Gas Laws and Relationships between P, V, and T Boyle’s Law Charles’s Law Gay-Lusaac’s Law How to use each

Gases and their Variables Four Variables when exploring gases: P, V, T, and n P: Pressure in atm’s V: Volume in L T: Temperature in K n: quantity of matter in moles How do they relate to one another? Let’s put on our PVT Cards and find out!

Gas Properties can be modeled using MATH! And again, Scientists have done all of the work for us!

Boyle’s Law States: The volume of a sample of gas is inversely proportional to its pressure, if temperature remains constant. Translation: At constant temperature and n, 1 α P V Robert Boyle EX. Volume Decreases, Pressure Increases Volume Increases, Pressure Decreases In an inverse relationship, the product of the two quantities is a constant. P 1 x V 1 = P 2 x V 2 P 1 V 1 = P 2 V 2

Boyle’s Law

Example 1 - A sample of gas collected in a 350 cm 3 container exerts a pressure of 103 kPa. What would be the volume of this gas at 150 kPa of pressure? (Assume that the temperature remains constant.) Solving: If temperature remains constant - use Boyle’s Law. Write the original formula: P 1 V 1 = P 2 V 2 Then list what is given and what is unknown. P 1 = 103 kPa V 1 = 350 cm 3 P 2 = 150 kPa V 2 = ? Then PLUG AND CHUG! Let’s work it together. 103 kPa (350 cm 3 ) = 150 kPa V kPa 240 cm 3 = V 2

Charles’s Law States: The volume of a sample of gas is directly proportional to its Kelvin temperature, if pressure remains constant. Translation: At constant pressure and n, V α T **Temperature ALWAYS in Kelvin** Jacques Charles EX. Volume Increases, Temperature Increases Volume Decreases, Temperature Decreases In an direct relationship, the quotient of the two quantities is a constant. V 1 / T 1 = V 2 / T 2 V 1 = V 2 T 1 T 2

Charles’s Law

Ex. If a gas occupies 733 cm 3 at 10.0 o C, at what temperature will it occupy 950 cm 3 ? Assume that pressure remains constant. Solving: If pressure remains constant - use Charles’s Law. Write the original formula:V 1 = V 2 T 1 T 2 Then list what is given and what is unknown. V 1 = 733 cm 3 T 1 = 10.0 o C V 2 = 950 cm 3 T 2 = ? Then PLUG AND CHUG! Let’s work it together. First convert o C to Kelvin: K = 10.0 o C = 283 K 733 cm 3 = 950 cm K T or 370 K= T 2

How does a hot air balloon work? TIP: Think about Charles’s Law.

Gay-Lussac’s Law States: The Kelvin temperature of a sample of gas is directly proportional to pressure, if volume remains constant. Translation: At constant volume and n, T α P **Temperature ALWAYS in Kelvin** Joseph Gay-Lussac EX. Temperature Increases, Pressure Increases Temperature Decreases, Pressure Decreases In an direct relationship, the quotient of the two quantities is a constant. P 1 / T 1 = P 2 / T 2 P 1 = P 2 T 1 T 2

Gay-Lussac’s Law

EX. If a gas is cooled from K to K and the volume is kept constant what final pressure would result if the original pressure was mm Hg? Solving: If volume remains constant - use Gay-Lussac’s Law. Write the original formula:P 1 = P 2 T 1 T 2 Then list what is given and what is unknown. P 1 = mm Hg T 1 = K P 2 = ? T 2 = K Then PLUG AND CHUG! Let’s work it together mm Hg = P K K mm Hg= P 2

ONE MORE FUN FACT! Standard Temperature and Pressure (STP) At STP: –Temperature = 273 K or 0 o C –Pressure = 1 atm = 760 mm Hg Gas Laws Song

Let’s Practice! 1. A sample of neon has a volume of 239 cm 3 at 2.00 atm of pressure. What would the pressure have to be in order for the gas to have a volume of 5.00 x 10 2 cm 3 ? 2. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased? 3. A sample of gas at 3.00 x 10 3 mm Hg inside a steel tank is cooled from °C to 0.00 °C. What is the final pressure of the gas in the steel tank? HOMEWORK = Practice problems with each law! FINISH LAB TO TURN IN AT END OF CLASS!