Gas Laws Ch 13.3

4 variables affecting gas behavior Pressure (P) Volume (V) Amount of gas (n) Temperature (T)

Boyle’s Law Pressure is inversely proportional to volume. (at the same temperature) P1V1 = P2V2 In a closed system, as volume goes down, pressure goes up. Marshmallow in syringe Cartesian diver

Boyle’s Law example 5 liters of gas at 1 atmosphere of pressure are compressed to 1.25 liters. What is the final pressure? P1V1 = P2V2 Solve for P2 P1V1/V2 = P2 (5 liters)(1 atm)/(1.25 liters)= 4 atm

Charles’s Law Temperature and volume are directly proportional. T1/V1 = T2/V2 or T1V2 = T2V1 In a closed, expandable system, as temperature goes up, volume goes up. Balloon in hot car

Charles’s Law Example A balloon with a volume of 0.5 liters at 20C is left in a 120 C car all day. What is the final volume of the balloon? Convert temperatures to Kelvin: 20C =293K, 120 C = 393 K T1V2 = T2V1 Solve for V2 V2 = T2V1 /T1 0.7 liters = (393 K)(0.5 liters)/293K

Combined Gas Law You can combine Charles and Boyles to solve problems with pressure, volume and temperature P1V1/T1= P2V2/T2 A fixed container has 5.5L of gas at STP. What is the pressure at 300K? If one of the variables doesn’t change, you can cancel it out. P1/T1= P2/T2 Solve for P2: P1T2 /T1=P2 (1 atm)(300K)/273K=1.1 atm              

Avogadro’s Law Amount is directly proportional to volume. How much gas, or number of moles of gas = n V1/n1 = V2/n2 The more gas you have, the more space it takes up at the same pressure. Therefore, we can compare the amount of gas in moles to calculate a final volume, if temperature and pressure remain constant.

Example 0.5 moles of O2 has a volume of 12.2 liters at standard temperature and pressure. If all the O2 is converted to O3 at the same temperature and pressure, what is the final volume? Given: 3O2(g) 2O3(g) V1/n1 = V2/n2 Solve for V2 n2V1/n1 = V2(n2)/n2 (2 moles)(12.2 L)/3 moles = 8.1 L

Ideal Gas Law The ideal Gas law defines how an “ideal gas” behaves. Most gas behaviors can be approximated with this formula at STP. PV = nRT n = moles R = .08206 L atm/K mol R is the universal gas constant

Example A sample of hydrogen gas at 1.5 atm and 273K has a volume of 8.56 L. how many moles of H2 are present? PV = nRT: solve for n PV/RT = n (1.5 atm)(8.56 L)/(.08206 Latm/Kmol)(273K) = 0.57 moles H2

Dalton’s Law of Partial Pressures The sum of the partial pressures of gases in a gas mixture is equal to the total pressure. Pt = Pa + Pb + Pc + … Example: Calculate the partial pressure in atmospheres of each gas in the mixture: Pt = 10 atm 1.0 mol CO2, 2.0 mol O2, 3.0 mol N2 10 = 1x + 2x + 3x = 6x, x= 1.67 1.67 atm + 3.33 atm + 5.00 atm = 10 atm

Graham’s Law The ratio of the rates of effusion of two gases is equal to the square root of the inverse ratio of their molecular masses or densities. The effusion rate of a gas is inversely proportional to the square root of its molecular mass. Mathematically, this can be represented as: Rate1 / Rate2 = square root of (Mass2 / Mass 1)

Calculating effusion rate Nitrogen, has a molecular mass of 28.0 g. O2, Oxygen, has a molecular mass of 32.0 g. Therefore, to find the ratio, the equation would be: RateN2/RateO2 = square root of (32.0 g/28.0 g): RateN2/RateO2 = 1.07 This tells us that N2 is 1.07 times as fast as O2.

Calculating molecular mass Gas A is 0.68 times as fast as Gas B. The mass of B is 17 g. What is the mass of A? 0.68 = square root of 17 g/Mass A. Squaring both sides gets: 0.4624 = 17 g/Mass A Mass A = 17g/0.4624 = 36.7647 g

Gas Law Summary Boyle’s Law P is inversely proportional to V P1V1=P2V2 Charles’s Law V is proportional to T V1/T1 = V2T2 Avogadro’s Law V is proportional to n V1/n1=V2/n2 Dalton’s Law Pt is sum of partial pressures Pt = Pa + Pb + Pc + …