2. Gas Laws

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

4. 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? 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? P 1 ·V 1 = P 2 ·V 2 T 1 T 2 (131.96 kPa)(46.0 mL) = (154 kPa) V 2 278 K 275.1 K 278 K 275.1 K V 2 = 39 mL

5. Boyle’s Law the volume of a gas is inversely proportional to pressure, at constant temperature the volume of a gas is inversely proportional to pressure, at constant temperature - if volume increases, then pressure decreases - if volume increases, then pressure decreases - if volume decreases, then pressure increases - if volume decreases, then pressure increases Equation Equation P 1 · V 1 = P 2 · V 2

6. 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? 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? P 1 · V 1 = P 2 · V 2 (210 kPa)(4.0 L) = P 2 (.250 L) P 2 = 3400 kPa

7. Charles’ Law volume of a gas is directly proportional to temperature (in Kelvin), at constant pressure volume of a gas is directly proportional to temperature (in Kelvin), at constant pressure –if temperature doubles, then volume also doubles Equation Equation V 1 = V 2 T 1 T 2

8. 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? 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? V 1 = V 2 T 1 T 2 0.620 L = V 2 298 K 273 K V 2 =.57 L

9. Gay-Lussac’s Law the pressure of a gas is directly proportional to absolute temperature (Kelvin), at constant volume 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: for a sample of gas, at constant mass and volume: P 1 = P 2 T 1 T 2

10. 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? 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? P 1 = P 2 T 1 T 2 1.88 atm = P 2 298 K 310 K 298 K 310 K P 2 = 1.96 atm

11. 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 VAVA = VBVB

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

13. The formula is pretty accurate for all gases at high temperatures and low pressures. 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. 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. The formula becomes less accurate as the gas becomes very compressed and as the temperature decreases.

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

15. Example… Calculate the number of moles of gas contained in a 3.0L vessel at 33°C and a pressure of 1.50 atm. 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

16. Variations… We don’t always know the number of moles, but we have the mass. 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

17. 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. 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

18. Examples… What is the density of NH 3 at What is the density of NH 3 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