Gas Laws

11.1 The Gas Laws Kinetic Theory Revisited 1. Particles are far apart and have negligible volume. 2. Move in rapid, random, straight-line motion. 3. Collide elastically. 4. No attractive or repulsive forces.

11.1 The Gas Laws Variables That Describe A Gas 1. Pressure (P) kPa 2. Volume (V) L 3. Temperature (T) K ○ T K = T ºC 4. Amount of gas (n) mol

Gases  Many of the chemicals we deal with are gases. They are difficult to weigh.  Need to know how many moles of gas we have.  Two things effect the volume of a gas Temperature and pressure  We need to compare them at the same temperature and pressure.

STP – Not Just a Motor Oil  STP – standard temperature and pressure 0ºC (273 K) 1 atm (101.3 kPa)

11.1 Effect of adding or removing gas Amount of a Gas  As the number of particles increases the pressure increases. Pn

11.2 Effect of changing the size of the container Volume  As volume increases the pressure decreases. PV

11.3 Boyle’s Law Boyle’s Law  At constant temperature and mass (moles), the pressure is inversely proportional to the volume. P 1 V 1 = P 2 V 2

Boyle’s Law Problem A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when rises to an altitude where the pressure is only 25.0 kPa? (assume the temp is constant) P 1 =P 2 = V 1 =V 2 = 103 kPa 30.0 L 25.0 kPa ? P 1 V 1 = P 2 V 2 P2P2 P2P2 P1V1P1V1 P2P2 V 2 = = (103 kPa)(30.0 L) 25.0 kPa = L 124 L

11.5 Charle’s Law Charles’s Law  At constant pressure and mass (moles), the volume of a gas is directly proportional to the Kelvin temperature. V 1 V 2 T 1 T 2 =

Charles’s Law Problem A balloon inflated in a room at 24 °C has a volume of 4.00 L. The balloon is then heated to a temperature of 58 °C. What is the new volume if pressure is constant? T 1 =T 2 = V 1 =V 2 = 24 °C 4.00 L 58 °C ? Temperature MUST be in Kelvin (add 273) 297 K331 K V 1 V 2 T 1 T 2 = T2T2 T2T2 V2V2 = T2V1T2V1 T1T1 = (331 K)(4.00 L) (297 K) = L 4.46 L

11.6 The effect of changing Temperature Temperature  As the temperature increases the pressure increases. PT

11.6 Temperature/Pressure Changes  At constant volume and mass (moles), the pressure of a gas is directly proportional to the Kelvin temperature. P 1 P 2 T 1 T 2 =

Problem The gas left in a used aerosol can is at a pressure of 103 kPa at 25 °C. If the can is thrown onto a fire and the pressure increases to 415 kPa, what is the temperature? P 1 =P 2 = T 1 =T 2 = 103 kPa 25 °C 415 kPa ?298 K P 1 P 2 T 1 T 2 = P2P2 P2P2 Do we want to solve for ? 1 T2T2

11.7 Combined Gas Law The 3 gas laws just discussed can be combined into a single expression: Combined Gas Law P1P1 V1V1 T1T1 = P2P2 V2V2 T2T2

Problem The gas left in a used aerosol can is at a pressure of 103 kPa at 25 °C. If the can is thrown onto a fire and the pressure increases to 415 kPa, what is the temperature? P 1 =P 2 = T 1 =T 2 = 103 kPa 25 °C 415 kPa ?298 K P 1 P 2 T 1 T 2 = T2T2 T2T2 P1P1 P1P1 T1T1 T1T1 T2T2 = P2P2 T1T1 P1P1 = (415 kPa)(298 K) 103 kPa = K 1.20  10 3 K

11.8 The Ideal Gas Law The ideal gas law is used to consider the amount of gas in a system Ideal Gas Law PV = nRT P = pressure (kPa, atm, or mmHg) V = volume (L) n = moles (mol) T = temperature (K) R = ideal gas constant LkPa mol K Latm mol K or LmmHg mol K or 62.4 R = 8.31

11.8 The Ideal Gas Law ALL OF THE OTHER GAS LAWS CAN BE DERIVED FROM THE IDEAL GAS LAW! PV = nRT Solve for the constant R! nT = R = P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2

Ideal Gas Law Problem What is the volume of 14.2 g of O 2(g) at STP? PV = nRT PP V (.444 mol)(8.31 )(273K) LkPa mol K = kPa = 9.94 L 14.2 g O 2 1 mol O g O 2 =.444 mol O 2

11.8 The Ideal Gas Law  Other expressions of the Ideal Gas Law molar mass (M) is: n = m/M

11.8 The Ideal Gas Law  Solving for molar mass (M)  Since density (D) = mass/volume  Solving for density

11.8 The Ideal Gas Law  Ideal Gases follow the assumptions of kinetic theory. No attractive forces between particles and negligible volume Under most temperatures and pressures most real gases behave like ideal gases  When do gases become less ideal and more real? Low temperature High pressure

11.12 Real Vs. Ideal Gases  Kinetic theory and gas laws assume that gases are ideal A gas that follows the gas laws at all conditions of pressure and temperature The kinetic theory assumes that particles of a gas have no volume and are not attracted to each other at all. There is no gas for which this is true~ Gases and vapors could not be liquefied if there were no attractions between molecules

This shows us how an ideal gas always = 1. Real gases can deviate from ideality! Real gases do have volumes and therefore must occupy a space!

11.9 Dalton’s Law of Partial Pressures Dalton’s Law At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the individual (partial) pressures. P total = P 1 + P 2 + P 3

Dalton’s Law ? 200 kPa kPa + x = 1100 kPa x = 1100 kPa – (500 kPa kPa)

11.10 Avogadro’s Hypothesis Avogadro’s Principle Equal volumes of gases at the same temperature and pressure contain an equal number of particles.

11.10 Avogadro’s Hypothesis  Solving for volume in the Ideal Gas Law for 1 mole at STP we obtain: 22.4 L = 1 mole for any ideal gas This known as the Standard Molar Volume As with all equalities~ We can use dimensional analysis!!

11.10 Gas Stoichiometry  The standard molar volume can be used as another conversion factor for stoichiometry problems Tools of Stoichiometry 1. Mole Ratio balanced equation coefficients moles A  moles B 2. Molar Mass periodic table mass = 1 mol mass A  1 mole A 3. Avogadro's # 6.02  particles = 1 mol particles particles A  1 mole A 4. Molar volume 22.4 L = 1 mol at STP volume A  mol A

Volume Stoichiometry How many grams of O 2 are needed to use up 13.6 L of H 2 at STP? 13.6 L H 2 = g O 2 mol H 2 mol O g O 2 mol O H 2(g) + O 2(g)  2H 2 O (g) 13.6 L? g 9.71 L H 2 mol H

 Diffusion – tendency of molecules to move from an area of high concentration to low concentration until uniformly mixed. If particles are in rapid straight-line motion, the will spread out. Perfume sprayed in a room Diffusion

 Effusion – rate at which gases escape from a small hole in a container Explaining the Behaviors of Gases Both diffusion and effusion depend upon the velocity of the particles!

11.11 Graham’s Law of Effusion  Most of the work on diffusion was done in the 1840’s by Thomas Graham. He noticed that lower molecular mass gases effuse faster than gases with a higher molecular mass.

11.11 Graham’s Law Graham’s Law The rate of effusion of a gas is inversely proportional to the square root of the gases molar mass. Rate A Molar Mass B Rate B Molar Mass A =

Graham’s Law Derived If both gases are in the same room, then they have the same temperature. Would they be moving at the same velocity? No, the lighter gas is faster! And we know this by looking at the kinetic energy equation!

What happens if we play with this equation? You get Graham’s law! Multiply both sides by 2 and get the velocities and masses on the same side. Get rid of the squared values by taking the square root of both sides.

11.11 Graham’s Law How much faster does oxygen diffuse (or effuse) than nitrogen. Rate A Molar Mass B Rate B Molar Mass A = Which is A?O2O2

11.11 Graham’s Law If O 2 is A, then N 2 would be B. The rate of oxygen to nitrogen would be the x. X =.935 Oxygen is.935 times faster than nitrogen (less than 1 means it’s slower)!