Gases

The Nature of Gases Regardless of their chemical identity, gases tend to exhibit similar physical behaviors Gas particles can be monatomic (Ne), diatomic (N2), or polyatomic (CH4) – but they all have some common characteristics: Gases have mass. Gases are compressible. Gases fill their containers. Gases diffuse. Gases exert pressure. Pressure is related to temperature

Kinetic Molecular Theory Theory used to explain the behaviors and experimental characteristics of ideal gases – The theory states that the tiny particles in all forms of matter are in continuous motion. There are 3 basic assumptions of the KMT as it applies to ideal gases.

KMT Assumption #1 A gas is composed of small particles. The particles have an insignificant volume and are relatively far apart from one another. There is empty space between particles. No attractive or repulsive forces between particles.

KMT Assumption #2 The particles in a gas move in constant random motion. Particles move in straight paths and are completely independent of each other Particles path is only changed by colliding with another particle or the sides of its container.

KMT Assumption #3 elastic collisions inelastic collisions All collisions a gas particle undergoes are perfectly elastic. They exert a pressure but don’t lose any energy during the collisions. elastic collisions inelastic collisions

Gases have mass. Gases are classified as matter, therefore, they must have mass.

Gases are squeezable The gas particles empty space can be compressed by added pressure giving the gas particles less room to bounce around thus decreasing the overall volume.

Gases are squeezable There are a huge number of applications Storm door closers Pneumatic tube delivery devices Tires Air tanks

Gases fill their containers Gases expand until they take up as much room as they possibly can.

Gases fill their containers The random bouncing motion of gases allows for the mixing up and spreading of the particles until they are uniform throughout the entire container.

Gases diffuse Gases can move through each other rapidly. The movement of one substance through another is called diffusion. Because of all of the empty space between gas molecules, gas molecules can pass between each other until the gases mix uniformly.

Gases diffuse

Gases diffuse

Gases diffuse

Gases diffuse This doesn’t happen at the same speeds for all gases though. Some gases diffuse more rapidly then other gases based on their size and their energy. Diffusion explains why gases are able to spread out to fill their containers. It’s why we can all breathe oxygen anywhere in the room. It also helps us avoid potential odoriferous problems.

Gases exert pressure Gas particles exert pressure by colliding with objects in their path. The definition of pressure is the force per unit area – so the total of all of the tiny collisions makes up the pressure exer- ted by the gas

Gases exert pressure It’s the pressure exerted by the gases that hold the walls of a container out The pressure of gases is what keeps our tires inflated, makes our basketballs bounce, makes hairspray come out of the can, helps our lungs inflate, allow vacuum cleaners to work, etc.

Pressure depends on Temp Temperature measures the average kinetic energy of the particles in an object. Therefore, the higher the temperature the more energy the gas particle has. So the collisions are more often and with a higher force. Think about the pressure of a set of tires on a car.

Pressure depends on Temp Today’s temp: 35°F Pressure Gauge

Pressure depends on Temp Today’s temp: 85°F Pressure Gauge

Measuring Gases Variables that are very important to studying the behavior of gases: Volume: generally in Liters (1L = 1000 mL) Temperature :given in Celsius but must be converted to Kelvin for gas law problems Kelvin = °C + 273 Pressure 1 atm=760 mmHg=760 Torr = 14.7 psi = 101.3 kPa amount generally given in moles

S T P Since the behavior of a gas is dependent on temperature and pressure, it is convenient to designate a set of standard conditions, called STP in order to study gas behavior. Standard Temperature = 0°C or 273K Standard Pressure = 1atm or 760mmHg or 101.3kPa (depending on the method of measure)

Atmospheric Pressure The gases in the air are exerting a pressure called atmospheric pressure Atmospheric pressure is a result of the fact that air has mass and is colliding with everything. Atmospheric pressure is measured with a barometer.

Atmospheric Pressure Atmospheric pressure varies with altitude The lower the altitude, the longer and heavier is the column of air above an area of the earth. Example: Recipes like the back of a cake box that describes how to cook a cake at a higher altitude

Boyle’s Law

Or to the volume if we changed the pressure? Boyle’s Mathematical Law: If we have a given amount of a gas at a starting pressure and volume, what would happen to the pressure if we changed the volume? Or to the volume if we changed the pressure? since PV equals a constant P1V1 = P2V2

Boyle’s Mathematical Law: Ex: A gas has a volume of 3.0 L at 2 atm. What will its volume be at 4 atm? List the variables or clues given: P1 = 2 atm V1 = 3.0 L P2 = 4 atm V2 = ? Plug in the variables & calculate: (2 atm) (3.0 L) = (4 atm) P1V1 = V2 P2 (V2) 1.5 L

Charles’ Law

Or to the temperature if we changed the volume? Charles’s Mathematical Law: If we have a given amount of a gas at a starting volume and temperature, what would happen to the volume if we changed the temperature? Or to the temperature if we changed the volume? since V/T = k = V1 V2 T1 T2

Charles’s Mathematical Law: Ex: A gas has a volume of 3.0 L at 400K. What is its volume at 500K List the variables or clues given: T1 = 400K V1 = 3.0 L T2 = 500K V2 = ? Plug in the variables & calculate: 3.0L X L = 500K 400K 3.8 L

Gay-Lussac’s Law

Or to the temp. if we changed the pressure? Gay-Lussac’s Mathematical Law: If we have a given amount of a gas at a starting temperature and pressure, what would happen to the pressure if we changed the temperature? Or to the temp. if we changed the pressure? since P/T = k P1 P2 T1 T2 =

Gay-Lussac’s Mathematical Law: Ex: A gas has a pressure of 3.0atm at 400K. What is its pressure at 500K? List the variables or clues given: T1 = 400K P1 = 3.0 atm T2 = 500K P2 = ? Plug in the variables & calculate: 3.0atm X atm = 3.8 atm 500K 400K

Combined Gas Law

Combined and ideal gas laws

Avogadro’s Law There is a lesser known law called Avogadro’s Law which relates Volume & moles (n). It turns out that they are directly related to each other. As number of moles increases then Volume increases. V/n = k

Ideal Gas Law PV = R nT PV = nRT If we combine all of the laws together including Avogadro’s Law we get: Where R is the universal gas constant PV nT = R Normally written as PV = nRT

Ideal Gas Constant L•atm R =.0821 mol•K L•mmHg R=62.4 mol•K L•kPa Because of the different pressure units there are 3 possibilities for our ideal gas constant R =.0821 L•atm mol•K If pressure is given in atm If pressure is given in mmHg or torr R=62.4 L•mmHg mol•K R=8.314 L•kPa mol•K If pressure is given in kPa

Practice Use the Ideal Gas Law to complete the following table for ammonia gas (NH3). Pressure Volume Temp Moles Grams 2.50 atm 0C 32.0 768 mmHg 6.0 L 100C

Variations of the Ideal Gas Law We need to know that the unit mole is equal to mass divided by molar mass. PV = nRT n = m/MM

Variations of the Ideal Gas Law We can then use the MM equation to derive a version that solves for the density of a gas. Remember that D = m/V MW = dRT or MW = mRT P VP Kitty cats say meow and they put dirt on their pee!!

Dalton’s Law of Partial Pressure States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. PT=P1+P2+P3+… What that means is that each gas involved in a mixture exerts an independent pressure on its container’s walls

Simple Dalton’s Law Calculation Three of the primary components of air are CO2, N2, and O2. In a sample containing a mixture of these gases at exactly 760 mmHg, the partial pressures of CO2 and N2 are given as PCO2= 0.285mmHg and PN2 = 593.525mmHg. What is the partial pressure of O2?

Simple Dalton’s Law Calculation PT = PCO2 + PN2 + PO2 760mmHg = .285mmHg + 593.525mmHg + PO2 PO2= 167mmHg

Dalton’s Law of Partial Pressure Partial pressures are also important when a gas is collected through water. Any time a gas is collected through water the gas is “contaminated” with water vapor. You can determine the pressure of the dry gas by subtracting out the water vapor

Ptot = Patmospheric pressure = Pgas + PH2O The water’s vapor pressure can be determined from a list and subtract-ed from the atmospheric pressure

WATER VAPOR PRESSURES Temp (°C) (mmHg) (kPa) 0.0 4.6 .61 5.0 6.5 .87 10.0 9.2 1.23 15.0 12.8 1.71 15.5 13.2 1.76 16.0 13.6 1.82 16.5 14.1 1.88 17.0 14.5 1.94 17.5 2.00 18.0 2.06 18.5 2.13

Simple Dalton’s Law Calculation Determine the partial pressure of oxygen collected by water displace-ment if the water temperature is 20.0°C and the total pressure of the gases in the collection bottle is 730 mmHg. PH2O at 20.0°C= 17.5 mmHg

Simple Dalton’s Law Calculation PT = PH2O + PO2 PH2O = 17.5 mmHg PT = 730 mmHg 730mmHg = 17.5 + PO2 PO2= 712.5 mmHg

Mole Fraction Moles of gasx x PT = Px Total moles A mixture of 4.00 moles of O2 and 3.00 moles of H2 exert a total pressure of 760 torr. What is the partial pressure of each gas? 4.00 moles of O2 x 760 torr = 434 torr 7.00 total moles 3.00 moles of H2 x 760 torr = 326 torr

Graham’s Law Thomas Graham studied the effusion and diffusion of gases. Diffusion is the mixing of gases through each other. Effusion is the process whereby the molecules of a gas escape from its container through a tiny hole

Diffusion Effusion

Graham’s Law Graham’s Law states that the rates of effusion and diffusion of gases at the same temperature and pressure is dependent on the size of the molecule. The bigger the molecule the slower it moves the slower it mixes and escapes.

The velocities of two different gases are inversely proportional to the square roots of their molar masses. Rate of effusion of A = Rate of effusion of B MMB MMA

Graham’s Law Example Calc. If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster? Rate of effusion of A = Rate of effusion of B MB MA

Graham’s Law Example Calc. Rate of effusion of He 40 g = Rate of effusion of Ar 4 g Helium is 3.16 times faster than Argon.