Gas Laws Chapter 14

Properties of Gases  Gases are easily compressed because of the space between the particles in the gas.

Properties of Gases  The amount of gas, volume, and temperature affect the pressure of a gas.

Properties of Gases  Doubling the number of particles in the container would double the pressure on a contained gas at constant temperature.  Boyle’s Law P 1 V 1 = P 2 V 2

 A) 50 kPa  B) 33 kPa  C) inverse proportion - as one increases the other decreases

 Reducing the volume of a contained gas to one third, while holding temperature constant, causes pressure to become tripled.

 A gas occupies a volume of 2.50 L at a pressure of kPa. If the temperature remains constant, what volume would the gas occupy at 1750 kPa?  V V 2 = ? V 1 = 2.50 L P 1 = kPa P 2 = 1750 kPa  F P 1 V 1 = P 2 V 2  S (350.0 kPa)(2.50 L) = (1750 kPa)V 2  A V 2 = L

.  If the volume of a container of gas is reduced, the pressure inside the container will increase.  Boyles Law & Temperature 14B

 The graph of several pressure-volume readings on a contained gas at constant temperature would be a curved line.

 If a balloon is squeezed, the pressure of the gas inside the balloon increases.

 Temperature is directly proportional to the average kinetic energy of the particles in a substance.

 As the temperature of the gas in a balloon decreases, the average kinetic energy of the gas decreases.

 Absolute zero is the temperature at which the average kinetic energy of particles would theoretically be zero.  This is the lowest possible temperature. ( °C)

 To get kelvins add 273 to the °C.  To get °C subtract 273 from the kelvins.

 A temperature of -25°C is equivalent to 248 K. [ =248]  A temperature of 295 K is equivalent to 22 °C. [ =22C]

 When the Kelvin temperature of an enclosed gas doubles, the particles of the gas move faster.  The Kelvin temperature must be used when working with proportions.

 Problem:  A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the temperature stays the same but the pressure is only 25.0 kPa?

 V V 2 = ? P 1 =103 kPa V 1 = 30.0 LP 1 =103 kPa V 1 = 30.0 L P 2 = 25.0 kPa V 2 = ?P 2 = 25.0 kPa V 2 = ?  F P 1 V 1 = P 2 V 2  S (103 kPa)(30.0 L) = (25.0 kPa)V 2  A V 2 = 124 L

Charles’ Law Notes 14C   when the temperature is expressed in kelvins.

 At constant pressure, the volume of a fixed mass of gas and its Kelvin temperature are said to be directly related.

 If a balloon is heated, the volume of the air in the balloon increases if the pressure is constant.

 The temperature of 6.24 L of a gas is increased from 125 K to 250 k at constant pressure. What is the new volume of the gas?

 V V 2 =? V 1 = 6.24 L  T 1 = 125 K T 2 = 250. K  F   S   A V 2 = L

 A) kelvins B) increases C) 0 L

Combined Gas Law Notes 14 D  Gay-Lussac’s Law

 If the Kelvin temperature of a gas in a closed container increases the pressure of the gas increases proportionally.

 As the temperature of a fixed volume of a gas increases, the pressure will increase.

 A sample of chlorine gas has a pressure of 9.99 kPa at 27°C. What will its pressure be at 627°C if its volume remains constant?

 V P 2 = ? P 1 = 9.99 kPa  T 1 = 300. K T 2 = 900. K  F   S   A P 2 = kPa

 The Combined Gas Law The combined gas law relates temperature, pressure, and volume.

 If a sample of oxygen occupies a volume of 6.00 L at a pressure of 68.0 kPa and a temperature of 264 K, what volume would this sample occupy at 204 kPa and 528 K?

 V V 2 =? V 1 = 6.00 L  P 1 = 68.0 kPa T 1 = 264 K  P 2 = 204 kPa T 2 = 528 K  F  S  A V 2 = 4.00 L

Ideal Gas Law Notes 14E  Ideal Gas Law PV = nRT  Where n = the number of moles R is the Ideal Gas Constant R is the Ideal Gas Constant  The ideal gas law can be used to calculate the number of moles of a contained gas.

 What is the volume (in L) that would be occupied by 1.00 mol of O 2 at STP?  V V = ? n = 1.00 mol T = 273K P = kPa  F PV = nRT  S (101.3 kPa)V = (1.00 mol) (273 K)  A Vol = 22.4 L

 How many moles of H 2 would be contained in 83.1 L of the gas at 137 kPa and 1.0°C?

 V n = ? V = 83.1 L P = 137 kPa T = 1.0°C = 274 K   F PV = nRT  S (137 kPa)(83.1 L) = n (274. K)  A n = 5.00 mol

Gas Mixtures 14 F  Dalton’s Law - In a mixture of gases, the total pressure is the sum of the partial pressures of the gases.  P Total = P 1 + P 2 + P 3 + …

 A sample of H 2 is collected over water such that the combined hydrogen- water vapor sample is held at a pressure of 1 standard atmosphere. What is the partial pressure of the H 2 if that of the water vapor is 2.5 kPa?  1 atm = kPa  kPa – 2.5 kPa = 98.8 kPa

 The tendency of molecules to move toward areas of lower concentration is called diffusion.  The gas propellant in an aerosol can moves from a region of high pressure to a region of lower pressure.

 The process that occurs when a gas escapes through a tiny hole in the container is called effusion.  The substance with the smallest molar mass would have the fastest rate of effusion.  So CH 4 effuses faster than NO 2.  16g/mol 46g/mol