The Gas Laws The density of a gas decreases as its temperature increases.

Ideal Gas Law PV = nRT Brings together gas properties. Can be derived from experiment and theory.

At the macroscopic level, a complete physical description of a sample of a gas requires four quantities: 1. Temperature (expressed in K) 1. Temperature (expressed in K) 2. Volume (expressed in liters) 2. Volume (expressed in liters) 3. Amount (expressed in moles) 3. Amount (expressed in moles) 4. Pressure (given in atmospheres) 4. Pressure (given in atmospheres) These variables are not independent — if the values of any three of these quantities are known, the fourth can be calculated. Ideal Gas Law

Ideal Gas Equation P V = n R T Universal Gas Constant Volume No. of moles Temperature Pressure R = atm L / mol K R = kPa L / mol K Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 366

PV = nRT P = pressure V = volume T = temperature (Kelvin) n = number of moles R = gas constant Standard Temperature and Pressure (STP) T = 0 o C or 273 K P = 1 atm = kPa = 760 mm Hg Solve for constant (R) PV nT = R Substitute values: (1 atm) (22.4 L) (1 mole)(273 K) R = atm L / mol K or R = 8.31 kPa L / mol K R = atm L mol K Recall: 1 atm = kPa (101.3 kPa) ( 1 atm) = 8.31 kPa L mol K 1 mol = 22.4 STP

Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300 o C and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine T = 300 o C P = 740 mm Hg R = atm. L / mol. K Step 2) Equation: V= nRT P V (500 g)( atm. L / mol. K)(300 o C) 740 mm Hg = Step 3) Solve for variable Step 4) Substitute in numbers and solve V = What MISTAKES did we make in this problem? PV = nRT

What mistakes did we make in this problem? What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300 o C and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine  Convert mass to gram; recall iodine is diatomic (I 2 ) x mol I 2 = 500 g I 2 (1mol I 2 / 254 g I 2 ) n = mol I 2 T = 300 o C  Temperature must be converted to Kelvin T = 300 o C T = 573 K P = 740 mm Hg  Pressure needs to have same unit as R; therefore, convert pressure from mm Hg to atm. x atm = 740 mm Hg (1 atm / 760 mm Hg) P = atm R = atm. L / mol. K

Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300 o C and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine n = mol I 2 T = 573 K (300 o C) P = atm (740 mm Hg) R = atm. L / mol. K V = ? L Step 2) Equation: PV = nRT V= nRT P V ( mol)( atm. L / mol. K)(573 K) atm = Step 3) Solve for variable Step 4) Substitute in numbers and solve V = 95.1 L I 2