Gas Laws Chapter 10

Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. 2

P and V are inversely proportional A plot of V versus P results in an inverse graph Therefore is the pressure is doubled, the volume will be halved. 3

Boyle’s Law Practice Problem If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be? P1V1 = P2V2 (1.5 atm)(5.6 L) = (x)(4.8 L) x = 1.8 atm

Charles’s Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature in Kelvins. i.e., V T = k A plot of V versus T will be a straight line. 5

Charles’s Law Practice Problem If I have 45 liters of helium in a balloon at 250 C and increase the temperature of the balloon to 550 C, what will the new volume of the balloon be?

Avogadro’s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. V1/n1 = V2/n2 Mathematically, this means V = kn 7

Avagadro’s Law Practice Problem A 6.0 L sample at 25 °C and 2.00 atm of pressure contains 0.5 moles of a gas. If an additional 0.25 moles of gas at the same pressure and temperature are added, what is the final total volume of the gas? Vf = (6.0 L x 0.75 moles)/0.5 moles Vf = 4.5 L/0.5 Vf = 9 L

Ideal-Gas Equation V  nT P So far we’ve seen that V  1/P (Boyle’s law) V  T (Charles’s law) V  n (Avogadro’s law) Combining these, we get V  nT P 9

Ideal-Gas Equation The constant of proportionality is known as R, the gas constant. 10

Ideal-Gas Equation PV = nRT nT P V  nT P V = R or The relationship then becomes or PV = nRT 11

Ideal Gas Law Practice Problem If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 L, what is the temperature? PV=nRT 205 K

Densities of Gases n P V = RT If we divide both sides of the ideal-gas equation by V and by RT, we get n V P RT = 13

Densities of Gases n   = m P RT m V = We know that moles  molecular mass = mass n   = m So multiplying both sides by the molecular mass ( ) gives P RT m V = 14

Densities of Gases P RT m V = d = Mass  volume = density So, P RT m V = d = Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas. 15

Molecular Mass P d = RT dRT P  = We can manipulate the density equation to enable us to find the molecular mass of a gas: P RT d = Becomes dRT P  = 16