Gases Chapter 13

13.1 – The Gas Laws Kinetic Theory = assumes that gas particles:  do not repel or attract each other  are much smaller than the distances between them (particles have no volume)  are in constant, random motion (straight lines)  have completely elastic collisions (no loss of KE)  Have the same average KE at a given temp. Nature of gases = determined by pressure, temperature, volume, and number of particles

13.1 – The Gas Laws Our variables =  Pressure (P)  Temperature (T) – Must be in Kelvin!  Volume (V)  Amount of particles/number of moles (n)  Gas constant - R

13.1 – The Gas Laws Boyle’s Law =  Studied relationship between pressure and volume of a gas  At a given temp., volume and pressure are inversely related P 1 V 1 = P 2 V 2  P 1 and V 1 are initial conditions  P 2 and V 2 are new conditions

13.1 – The Gas Laws (Boyle’s Law)Boyle’s Law

13.1 – The Gas Laws Steps for solving gas law problems = 1.Identify all variables. 2.Analyze the problem. “Which equation should I use?” 3.Rearrange the equation to solve for the unknown variable. 4.Plug in the numbers from step 1 into the equation from step 3  solve!

13.1 – The Gas Laws (Boyle’s Law) A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L?

13.1 – The Gas Laws (Boyle’s Law) 1.Identify all variables. T=constantP 1 =210 kPa V 1 =4.0 LP 2 =? V 2 =2.5 L 2.Which equation should I use? We know P and V. Boyle’s Law: P 1 V 1 =P 2 V 2 3.Rearrange the equation. To solve for P 2, divide both sides by V 2 : 4.Plug in numbers from #1 into equation from #3:

13.1 – The Gas Laws (Boyle’s Law) The volume of a gas at 99.0 kPa is mL. If the pressure is increased to 188 kPa, what will be the new volume? Air trapped in a cylinder fitted with a piston occupies mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston?

13.1 – The Gas Laws (Charles’s Law)Charles’s Law Charles’s Law =  Studied relationship between volume and temperature of a gas  At a given pressure, volume and temperature are directly related.  V 1 and T 1 = initial cond.  V 2 and T 2 = new cond.

13.1 – The Gas Laws (Charles’s Law)Charles’s Law

13.1 – The Gas Laws (Charles’s Law) A gas sample at 40.0°C occupies a volume of 2.32 L. If the temperature is raised to 75.0°C, what will the volume be, assuming the pressure remains constant? A gas at 89°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L?

13.1 – The Gas Laws (Gay-Lussac’s Law) Gay-Lussac’s Law =  Studied the relationship between temperature and pressure of a gas  At a given volume, temperature and pressure are directly related.  P 1 and T 1 = initial cond.  P 2 and T 2 = new cond.

13.1 – The Gas Laws (Gay-Lussac’s Law)Gay-Lussac’s Law

13.1 – The Gas Laws (Gay-Lussac’s Law) The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank? A gas in a sealed container has a pressure of 125 kPa at a temperature of 30.0°C. If the pressure in the container is increased to 201 kPa, what is the new temperature?

13.2 – The Combined Gas Law Combines all four equations together into one that relates temperature, pressure, and volume:

13.2 – The Combined Gas Law A gas at 110 kPa and 30.0°C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0°C and the pressure increased to 440 kPa, what is the new volume? At 0.00°C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0°C and the entire gas sample is transferred to a 20.0-mL container, what will be the gas pressure inside the container?

13.2 – Avogadro’s PrincipleAvogadro’s Principle Avogadro’s principle = equal volumes of gases at the same temperature and pressure contain equal numbers of particles  STP=Standard Temperature and Pressure, 0°C (273 K) and 1 atm  **One mole of any gas will occupy 22.4 L at STP  Now we can convert from liters to moles!

13.2 – Avogadro’s Principle Calculate the volume that mol of a gas at standard temperature and pressure (STP) will occupy. How many moles of nitrogen gas will be contained in a 2.00-L flask at STP?

13.2 – The Ideal Gas Law All other gas laws apply to “a fixed mass” or “a given amount” Changing the number of gas particles affects other variables  Increasing the number of particles will… Increase P (if T and V are constant) Increase V (if T and P are constant) We need a new equation that includes amount of gas present

13.2 – The Ideal Gas LawThe Ideal Gas Law PV = nRT P = pressure V = volume n = number of moles of gas present R = ideal gas constant (depends on units of P) T = temperature UNITS of P UNITS of R VALUE of R atmL·atm mol·K kPaL·kPa mol·K mm HgL·mm Hg mol·K 62.4

13.2 – The Ideal Gas Law Calculate the number of moles of gas contained in a 3.0-L vessel at 3.00 x 10 2 K with a pressure of 1.50 atm. Determine the kelvin temperature required for mol of gas to fill a balloon to 1.20 L under atm pressure.

13.2 – The Ideal Gas Law So what’s an ideal gas anyway?  Its particles don’t take up space and have no intermolecular attractive forces  Follows the gas laws under all conditions of T and P  **In the real world, NO gas is truly ideal!  When do real gases not behave as “ideal” gases?  At high P and low T  we can compress them into liquids Ex. Propane and liquid nitrogen

13.3 – Gas Stoichiometry Volume – Volume  Solve same as mole-mole problems  How many liters of propane gas (C 3 H 8 ) will undergo complete combustion with 34.0L of oxygen gas? C 3 H 8 + O 2 → H 2 O + CO 2

13.3 – Gas Stoichiometry Volume – Mass If 5.00L of nitrogen reacts completely with hydrogen at a pressure of 3.00atm and a temperature of 298K, how much ammonia, in grams, is produced? N 2 + 3H 2 → 2NH 3 Calculate as a Volume to Volume Problem 5.00L N 2 x (2NH 3 /1N 2 ) = 10.00L NH 3 Utilize PV=nRT to solve for the number of moles (n) (3.00atm)(10.00L) = n (0.0821L∙amt/mol∙K)(298K)n=1.23mol NH 3 Convert moles to mass 1.23mol NH 3 ÷ 17.04g/mol NH 3 = 21.0g NH 3

13.3 Gas Stoichiometry When 3.00L of propane gas is completely combusted to form water vapor and carbon dioxide at 350°C and 0.990atm, what mass of water vapor results? C 3 H 8 + O 2 → H 2 O + CO 2