The Properties of Gases Chapter 12

Properties of Gases (not in Notes) Gases are fluids… Fluid: (not just to describe liquids)  can describe substances that can “flow.” Gases flow because the molecules are far apart and therefore are able to move past one another. Gases have low density… Because of the distance between molecules, gases have a lower density due to the space between molecules. (Unlike solid and liquid)

Properties of Gases Compressibility… Is the measure of how much the volume of matter decreases under pressure. Gases unlike solids have space between the particles so they are more easily compressible.

Three properties to Gases assumed by the kinetic theory First it assumes that gases consist of hard, spherical particles that are small. They are so small in relation to the distance between them that their individual volumes can be assumed to be insignificant. The large distance between gas particles means that there is considerable space between them leading to the explanation that gases are easily compressible

Three properties to Gases assumed by the kinetic theory Second property of gas particles is that there is no attractive or repulsive forces exist between the particles. Gases are free to move with in their confined spaces A gas will expand until it takes the shape and volume of its container

Three properties to Gases assumed by the kinetic theory Third assumption is that gas particles more rapidly in constant random motion. The particles travel in a straight path and independently of one another. When they collide the kinetic energy is transferred with out lose, therefore the particles are thought to be elastic.

STP STP: Standard Temperature and Pressure Standard Temperature = 0  C or 273K Kelvin Temperature is an alternate form of temperature that can measure the theoretical form of absolute zero and does not use negative temperature. Standard Pressure = 101.3KPa, 1 atm, or 760 mmHg KPa = 1 atm = 760mm Hg

Practice Problem with STP The critical pressure of carbon dioxide is 72.7atm. What is this value in KPa and Pa? 72.7 atm = 1 atmcross multiple x KPa KPa x = 7,364.5 KPa OR… 72.7atm x KPa = 7,364.5 KPa 1 atm

Practice Problem with STP 2 The vapor pressure of water at 50.0  C is 12.33KPa. What is this value in millimeters of mercury? 12.33KPa x 760mm Hg = mm Hg KPa

Gas Laws: Boyle’s Law (P vs V) Boyle’s Law: states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with the pressure. P 1 x V 1 = P 2 x V 2

Sample Question: Boyle’s Law A high-altitude balloon contains 30.0L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (temperature is constant V 2 = 30.0 L x 103kPa = 1.24 x 10 2 L 25.0 kPa

Gas Laws: Charles’s Law (T vs V) Charles’s Law states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. V 1 = V 2 Solve as a proportion T 1 T 2

Sample Questions: Charle’s Law A balloon is inflated in a room at 24°C has a volume of 4.00L. The balloon is then heated to a temperature of 58°C. What is the new volume if the pressure remains constant? V 2 = 4.00L x 331K 297K V 2 = 4.46L

Gas Laws: Gay-Lussac’s Law(T vs P) Gay-Lussac’s Law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant. P 1 = P 2 Solve as a proportion T 1 T 2

Sample Qustn: Gay-Lussac’s Law The gas left in a used aerosol can is at a pressure of 103kPa at 25°C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928°C? P 2 = 103kPa x 1201K 298K P 2 = 4.15 x 10 2 kPa

Gas Laws: Combined Gas Law In the combined gas law, there is nothing left constant P 1 x V 1 = P 2 x V 2 T 1 T 2 Using this equation you can still hold temperature, volume or pressure constant, you just need to know how to maneuver the variables.

Sample Question: Combined Gas The volume of a gas-filled balloon is 30.0L at 40°C and 153kPa pressure. What volume will the balloon have at standard temperature and pressure (STP)? V 2 = 30.0L x 153kPa x 273K 101.3kPa x 313K V 2 = 39.5L

Ideal Gas Law The ideal gas law allows you to solve for the number of moles of a contained gas when P, V and T are know. (R) has the value of 8.31(L x kPa)/(K x mol) P x V =n x R x T

Sample Question:Ideal Gas Law You fill a rigid steel cylinder that has a volume of 20.0L with nitrogen gas (N 2(g) ) to a final pressure of 2.00 x 10 4 kPa at 28°C. How many moles of N 2(g) does the cylinder contain? PV = nRT  rearrange the equation to solve for n n= 2.00 x 10 4 kPa x 20.0L 8.31 x 301K n= 1.60 x 10 2 mol N 2(g)

Sample Question: Ideal Gas Law A deep underground cavern contains 2.24 x 10 4 L of methane gas (CH 4(g) ) at a pressure of 1.50 x10 3 kPa and a temperature of 42°C. How many kilograms of CH 4 does this natura1-gas deposit contain? n=1.50 x 10 3 kPa x (2.24 x 10 4 L) / 8.31 (L x kPa / K x mol) x 315K n= 1.28 x 10 4 mol CH 4 Molar mass= 16.0g CH 4 /mol CH x 10 4 mol CH 4 x 16gCH 4 /mol =20.5 x 10 4 g CH x 10 5 g CH 4 x 1kg/1000g = 2.05 x10 2 kg CH 4

Avogadro’s Hypothesis Equal volumes of gases at the same temperature and pressure contain equal numbers of particles. Basically, saying that two rooms the same size one could be filled with basketballs and the other would be filled with the same number of tennis balls. At STP, 1 mol (6.02 x ) of particles of any gas, regardless of the size of the particles, occupies a volume of 22.4 L

Sample Question: 1 Determine the volume (in L) occupied by mol of a gas at standard temperature and pressure (STP) mol x 22.4L = 4.52 L 1 mol

Sample Question: 2 How many oxygen molecules are in 3.36L of oxygen gas at standard temperature and pressure? 3.36L O 2 x 1 mol O 2 x 6.02 x mlcls O L O 2 1 mol O 2 = 9.03 x mlcls O 2

Sample Question: 3 Determine the volume (in L) occupied by 14.0g of Nitrogen gas at STP. Molar mass N 2 = 2 mol N x 14.0g N 1 mol N Molar mass N 2 = 28 g 14.0g N 2 x 1 mol N 2 x 22.4L N 2 = 11.2 L N g N 2 x 1 mol N 2

Dalton’s Law of Partial Pressure Partial Pressure: the contribution of each gas in a mixture makes to the total pressure that is exerted by the gas. Dalton’s Law of Partial Pressure: at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressure of the component gases P total = P 1 + P 2 + P 3 +…

Sample Question: Dalton’s Law Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (P O2 )at 101.3kPa of total pressure if the partial pressures of nitrogen, carbon dioxide, and other gases are 78.10kPa, 0.040kPa and 0.94kPa, respectively. P O2 = P total –(P N2 + P CO2 + P others ) P O2 = 101.3kPa – (79.10kPa kPa kPa) P O2 = 21.22kPa