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The Gas Laws

The Gas Laws. Pressure Volume & Temperature. States of Matter. In liquids and solids, the primary particles (atoms or molecules) are always in contact with each other. In gases, particles move independently.

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The Gas Laws

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  1. The Gas Laws Pressure Volume & Temperature

  2. States of Matter In liquids and solids, the primary particles (atoms or molecules) are always in contact with each other. In gases, particles move independently. Because the atoms of gases are far apart they are very compressible.

  3. Compressibility Whenpressure is applied, the volume occupied by a gas can be decreased. Gases fill all of the space available to them. In a contained sample (e.g. balloon) gases expand to fill the total volume of the balloon. If not contained: Gases expand indefinitely.

  4. Volume If the cubic meter were divided into 1000 equal smaller parts,each part would be equal to 1 Liter (slightly larger than a quart) 1 qt = 1.057 L If each liter were divided into 1000 equal smaller parts, each part would be equal to 1 milliliter (mL) or 1 cubic centimeter (cc)  1 mL = 1 cc 

  5. Volume Space occupied in 3 dimensions. Units: liters One liter is similar in volume to a quart 1 qt = 1.057 L One liter is equal to 1,000 cubic centimeters

  6. Pressure Force per unit of area. Force / area  =  lbs / sq.in   Pounds per square inch =  psi 14.7 psi = 1 atm 1 mm Hg = 1 torr 1 atm = 760 mm Hg

  7. Boyle’s Law Increase the pressure  Volume decreases proportionally   Pressure x Volume = constant Product of pressure and volume is fixed. P x V = constant  P1 x V1 = P2 x V2

  8. Example Compressed gas cylinder Pressure = 135 atm Volume = 15.0 liters What volume the gas will occupy at 1.00 atm ? P1 = 135 atm V1 = 15.0 L P2 = 1.00 atm

  9. Example (cont.) Determine V2 P1 x V1 = P2 x V2 V2 = ( P1 x V1 ) / P2 V2 = ( 135 atm ) ( 15.0 L ) / 1.00 atm = 2,030 liters

  10. Charles’ Law Increase the temperature  Volume will increase proportionally. The volume of a sample divided by the temperature is equal to a constant. V / T = constant V1 / T1 = V2 / T2

  11. Example Determine the final volume of a 0.35 liter balloon which is heated from room temperature to 100 degrees C. V1 / T1 = V2 / T2 Convert all temperatures to Kelvins. T1 = 25 °C + 273 = 298 K T2 = 100 °C + 273 = 373 K

  12. Example (cont.) V1 / T1 = V2 / T2 V2 = ( V1 x T2 ) / T1 = V1 x ( T2 / T1 ) = ( 0.35 L ) ( 373 K / 298 K) = ( 0.35 ) ( 1.25) = 0.44 liters

  13. Gay Lussac’s Law Pressure is proportional to the temperature The ratio of the absolute temperature to the pressure is always constant. P1 / T1 = P2 / T2

  14. Example The pressure inside a compressed gas cylinder is 134 atm @ 25 °C. Calculate the new pressure inside the cylinder if it is heated to 48 °C. P1 = 134 atm T1 = 25 + 273 = 298 K   T2 = 48 + 273 = 321 K Determine P2

  15. Example (cont.) P1 / T1 = P2 / T2 P2 = ( P1 x T2 ) / T1 P2 = ( 134 atm ) ( 321 K ) / 298 K = 144 atm

  16. Example The pressure of CO2 inside a bottle of carbonated soda pop is approximately 1.35 atm @ 25 °C (298 K). Determine the pressure inside the bottle if it is chilled to 0 °C (273 K) .

  17. Example (cont.) P1 / T1 = P2 / T2 P2 = ( P1 x T2 ) / T1 P2 = ( 1.34 atm ) ( 273 K ) / 298 K = 1.23 atm

  18. Combined Gas Law We can combine all of these laws to get a combined gas law: P V / T = constant P1 x V1 / T1 = P2 x V2 / T2 This law holds for a fixed amount of gas (or a fixed number of moles, n ).

  19. Example Start with 2.37 liters of gas @ 25.0 °C ( 298 K ) and 1 atmosphere Heat it to 297 °C ( 570 K ). Increase the pressure to 10 atmospheres. What is the final volume? *Note: Upon heating, volume will increase. But on compression, volume will decrease.  Opposing forces

  20. Example (cont.) P1 x V1 / T1 = P2 x V2 / T2 Solve for V2 (isolate the variable):  V2 = [ P1 x V1 / T1 ] x ( T2 / P2 ) Express as a product of ratios: V2 = V1 x [ P1 / P2 ] x [ T2 / T1 ]

  21. Example (cont.) P1 / P2 = 1 / 10  T2 / T1 = 570 / 298  V2 = ( 2.37 ) ( 1 / 10 ) ( 570 / 298) V2 = ( 2.37 ) ( 0.19 ) = 0.453 L *Note: Ratio of pressures = 0.10 < 1 Ratio of temps = 1.91 > 1 They offset each other.

  22. Ideal Gas Law P V = n R T n = # of moles of gas R = 0.0821 liter * atm / mol * K PV / nT = constant ( P1 x V1 ) / ( n1 x T1 ) = ( P2 x V2 ) / ( n2 x T2 )

  23. Example Calculate the volume of 1 mole of Ideal gas @ Room temp (298 K) and pressure (1 atm). P V = n R T V = n R T / PV = ( 1.0 ) ( 0.0821 ) ( 273 ) / 1.0 = 22.4 L

  24. Example 11.2 L tank of gas is found in the coldest part of the refrigerator (0 °C = 273 K).   It contains 4 moles of gas: (1 mole of oxygen and 3 moles of neon). What is the pressure in the tank? P = nRT / V = = ( 4.0 )( 0.0812 ) ( 273 ) / 11.2 = 7.91 atm 

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