1. The Combined and Ideal Gas Laws Honors Chemistry

2. The Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2  Example: A balloon with Helium has a volume of 50.L at 25 o C and a pressure of 1.08 atm. What volume will it have at 0.855 atm and 10 o C ? **Convert Temps to Kelvin T 1 = 25 + 273 = 298K T 2 = 10 + 273 = 283K Plug in and Solve: (1.08atm)(50L) = (0.855atm)V 2 298K283 K (1.08)(50)(283) = (0.855)(298) V 2 V2 = 60. Liters

3. Standard Temperature and Pressure (STP)  Refers to a Temperature of 0 o C, and a pressure of 1 atm.  Keep in mind the STP values in other units: Standard Temp = 0 o C or 273 Kelvin Standard Pressure = 1atm = 760 mmHg Standard Pressure = 1atm = 760 mmHg = 760 torr = 101.325kPa = 760 torr = 101.325kPa

4. Ideal Gas  An imaginary gas that fits all the assumptions of the kinetic molecular theory of gases.  In reality, the noble gases (He, Neon, etc) are the only gases that most closely behave like an ideal gas.  All gases behave like an ideal gas at high temperatures and low pressure. This is because they cannot be attracted to each other if they are moving to fast and are very far apart from each other.

5. Avagadro’s Law  Volume is directly proportional to the moles of a gas at the same temperature and pressure.  V 1 = V 2 n = molesn 1 n 2  Remember 1 mole of a gas at STP occupies 22.4Liters

6. The Ideal Gas Law  Mathematical formula that relates the pressure, volume, temperature and the number of particles(or moles) of a gas. PV = nRT  “R” is the gas constant, R = 0.0821 atm.L R = 0.0821 atm.L K.mole K.mole The R value comes from using 1atm, 1 mole, 273K and 22.4 Liters as our comparison values.

7. Ideal gas law example What is the pressure in atmospheres exerted by a 0.500gram sample of nitrogen (N 2 ) gas in 10. Liter container at 25 o C ? What is the pressure in atmospheres exerted by a 0.500gram sample of nitrogen (N 2 ) gas in 10. Liter container at 25 o C ? PV = nRT P (10L) = (0.0178moles)(0.0821 L.atm/K.mole) (298K) P = 0.044 atm You must use Liters, atm and kelvin in this formula!!!!! *you must convert grams to moles (and Celcius to Kelvin!!) 0.500g 1 mole = 0.0178 moles 28.02 grams

8. Other Derivations of the ideal gas law  To calculate the molar mass (M) of a gas PV = nRT Moles = n = mass (m) Molar Mass(M) PV = m R T M Rearranging Molar Mass = M = m R T PV

9. Example:  A 6.39 gram sample of a gas occupies a 2.00L container at 117kPa and 35.1 o C. What is the molar mass of the gas? Molar Mass = M = m R T PV M = (6.39g)(0.0821)(308.1) (117/101.325)(2.00) = 70.0 g/mole

10. To calculate the density of a gas PV = nRT Density = mass/volume or m/V PV = m R T M Rearrange PM = mRT V Density = m = PM V RT

11. Ideal Gas law Tutorial - Online