1. Physical Chemistry I (TKK-2246) 13/14 Semester 2 Instructor: Rama Oktavian Email: rama.oktavian86@gmail.com Office Hr.: M.13-15, Tu. 13-15, W. 13-15, Th. 13-15, F. 09-11

2. Outlines 1. Gas: Properties 2. Gas laws: Boyle and Charles law 3. Ideal gas law 4. Equation of state

3. Gas state State of matter There are 4 states of matter

4. Gas state State of matter Properties of solid Microscopic view of a solid retains a fixed volume and shape rigid - particles locked into place not easily compressible little free space between particles does not flow easily rigid - particles cannot move/slide past one another

5. Gas state State of matter Properties of liquid Microscopic view of a solid assumes the shape of the part of the container which it occupies particles can move/slide past one another not easily compressible little free space between particles flows easily particles can move/slide past one another

6. Gas state State of matter Properties of gas Microscopic view of a solid assumes the shape and volume of its container particles can move past one another compressible lots of free space between particles flows easily particles can move past one another Microscopic properties

7. Gas state State of matter Properties of gas Microscopic view of a solid Properties that can be observed and measured Macroscopic properties Properties of bulk gases Observable – Pressure, volume, mass, temperature.. How to make relation between those macroscopic properties of gas?? The general form of an equation of state is p=f(T,V,n)

8. Gas state

9. Gas properties Gases Exert Pressure: What is Pressure? Pressure is defined as the force exerted divided by the area it acts over Pressure = Force/Area The SI unit of pressure, the pascal(Pa), is defined as 1 newton per metre-squared: 1 Pa =1 N m −2 1 Pa =1 kg m −1 s −2 1 atm =1.013 25 ×10 5 Pa exactly 1 bar =10 5 Pa

10. Gas properties Gases Exert Pressure: What is Pressure? Pressure is defined as the force exerted divided by the area it acts over Pressure = Force/Area Self-test 1.1 Calculate the pressure (in pascals and atmospheres) exerted by a mass of 1.0 kg pressing through the point of a pin of area 1.0 ×10 −2 mm 2 at the surface of the Earth.

11. Gas properties Pressure measurement Barometer – device that measures atmospheric pressure Invented by Evangelista Torricelli in 1643 the height of the mercury column is proportional to the external pressure

12. Gas properties Pressure measurement Derive an equation for the pressure at the base of a column of liquid of mass density ρ(rho) and height h at the surface of the Earth. p=F/AF = mgm = ρV V = Ahm = ρAhF = mg = ρAhg the pressure is independent of the shape and cross-sectional area of the column.

13. Gas properties Pressure measurement

14. Gas properties Pressure measurement  A manometer measures the pressure of a gas in a container  Gas pressure is the force exerted by the collisions of gas particles with a surface

15. Gas properties Pressure measurement Atmospheric pressure By definition the average pressure at sea level will support a column of 760 mm of mercury. (760 torr) g = 9.81 m.s -2 h = 0.76 m ρ = 13.6 g.cm -3 = 13.6 kg.L -1 = 13.6x10 3 kg.m -3 P = 9.81x0.76x13.6x10 3 = 1.013x10 5 Pa (N.m -2 )

16. Gas properties Pressure measurement Atmospheric pressure problem If we made a barometer out of water, what would be the height of the water column if the pressure is 745 torr?

17. Gas properties Pressure measurement Variation in atmospheric pressure – Changing weather conditions

18. Gas properties Pressure measurement Variation in atmospheric pressure – Changing altitude

19. Gas laws Boyle’s law Boyle’s Law is one of the laws in physics that concern the behaviour of gases When a gas is under pressure it takes up less space: The higher the pressure, the smaller the volume Boyles Law tells us about the relationship between the volume of a gas and its pressure at a constant temperature The law states that pressure is inversely proportional to the volume

20. Gas laws Boyle’s law Pressure-volume relationship pressure-volume behavior of gases were made by Robert Boyle in 1662 P  1/V P x V = constant P 1 x V 1 = P 2 x V 2

21. Gas laws Pressure and Volume: Boyle’s Law Volume and pressure are inversely proportional. –If one increases the other decreases.

22. Gas laws Pressure and Volume: Boyle’s Law Boyle’s Law and Breathing: Inhalation During inhalation,  the lungs expand  the pressure in the lungs decreases  air flows towards the lower pressure in the lungs

23. Gas laws Pressure and Volume: Boyle’s Law Boyle’s Law and Breathing: Inhalation During exhalation lung volume decreases pressure within the lungs increases air flows from the higher pressure in the lungs to the outside

24. Gas laws Pressure and Volume: Boyle’s Law A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg

25. Gas laws Pressure and Volume: Boyle’s Law Guide to Calculations with Gas Laws

26. Gas laws Pressure and Volume: Boyle’s Law Problem A deep sea diver is working at a depth where the pressure is 3.0 atmospheres. He is breathing out air bubbles. The volume of each air bubble is 2 cm 2. At the surface the pressure is 1 atmosphere. What is the volume of each bubble when it reaches the surface?

27. Gas laws Charles’s law Jacques Charles French chemist Jacques Charles discovered that the volume of a gas at constant pressure changes with temperature. As the temperature of the gas increases, so does its volume, and as its temperature decreases, so does its volume. The law says that at constant pressure, the volume of a fixed number of particles of gas is directly proportional to the absolute (Kelvin) temperature

28. Gas laws Charles’s law Volume-temperature relationship Variation of gas volume with temperature at constant pressure V  TV  T V = constant x T V 1 /T 1 = V 2 /T 2

29. Gas laws Charles’s law Volume-temperature relationship  For two conditions, Charles’s law is written V 1 = V 2 (P and n constant) T 1 T 2  Rearranging Charles’s law to solve for V 2 gives T 2 x V 1 = V 2 x T 2 T 1 T 2 V 2 = V 1 x T 2 T 1

30. Gas laws Charles’s law Example problem A balloon has a volume of 785 mL at 21 °C. If the temperature drops to 0 °C, what is the new volume of the balloon (P constant)? STEP 1 Set up data table: Conditions 1 Conditions 2 Know Predict V 1 = 785 mL V 2 = ? V decreases T 1 = 21 °C T 2 = 0 °C = 294 K = 273 K T decreases Be sure to use the Kelvin (K) temperature in gas calculations.

31. Gas laws Charles’s law STEP 2 Solve Charles’s law for V 2 : V 1 = V 2 T 1 T 2 V 2 = V 1 x T 2 T 1 Temperature factor decreases T STEP 3 Set up calculation with data: V 2 = 785 mL x 273 K = 729 mL 294 K

32. Gas laws Charles’s law A sample of oxygen gas has a volume of 420 mL at a temperature of 18 °C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443 °C 2) 170 °C 3) –82 °C

33. Gas laws Avogadro’s law Avogadro’s law states that  the volume of a gas is directly related to the number of moles (n) of gas  T and P are constant V 1 = V 2 n 1 n 2

34. Ideal Gas law The combination of those laws gives Usually written as: R is gas constant

35. Ideal Gas law R is known as universal gas constant Using STP conditions

36. Ideal Gas law R is known as universal gas constant Learning check What is the value of R when the STP value for P is 760 mmHg?

37. Ideal Gas law Problem Learning check A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

38. Ideal Gas law Problem Learning check In an industrial process, nitrogen is heated to 500 K in a vessel of constant volume. If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the working temperature if it behaved as a perfect gas?

39. Equation of state The general form of an equation of state is p=f(T,V,n) Ideal gas equation is equation of state

40. Equation of state P, V, n, T are properties Intensive properties – independent on the quantity of material P, T Extensive properties – dependent on the quantity of material n, V Intensive properties The ratio of any two extensive variables is always an intensive variable

41. Equation of state Expressed in intensive properties gives