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Gas Laws

Gas Laws. Combined Gas Law. relationship of pressure, volume, and temperature of a sample of gas with constant mass can be expressed: P 1 ·V 1 = P 2 ·V 2 T 1 T 2 to use the equation, solve for the unknown temperature must be in Kelvin. Example….

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Gas Laws

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  1. Gas Laws

  2. Combined Gas Law • relationship of pressure, volume, and temperature of a sample of gas with constant mass • can be expressed: P1·V1= P2·V2 T1 T2 • to use the equation, solve for the unknown • temperature must be in Kelvin

  3. Example… • An unopened, cold 2.00 L bottle of soda contains 46.0 mL of gas confined at a pressure of 131.96 kPa at a temperature of 5.0°C. If the bottle is dropped into a lake and sinks to a depth at which the pressure is 154 kPa and a temperature of 2.1°C, what will be the new volume of the gas? P1·V1= P2·V2 T1 T2 (131.96 kPa)(46.0 mL) = (154 kPa) V2 278 K 275.1 K V2 = 39 mL

  4. Boyle’s Law • the volume of a gas is inversely proportional to pressure , at constant temperature - if volume increases, then pressure decreases - if volume decreases, then pressure increases • Equation P1· V1 = P2· V2

  5. Example - • A sample of helium gas in a balloon is compressed from 4.0L to 250mL at a constant temperature. If the original pressure of the gas is 210kPa, what will be the pressure of the compressed gas? P1· V1 = P2· V2 (210 kPa)(4.0 L) = P2(.250 L) P2 = 3400 kPa

  6. Charles’ Law • volume of a gas is directly proportional to temperature (in Kelvin), at constant pressure • if temperature doubles, then volume also doubles • Equation V1 = V2 T1 T2

  7. Example… • What is the volume of the air in a balloon that occupies 0.620 L at 25°C if the temperature is lowered to 0.0°C? V1 = V2 T1 T2 0.620 L = V2 298 K 273 K V2 = .57 L

  8. Gay-Lussac’s Law • the pressure of a gas is directly proportional to absolute temperature (Kelvin), at constant volume • if temperature doubles, then pressure also doubles • for a sample of gas, at constant mass and volume: P1 = P2 T1 T2

  9. Example… • The pressure in an automobile tire is 1.88 atm at 25°C. What will be the pressure in psi if the temperature warms up to 37.0°C? P1 = P2 T1 T2 1.88 atm= P2 298 K 310 K P2 = 1.96 atm

  10. Graham’s Law Graham’s Law relates the relative velocities of two gases in terms of the square root of the inverse of their masses ( the square root of mass of gas B to gas A). Usually the heavier gas is assigned as gas B. The heavier gas will move slower than the lighter one at the same temperature. EXAMPLE: What is the ratio of the rates of diffusion of hydrogen gas to ethane gas, C2H6? = √(30g/mol)/(2g/mol) = 3.9

  11. Ideal Gas Law : PV = nRT Where P = pressure V = volume in Liters n = number of mols R = ideal gas constant T = temperature (Kelvin)

  12. The formula is pretty accurate for all gases at high temperatures and low pressures. • We assume that the gas molecules are little, hard particles and the collisions of the molecules are totally elastic. • A completely elastic collision means that there is no loss of energy in the collision. • The formula becomes less accurate as the gas becomes very compressed and as the temperature decreases.

  13. Calculate R, the Ideal Gas Constant… • Using 1 mol of gas at STP and molar volume • R = 0.0821 atm·L mol·K = 62.4 mmHg·L mol·K = 8.314 kPa·L mol·K

  14. Example… • Calculate the number of moles of gas contained in a 3.0L vessel at 33°C and a pressure of 1.50 atm. PV=nRT (1.50 atm) (3.0 L) = n (0.0821atm·L/mol·K)(306K) n = 0.18 moles

  15. Variations… • We don’t always know the number of moles, but we have the mass. • n = m/M so… PVM = mRT • d= m/V so… PM = dRT

  16. Example… • Calculate the grams of oxygen gas present in a 2.50L sample kept at 1.66 atm pressure and a temperature of 10.0°C. PVM = mRT (1.66 atm)(2.50 L)(32 g/mol) = m (0.0821atm·L/mol·K)(283 K) m = 5.72 g

  17. Examples… • What is the density of NH3 at 800 mmHg and 25°C? PM = dRT (800 mmHg)(17 g/mol) = d (62.4 mmHg·L/mol·K)(298 K) d = 0.7 g/L

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