Molecules in Motion A.the kinetic theory all matter is composed of small particles (atoms, ions, molecules) these small particles are in constant motion -many properties of matter are the result of this motion all collisions between particles are perfectly elastic -there is no change in the total kinetic energy of the colliding particles B. molecules are moving so fast that they collide often

Pressure A. gas molecules collide with each other, but also with their container walls B. when a gas molecule collides with its container, it exerts a force on that container it is the force of collision and the number of collisions (with the walls) that cause gas pressure -this pressure is measured in terms of force per unit area (units)

Air pressure varies from place to place and time to time therefore, scientists have come up with a standard of pressure a. it is representative of the average air pressure at sea level b kPa or 760mmHg c. known as standard atmospheric pressure a pascal (Pa) is a derived SI unit 1 kPa = 7.501mmHg (conversion factor) equal to 1 newton (force) per square meter (unit area) a. N/m2

Measuring Pressure 1.to measure pressure we us an instrument called a manometer 2 types a. “open” type – open to the atmosphere, so you must subtract the gas pressure from atmospheric pressure b. “closed” type – in a closed tube, usually calibrated in millimeters of mercury (mmHg), when used to measure atmospheric pressure is called a barometer we use conversion factors to convert between mmHg and kPa see page 419

Kinetic Theory of Gases A. when the temperature of a gas is raised or lowered, the change in its volume is significant B. a gas has no particular volume, instead it occupies the entire volume of its container C. an ideal gas is a gas composed of molecules with mass but no volume and no mutual attraction between molecules D. the volume, number of particles, pressure and temperature for a gas are all variables that depend on each other i. therefore when we discuss gases we have to specify not only the volume, but also the temperature and pressure

E. standard atmospheric pressure = kPa (760mmHg) and standard temperature = 0˚C F. we indicate that a gas has been measured at standard conditions (temperature and pressure) by using the letters STP (standard temp & press.) G. gas exerts pressure on the walls of its container because of collisions with the walls i. the pressure exerted depends on 3 factors a. the number of molecules b. volume c. the average kinetic energy, which in turn depends on the temperature d. a change in any of these will change the pressure

Boyle’s Law A. the relationship between pressure and volume (keeping the number of molecules and the average kinetic energy constant) B. states: if the amount and the temperature of a gas remain constant, the pressure exerted by a gas varies inversely with the volume V2/V1 = P1/P2 or V2 = V1P1/ P2 don’t just plug numbers into the equations, visualize the change a. if volume is increased then the pressure will decrease and vice versa

Dalton’s Law for Partial Pressure A. if more than 1 gas occupies a container each gas contributes to the total pressure i. each gas exerts a partial pressure B. states: the total pressure in a container is the sum of all the partial pressures of all the gases in the container C. each gas exerts the same pressure as if it were alone at the same temperature D. Ptot = P1 + P2 + P3... i. if given % of gases and total pressure, to find each partial pressure take the % and convert it to a decimal, then multiply by the total for each gas ii. sometimes we use this to find a total pressure and then use Boyles Law to standardize it

Charles’ Law and Applying it A. volume – temperature relationship (at constant pressure) B. states: the volume of a quantity of gas, held at constant pressure, varies directly with temperature (˚K) i. remember to change to kelvin if given Celsius (+273) ii. V2/V1 = T2/T1 or V2 = V1T2/T1

Combined Gas Law A. this law looks at varying temperature and pressure and solving for volume i. V2 = V1P1T2/P2T1 ii. combines Boyles and Charles’ Laws

Gay-Lussac’s Law A. the relationship between temperature and pressure is direct B. P1/P2 = T1/T2

Diffusion and Graham’s Law A. diffusion is the random scattering of gas molecules to become more evenly distributes B. all gases do not diffuse at the same rate i. the rate of diffusion varies directly with the velocity of the molecules ii. at the same temperature, molecules of smaller masses diffuse faster than molecules of larger masses because they travel faster a. they will effuse (pass through small holes) faster C. Graham’s Law states: the relative rates at which 2 gases, under identical conditions of temperature and pressure, will diffuse vary inversely with the square roots of the molecular masses of the gases i. V1/V2 = √m2/m1 -when the temperatures of the gases are the same and they diffuse in vacuum or into each other

Avogadro’s Principle 1. Molar Volume A. Avogadro’s Principle states: at equal temperatures and pressures, equal volumes of gases contain the same number of molecules i. n represents the number of moles of a gas and V represents volume ii. therefore, for gases at the same temperature and pressure, if V1=V2 then n1=n2 iii. 1 mole (6.02 x 1023 molecules) of any gas at STP will occupy the same volume (22.4dm3 or L) a. this is called the molar volume of the gas at STP

The Ideal Gas Equation A. now we can combine all 4 variables (P, V, T, and n) into one equation i. PV=nRT is the ideal gas equation a. R is a constant, the ideal gas constant -we got the value for it by using all of our standards in the ideal gas equation and solving for R -R = 8.31 dm3 kPa / mol K˚ b. the equation can be rearranged to solve for any variable

Molecular Mass Determination A. we can readjust the ideal gas equation to solve for the molecular masses i. M(molecular mass) = m(mass)RT/PV B. we can also readjust the equation to solve for density i. D=m/V=PM/RT

Gas Stoichiometry 1. We now know that the number of moles relates to the volume of a gas, so we can now use this relationship to perform stoichiometry problems with gases A. remember, in stoichiometry problems we convert our given to moles, then use the mole ratio and finally convert to the units you are asked for

Mass-Gas Volume Relationship (similar to mass-mass) A. follow these steps i. write a balanced equation ii. draw a factor label 4 step grid iii. write the grams given iv. find the number of moles of the given substance (convert grams to moles using molar mass of given) v. use the ratio of the moles given to the moles of required (from the equation, unknown on top and given on bottom) vi. express the moles of gas in terms of volume (convert from moles to dm3 using the conversion factor of 1 mole of a gas = 22.4dm3 at STP)

Gas Volume – Mass Relationships (opposite of mass – gas vol.) A. follow these steps i. write a balanced equation ii. draw a factor label 4 step grid iii. write down the given volume of the gas iv. convert the volume to moles using the conversion factor of 1 mole of a gas = 22.4dm3 at STP) v. use the ratio of the moles given to the moles of required (from the equation, unknown on top and given on bottom) vi. convert moles of required substance to grams using the molar mass of the unknown substance

Volume – Volume Relationship (dm3 to dm3 or L to L) A. follow these steps i. write a balanced equation ii. draw a factor label 3 step grid iii. write down the given volume of the gas iv. use the ratio from the equation (unknown on top and known on bottom), but we can interchange the moles for volume (dm3) a. Example if the mole ratio in the equation is 2 moles to 1 mole, then we can substitute the units for volume in and we have the ratio 2 dm3 to 1 dm3 (this is the ratio we use in our calculation)

-In this chapter we have assumed that gas molecules have no volume and no attraction to each other. This is true only for ideal gases (not real). But, for most common gases the ideal gas laws are accurate to 1%, so we still use them.