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The Ideal Gas: Boyles Law

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Objectives n To study and verify Boyles Law. n Illustrate that pressure, P is directly proportional to the inverse height of air, 1/h.

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Equipments n Boyles law apparatus n Barometer

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Theory n There are three major concepts in measuring the behavior of various gases. n The volume V is proportional to the number of moles, n when the temperature and pressure are constant. If we are to double the number of moles, keeping the temperature and pressure constant, the volumes will double. n The volume changes inversely with the pressure when the temperature and quantity of air are constant.

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Theory (contd) n If once again, we double the pressure while the temperature and number of moles are constant, the air is compressed to one half of its initial volume. n The pressure is proportional to absolute temperature for the given amount of gas, when volume is constant. If the absolute temperature is doubled, keeping the volume and number of moles constant the pressure doubles.

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Theory (contd) n These three details are all connected into a single equation which is called the ideal gas equation: pV = nRT(1) where R is the proportionality constant or universal gas constant which in the SI system of units, R = 8.31 J/mole.K where R is the proportionality constant or universal gas constant which in the SI system of units, R = 8.31 J/mole.K

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Theory (contd) n If a constant number of moles or a constant mass of gas of an ideal gas the product nR is constant, so the quantity pV/T is also constant. The subscripts 1 and 2 refer to two states of the same mass of gas, but different pressures, volumes and absolute temperatures. p 1 V 1 = p 2 V 2 (2) T 1 T 2

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Theory (contd) n This equation is known as the ideal gas law which states that the product of pressure and volume of the given mass of gas divided by the absolute temperature of the gas is a constant n If the temperature T 1 and T 2 are the same, then p 1 V 1 = p 2 V 2 = const, T = const, m = const (3) p 1 V 1 = p 2 V 2 = const, T = const, m = const (3) or or pV = const, T = const, m = const (4) pV = const, T = const, m = const (4)

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Theory (contd) n Therefore the product of the pressure and volume of a given mass of a gas at constant temperature is equal to a constant, which has became to be known as the Boyles law, in honor of the British physicist and chemist Robert Boyle. It is also true that the pressure of the given mass of a gas is inversely proportional to the volume at a constant temperature. p is inversely proportional to 1 (5) V

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Theory (contd) n Meaning, if we were to measure the volume of gas for each pressure and graph the pressure of the gas as a reciprocal of its volume, we will derive a straight line. p 1 h 1 = p 2 h 2, T = const, m = const (6) p 1 h 1 = p 2 h 2, T = const, m = const (6) ph = constant, T = const, m = const (7)

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Procedure n Fill the ¾ of the larger cylinder with water. Using the barometer determine the current atmospheric pressure. n Record the length of the small cylinder, L 0 in the data table. n Invert the smaller cylinder and immerse it into the water column to a 100mm depth and measure the immersion depth (D) and height of the water (H) using the scale provided on the small column.

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Procedure (contd) n Record the data in the data table. n Then repeat step 4 for every 100 mm until the small cylinder is at the bottom of the large cylinder.

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Data Analysis Atmospheric pressure, P 0 = 765 mm Hg Height of large cylinder, L 0 = 920 mm Height of large cylinder, L 0 = 920 mm

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Data Analysis (contd)

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Calculations n Calculate the volume of the compressed gas at the depth x, h = L 0 - H h 1 = 920 mm Hg - 7 mm = 913 mm h 1 = 920 mm Hg - 7 mm = 913 mm h 2 = 920 mm Hg - 15 mm = 905 mm h 2 = 920 mm Hg - 15 mm = 905 mm h 3 = 920 mm Hg - 22 mm = 898 mm h 3 = 920 mm Hg - 22 mm = 898 mm h 4 = 920 mm Hg - 30 mm = 890 mm h 4 = 920 mm Hg - 30 mm = 890 mm h 5 = 920 mm Hg - 39 mm = 881 mm h 5 = 920 mm Hg - 39 mm = 881 mm h 6 = 920 mm Hg - 46 mm = 874 mm h 6 = 920 mm Hg - 46 mm = 874 mm h 7 = 920 mm Hg - 54 mm = 866 mm h 7 = 920 mm Hg - 54 mm = 866 mm h 8 = 920 mm Hg - 60 mm = 860 mm h 8 = 920 mm Hg - 60 mm = 860 mm

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Calculations (contd) n Calculate the absolute pressure using the equation P = P 0 + D-H 13.55 13.55 where 13.55 is the density of mercury at room temperature. where 13.55 is the density of mercury at room temperature. P = 765 mm Hg + (100 mm - 7 mm) = 772 mm Hg P = 765 mm Hg + (100 mm - 7 mm) = 772 mm Hg P = 765 mm Hg + (200 mm - 15 mm) = 779 mm Hg P = 765 mm Hg + (200 mm - 15 mm) = 779 mm Hg P = 765 mm Hg + (300 mm - 22 mm) = 786 mm Hg P = 765 mm Hg + (300 mm - 22 mm) = 786 mm Hg P = 765 mm Hg + (400 mm - 30 mm) = 792 mm Hg P = 765 mm Hg + (400 mm - 30 mm) = 792 mm Hg P = 765 mm Hg + (500 mm - 39 mm) = 799 mm Hg P = 765 mm Hg + (500 mm - 39 mm) = 799 mm Hg P = 765 mm Hg + (600 mm - 46 mm) = 806 mm Hg P = 765 mm Hg + (600 mm - 46 mm) = 806 mm Hg

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Calculations (contd) P = 765 mm Hg + (700 mm - 54 mm) = 813 mm Hg P = 765 mm Hg + (700 mm - 54 mm) = 813 mm Hg P = 765 mm Hg + (750 mm - 60 mm) = 816 mm Hg P = 765 mm Hg + (750 mm - 60 mm) = 816 mm Hg

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Questions n Boyles Law assumes that the gas is at constant temperature. What change in pressure might be expected if the gas temperature increased? If the gas temperature increases, the pressure will also increase. If the gas temperature increases, the pressure will also increase. n If 1 liter of gas at a pressure of 20 mm Hg is compressed to a volume of 10cm^3, what will the resulting pressure be? What would the volume be if the desired pressure is 760 mm Hg ( 1 atmosphere )?

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Questions(contd) PV = constant PV = constant 10 mm = 1 cm 10 mm = 1 cm (20 mm Hg)(100 mm^3) = 2000 mm Hg (20 mm Hg)(100 mm^3) = 2000 mm Hg when V = 760 mm Hg when V = 760 mm Hg (760 mm Hg)(100 mm^3) = 76000 mm Hg (760 mm Hg)(100 mm^3) = 76000 mm Hg

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Conclusion In closing, our main goal for this laboratory exercise was achieved. The graph of pressure, P versus inverse height of air, 1/h illustrates that the pressure of a fixed mass of air is directly proportional to the inverse height of air. Boyles law stated that the product of the pressure and volume of a given mass of air at constant temperature is equal to a constant. Our values for Ph was almost constant but we were are slightly off due to experimental errors. In closing, our main goal for this laboratory exercise was achieved. The graph of pressure, P versus inverse height of air, 1/h illustrates that the pressure of a fixed mass of air is directly proportional to the inverse height of air. Boyles law stated that the product of the pressure and volume of a given mass of air at constant temperature is equal to a constant. Our values for Ph was almost constant but we were are slightly off due to experimental errors.

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The End Thanks for listening

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