288 K V2 = 91.9 ml T2 = 273 K Class exercise 7 100 ml of an ideal gas was collected over water at 15° C and 750 mm of Hg pressure. The volume of the dry gas is 91.9 ml at NTP. Find the aqueous vapour pressure at 15° C. Solution:

  50. Solution p = 13.2 mm

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