# Gas Laws

Topics: Pressure, Ideal gas law, Gas Pages: 56 (17279 words) Published: October 30, 2010
CHAPTER 5 GASES AND THE KINETICMOLECULAR THEORY
5.1 Plan: Review the behavior of the gas phase vs. the liquid phase. Solution: a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger container. b) The volume of the container holding the gas sample increases when heated, but the volume of the container holding the liquid sample remains essentially constant when heated. c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced. The particles in a gas are further apart than those are in a liquid. a) The greater empty space between gas molecules allows gases to be more compressible than liquids. b) The greater empty space between gas molecules allows gases to flow with less resistance (hindrance) than liquids. c) The large empty space between gas molecules limits their interaction, allowing all mixtures of gases to be solutions. d) The large empty space between gas molecules increases the volume of the gas, therefore decreasing the density. The mercury column in the mercury barometer stays up due to the force exerted by the atmosphere on the mercury in the outer reservoir just balancing the gravitational force on the mercury in the tube. Its height adjusts according to the air pressure on the reservoir. The column of mercury is shorter on a mountaintop as there is less atmosphere to exert a force on the mercury reservoir. On a mountaintop, the air pressure is less, so the height of mercury it balances in the barometer is shorter than at sea level where there is more air pressure. The pressure of mercury is its weight (force) per unit area. The weight, and thus the pressure, of the mercury column is directly proportional to its height. When the mercury level in the arm attached to the flask is higher than the level in the other arm, the pressure in the flask is less than the pressure exerted in the other arm. This is an impossible situation for a closed-end manometer as the flask pressure cannot be less than the vacuum in the other arm. Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the densities of the two liquids. Solution: h H2O d Hg = h Hg d H2O h H2O = d Hg d H2O

⎛ 10−3 m ⎞ ⎛ 1 cm ⎞ ⎛ 13.5 g/mL ⎞ xh Hg = ⎜ = 985.5 = 990 cmH2O ( 730 mmHg ) ⎜ ⎟ ⎜ 1 mm ⎟ ⎜ 10−2 m ⎟ ⎟⎜ ⎟ ⎝ 1.00 g/mL ⎠ ⎠ ⎝ ⎠⎝ d Hg d H2O ⎛ 10−3 m ⎞⎛ 1 cm ⎞ ⎛ 13.5 g/mL ⎞ xh Hg = ⎜ = 1019.25 = 1.02 x 103 cmH2O ⎟ ( 755 mmHg ) ⎜ ⎜ 1 mm ⎟⎜ 10−2 m ⎟ ⎟⎜ ⎟ ⎝ 1.00 g/mL ⎠ ⎠ ⎝ ⎠⎝

5.2

5.3

5.4 5.5

5.6

5.7

P (mmH2O) = h H2O =

5-1

5.8

Plan: Use the conversion factors between pressure units: 1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar Solution: ⎛ 760 mmHg ⎞ a) ( 0.745 atm ) ⎜ ⎟ = 566.2 = 566 mmHg ⎝ 1 atm ⎠ ⎛ 1.01325 bar ⎞ b) ( 992 torr ) ⎜ ⎟ = 1.32256 = 1.32 bar ⎝ 760 torr ⎠ ⎛ ⎞ 1 atm c) ( 365 kPa ) ⎜ ⎟ = 3.60227 = 3.60 atm 101.325 kPa ⎠ ⎝ ⎛ 101.325 kPa ⎞ d) ( 804 mmHg ) ⎜ ⎟ = 107.191 = 107 kPa ⎝ 760 mmHg ⎠

5.9

⎛ 10−2 m ⎞ ⎛ 1 mm ⎞ ⎛ 1 atm ⎞ = 1.01083 = 1.01 atm a) ( 76.8 cmHg ) ⎜ ⎜ 1 cm ⎟ ⎜ 10−3 m ⎟ ⎜ 760 mmHg ⎟ ⎟⎜ ⎟ ⎠ ⎠⎝ ⎝ ⎠⎝ ⎛ 101.325 kPa ⎞ 3 3 b) ( 27.5 atm ) ⎜ ⎟ = 2.786 x 10 = 2.79 x 10 kPa 1 atm ⎝ ⎠ ⎛ 1.01325 bar ⎞ c) ( 6.50 atm ) ⎜ ⎟ = 6.5861 = 6.59 bar ⎝ 1 atm ⎠ ⎛ 760 torr ⎞ d) ( 0.937 kPa ) ⎜ ⎟ = 7.02808 = 7.03 torr ⎝ 101.325 kPa ⎠

5.10

Plan: Since the height of the mercury column in contact with the gas is higher than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the pressure corresponding to the height difference between the two arms is subtracted from the atmospheric pressure. Solution: ⎛ 10−2 m ⎞⎛ 1 mm ⎞ ⎛ 1 torr ⎞ = 23.5 torr ( 2.35 cm ) ⎜ ⎜ 1 cm ⎟⎜ 10−3 m ⎟ ⎜ 1 mmHg ⎟ ⎟⎜ ⎟ ⎠ ⎠⎝ ⎝ ⎠⎝ 738.5 torr - 23.5 torr = 715.0 torr ⎛ 1 atm ⎞ P(atm) = ( 715.0 torr ) ⎜ ⎟ = 0.940789 = 0.9408 atm ⎝ 760 torr ⎠ Since the height of the mercury column in contact with the gas is higher than the column in contact with the air,...