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CHAPTER 14. Gases. 14.2 The Gas Laws. Mostly empty space No interaction between gas atoms or molecules except in collisions Straight trajectories until collision occurs. Gases consist of widely separated atoms or molecules in constant, random motion. Mostly empty space
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CHAPTER 14 Gases 14.2 The Gas Laws
Mostly empty space No interaction between gas atoms or molecules except in collisions Straight trajectories until collision occurs Gases consist of widely separated atoms or molecules in constant, random motion
Mostly empty space No interaction between gas atoms or molecules except in collisions Straight trajectories until collision occurs Because gases can expand and contract they behave differently from solid and liquids. Gases consist of widely separated atoms or molecules in constant, random motion
Gas pressure is increased by more frequent and/or harder collisions
You can affect the gas pressure by changing: 1. the density More molecules means more impacts and a higher pressure.
You can affect the gas pressure by changing: 1. the density More molecules means more impacts and a higher pressure. 2. the volume of the container With less space to move around, there are more collisions and a higher pressure.
You can affect the gas pressure by changing: 1. the density More molecules means more impacts and a higher pressure. 2. the volume of the container With less space to move around, there are more collisions and a higher pressure. 3. the temperature With more kinetic energy, the molecules move faster. The collisions are harder and more frequent.
Boyle’s law: V versus P Robert Boyle’s experiment: • Mercury (Hg) was poured down a tube shaped like the letter “J.” • - The tube was closed on the lower end. • The gas inside the tube took up space (volume). • The temperature and number of gas molecules inside the tube stayed constant. • - Boyle observed the change in pressure in mmHg as a function of volume. He then graphed the relationship between pressure and volume.
Boyle’s law: V versus P Pressure versus volume • Temperature • Number of moles constant
Pressure versus volume • Temperature • Number of moles constant Boyle’s law: V versus P
Boyle’s law: V versus P At higher altitudes, the pressure outside the balloon is lower, so the volume of the balloon increases. The pressure inside the balloon is equal to the pressure outside
Boyle’s law: V versus P If a weather balloon is released on the ground with a volume of 3.0 m3 and a pressure of 1.00 atm, how large will it get when it reaches an altitude of 100,000 ft, where the pressure is 0.0100 atm?
Boyle’s law: V versus P If a weather balloon is released on the ground with a volume of 3.0 m3 and a pressure of 1.00 atm, how large will it get when it reaches an altitude of 100,000 ft, where the pressure is 0.0100 atm? Asked: Volume of the balloon when it reaches 100,000 ft Given: Relationships:
Boyle’s law: V versus P If a weather balloon is released on the ground with a volume of 3.0 m3 and a pressure of 1.00 atm, how large will it get when it reaches an altitude of 100,000 ft, where the pressure is 0.0100 atm? Asked: Volume of the balloon when it reaches 100,000 ft Given: Relationships: Solve: Answer:The balloon will have a volume of 300 m3.
Gases can only push Three states of inhalation
Gases can only push Three states of inhalation When you inhale, you create a lower pressure in your lungs, and the air outside pushes in
Gases can only push Three states of inhalation There is no such thing as suction
Charles’s law: V versus T The volume decreases as the temperature decreases
Charles’s law: V versus T The volume increases as the temperature increases
Volume versus temperature • Pressure • Number of moles constant Charles’s law: V versus T
Charles’s law: V versus T Doubling the Kelvin temperature will double the volume of a gas Kelvin temperatures simplify the V versus T relationship
Charles’s law: V vs. T If you inflate a balloon to a size of 8.0 L inside where the temperature is 23oC, what will be the new size of the balloon when you go outside where it is 3oC?
Charles’s law: V vs. T If you inflate a balloon to a size of 8.0 L inside where the temperature is 23oC, what will be the new size of the balloon when you go outside where it is 3oC? Asked:Volume of the balloon when the temperature drops to 3oC Given: Relationships:
Charles’s law: V vs. T If you inflate a balloon to a size of 8.0 L inside where the temperature is 23oC, what will be the new size of the balloon when you go outside where it is 3oC? Asked: Volume of the balloon when the temperature drops to 3oC Given: Relationships: Solve:
Charles’s law: V vs. T If you inflate a balloon to a size of 8.0 L inside where the temperature is 23oC, what will be the new size of the balloon when you go outside where it is 3oC? Asked: Volume of the balloon when the temperature drops to 3oC Given: Relationships: Solve:
Imagine you were to hitch a ride on a high-altitude research balloon that reaches and altitude of 100,000 ft. At sea level, where the pressure is 1.00 atm and the temperature is 20oC, you’ll need 18 m3 of helium to fill the balloon. What will be the new volume of the gas when you reach altitude, where the pressure is 0.0100 atm and the temperature is –50oC?
Imagine you were to hitch a ride on a high-altitude research balloon that reaches and altitude of 100,000 ft. At sea level, where the pressure is 1.00 atm and the temperature is 20oC, you’ll need 18 m3 of helium to fill the balloon. What will be the new volume of the gas when you reach altitude, where the pressure is 0.0100 atm and the temperature is –50oC? Asked: Volume of the balloon when it reaches 100,000 ft Given: Relationships:
Imagine you were to hitch a ride on a high-altitude research balloon that reaches and altitude of 100,000 ft. At sea level, where the pressure is 1.00 atm and the temperature is 20oC, you’ll need 18 m3 of helium to fill the balloon. What will be the new volume of the gas when you reach altitude, where the pressure is 0.0100 atm and the temperature is –50oC? Asked: Volume of the balloon when it reaches 100,000 ft Given: Relationships: Solve:
Avogadro’s law: V versus moles Two equal volumes of hydrogen react with one volume of oxygen to produce two volumes of water vapor. Gas volumes act like moles because the same size container has the same number of molecules (at the same temperature and pressure).
Moles versus volume • Temperature • Pressure constant Avogadro’s law: V versus moles
Moles versus volume • Temperature • Pressure constant Avogadro’s law: V versus moles
Moles versus volume • Temperature • Pressure constant Avogadro’s law: V versus moles Twice the volume, twice the number of moleculesassuming the temperature and pressure are constant
Moles versus volume • Temperature • Pressure constant Avogadro’s law: V versus moles Liquid volumes result from the number of molecules, whereas gas volumes result from the size of the container
Moles versus volume • Temperature • Pressure constant 14.2 The Gas Laws Avogadro’s law: V versus moles Each molecule here has the same kinetic energy. In gases with the same temperature, lighter molecules move faster, so the force from the impact is the same as for the slower molecules that have more mass.
The ideal gas law Combining the previous gas laws, we obtain the ideal gas law In reality, the ideal gas law is an approximation which is accurate for many gases over a wide range of conditions. The ideal gas law is not accurate at very high density or at very low temperature.
The ideal gas law R is the only constant The universal gas constant
The ideal gas law Calculating the universal gas constant using various units Watch out for the units!
The ideal gas law Limitations of the ideal gas law In an ideal gas, we assume that: 1. individual gas molecules take up no space For very small volumes, the size of gas atoms or molecules becomes significant
The ideal gas law Limitations of the ideal gas law In an ideal gas, we assume that: 1. individual gas molecules take up no space 2. gas molecules do not interact with each other For very small volumes or very low temperatures, gas atoms and molecules are very close together, and van der Waals attractions are no longer negligible
PV = nRT What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20oC?
PV = nRT What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20oC? Asked: Tank pressure Given: Relationships:
PV = nRT What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20oC? Asked: Tank pressure Given: Relationships: Solve: