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Gases Chapter 14

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Properties of Gases as demonstrated by experiments Gases have mass Gases exert pressure Gases can expand to take up large volumes Gases can be compressed to fit in small volumes

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Kinetic Molecular Theory The gas consists of very small particles, all with non-zero mass.mass These molecules are in constant, random motion. The rapidly moving particles constantly collide with the walls of the container.random The collisions of gas particles with the walls of the container holding them are perfectly elastic. The interactions among molecules are negligible. They exert no forces on one another except during collisions.interactions negligibleforces

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Boyle’s Law Boyle’s law states that at constant temperature, the absolute pressure and the volume of a gas are inversely proportional. Meaning, as pressure increases, volume decreases. For Example: http://www.grc.nasa.gov/WWW/K12/airplane/aboyle.html http://www.grc.nasa.gov/WWW/K12/airplane/aboyle.html

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Symbolic Representation of Boyle’s Law As pressure increases, volume decreases. P * V = constant So, P 1 V 1 = P 2 V 2 Pressure (kPa) Volume (L)

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Measuring pressure Many ways to measure pressure of a gas. –Atmospheres (atm) force per unit area exerted against a surface by the weight of air above that surface in the Earth’s atmosphere. –mm Hg (mmHg) pressure of 1 mm of mercury in a column of mercury. –torr equal to 1 mmHg –pascals or kilopascals

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Measuring pressure in Chemistry class Pascals (Pa) or kilopascals (kPa) 1 Pa = the amount of force 1 kg exerts on 1 m 2 for every 1 s 2. 1 Pa = 1 x 10 -6 atm 1 kPa = the amount of force 1000 kg exerts on 1 m 2 for every 1 s 2. 1 kPa = 1 x 10 -3 atm

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Boyle’s Law Problem A balloon contains 20.0 L of helium gas at 103kPa. The balloon rises to an altitude where the pressure is only 15.0 kPa. What happens to the volume of the balloon? P 1 V 1 = P 2 V 2 P 1 = 103 kPa V 1 = 20.0 L P 2 = 15.0 kPa V 2 = ? (103kPa)(20.0L) = (15.0kPA)V2

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Charles’ Law When pressure is held constant, the volume and temperature of a gas are directly proportional. Meaning as temperature increases, volume increases. For example, http://www.wisc-online.com/objects/index_tj.asp?objID=GCH1704http://www.wisc-online.com/objects/index_tj.asp?objID=GCH1704 http://www.youtube.com/watch?v=ZvrJgGhnmJo

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Symbolic Representation of Charles’ Law As temperature increases, volume increases. So, V/T = constant Temperature (K) Volume (L)

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Measuring Temperature Fahrenheit –(based on the freezing and melting point of water) Celsius –(based on the freezing and melting point of water but in base 10) Kelvin –(based on the idea that at 0 Kelvin or absolute zero, atoms are NOT in motion)

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Converting between Celsius and Kelvin Kelvin, K = o C + 273 So, 32 o C = 32 + 273 = 305 K 100 o C = 100 + 273 = 373 K

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Charles’ Law Practice A balloon has a volume of 1.00 L. The temperature of the room where the balloon is located is 273 K. If the room is heated to 300 K, what happens to the volume of the balloon? Make a qualitative assessment of what is going to happen to the volume of the balloon after an increase in temperature.

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Gay-Lussac’s Law If the volume is constant, as the temperature of the gas increases, the pressure increases. Temperature and pressure are directly proportional. For example, http://www.marymount.k12.ny.us/marynet/06stwbwrk/06gas/1amcslussac/amcsgaylussac.ht ml http://www.marymount.k12.ny.us/marynet/06stwbwrk/06gas/1amcslussac/amcsgaylussac.ht ml

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Symbolic Representation of Gay-Lussac’s Law As temperature increases, pressure increases. So, P/T = constant Temperature (K) Pressure (kPa)

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Gay-Lussac’s Practice A sealed cylinder of gas contains nitrogen gas at 1,000 kPa and a temperature of 293K. When the cylinder is left in the sun, the temperature of the gas increases to 323K. What happens to the pressure in the cylinder? Step #1: Qualitative Assessment of the Problem Step #2: Quantitative Assessment of the Problem

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Combined Gas Law Assume the number of gas molecules remains constant, temperature, pressure and volume are related.

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Combined Gas Law Example A sample of gas has a volume of 283 mL at 25 o C and 0.500 atm pressure. What is the volume at 100 o C and 1.00 atm? V 2 = 177 mL

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Avogadro’s Law If the temperature and pressure are constant, as the number of particles of the gas increases, the volume increases. Number of particles and volume are directly proportional.

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Symbolic Representation of Avogadro’s Law As number of particles increases, volume increases. So, V/n = constant Number of Particles Volume (L)

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Ideal Gas Law An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and in which there are no intermolecular attractive forces. Think of a gas as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other. Like Billiards!!

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More about Ideal Gas Law An ideal gas can be characterized by three variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them may be deduced from kinetic theory.

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Exceptions to the Ideal Gas Law at low temperatures (close to 0 K) the gas molecules have less kinetic energy (move around less) so they do attract each other. at high pressures (like many, many kPa) the gas molecules are forced closer together so that the volume of the gas molecules becomes significant compared to the volume the gas occupies.

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Symbolic Representation Ideal Gas Law PV = nRT n = number of molesmoles R = universal gas constant = 8.3145 J/mol*K One mole of an ideal gas at STP (Standard Temperature and Pressure) occupies 22.4 L.STP

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Ideal Gas Law Example What volume is needed to store 0.050 moles of helium gas at 202.6 kPa and 400 K? P = 202.6 kPa n = 0.050 mol T = 400K V = ? L R = 8.314 J/mol*K PV = nRT (202.6)(V) = (0.050)(8.314)(400) (202.6)(V) = 166.28 V=166.28/202.6 V = 0.821

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Another Example What pressure will be exerted by 20.16 g hydrogen gas in a 7.5 L cylinder at 20 o C? What are we trying to find? Pressure! What do we know? V= 7.5 L T= 20 Celsius R= 8.314 J/mol*K n= 20.16 g Do we need to change any units? Celsius Kelvin 20 C + 273 = 293 Mass: Grams moles mass = 20.16 g molar mass of H 2 = 2 x 1.008 =2.016 g/mol NOW SOLVE!

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