Gases KMS 8 th Grade Science Ms. Bormann

The Nature of Gases The first gas to be studied was air & it was a long time before it was discovered that air was actually a mixture of particles rather than a single gas. The first gas to be studied was air & it was a long time before it was discovered that air was actually a mixture of particles rather than a single gas. Although air is a mixture of several different gases, it behaves much the same as any single gas. Although air is a mixture of several different gases, it behaves much the same as any single gas. Regardless of their chemical identity, gases tend to exhibit similar physical behaviors Regardless of their chemical identity, gases tend to exhibit similar physical behaviors

Characteristics of Gases

The Kinetic Molecular Theory Kinetic = Kinetic = Molecular = Molecular = The theory states The theory states Motion Molecules the tiny particles in all forms of matter are in constant motion.

Boyle’s Law Robert Boyle was among the first to note the relationship between of a gas. Robert Boyle was among the first to note the relationship between of a gas. During his experiments and were NOT allowed to change. During his experiments and were NOT allowed to change. Boyle’s Law states Boyle’s Law states For example: If the volume is halved, the pressure would be For example: If the volume is halved, the pressure would be pressure and volume temperature the amount of gas The pressure exerted by a gas held at constant temperature varies inversely with the volume of the gas. doubled

Pressure Pressure – amount of force exerted per unit of area. Pa = Pascal. SI unit for pressure. One Pascal of pressure is 1 Newton per square meter – therefore a Pascal is a very small unit. kPa = kilopascal. A kilopascal is 1,000 Pa. Example: At sea level atmospheric pressure is kPa. This means that at Earth’s surface, the atmosphere exerts a force of 101,300 Newton’s on every square meter (this is about the weight of a large truck) Atm = standard atmosphere kPa or 101,300 Pa

Boyle’s Mathematical Law A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm? A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm? P 1 V 1 = P 2 V 2

Boyle’s Law 1. Determine which variables you have 1. Determine which variables you have 2. Determine which law is being represented 2. Determine which law is being represented 3. Rearrange the equation for the unknown variable. 3. Rearrange the equation for the unknown variable. 4. Plug in the variables and solve. 4. Plug in the variables and solve.  P 1 = 2 atm  V 1 = 3.0 L  P 2 = 4 atm  V 2 = ?  P 1 = 2 atm  V 1 = 3.0 L  P 2 = 4 atm  V 2 = ? P and V = Boyle’s Law

Charles’s Law Jacques Charles was among the first to note the relationship between of a gas. Jacques Charles was among the first to note the relationship between of a gas. During his experiments and are NOT allowed to change. During his experiments and are NOT allowed to change. Charles’s Law states Charles’s Law states For example: If the temperature is increased, the pressure will For example: If the temperature is increased, the pressure will temperature and volume pressure the amount of gas At constant pressure, the volume of a fixed number of particles of gas is directly proportional to the absolute (Kelvin) temperature increase

Temperature Charles' Law must be used with the Kelvin temperature scale. This scale is an absolute temperature scale. At 0 K, there is no kinetic energy (Absolute Zero). According to Charles' Law, there would also be no volume at that temperature. This condition cannot be fulfilled because all known gases will liquefy or solidify before reaching 0 K.

Charles’s Mathematical Law Eg: A gas has a volume of 3.0 L at 127°C. What is its volume at 227 °C? Eg: A gas has a volume of 3.0 L at 127°C. What is its volume at 227 °C?

Charles’s Law 1. Determine which variables you have 1. Determine which variables you have 2. Determine which law is being represented. 2. Determine which law is being represented. 3. Plug in variables 3. Plug in variables 4. Cross multiply and solve 4. Cross multiply and solve  T 1 = 127°C = 400K  V 1 = 3.0 L  T 2 = 227°C = 5ooK  V 2 = ?  T 1 = 127°C = 400K  V 1 = 3.0 L  T 2 = 227°C = 5ooK  V 2 = ?