Ideal vs. Real Gases

No gas is ideal. As the temperature of a gas increases and the pressure on the gas decreases the gas acts more ideally.

Equation of State of an Ideal Gas Robert Boyle ( 1662 ) found that at fixed temperature –Pressure and volume of a gas is inversely proportional PV = constantBoyle’s Law J. Charles and Gay-Lussac ( circa 1800 ) found that at fixed pressure –Volume of gas is proportional to change in temperature Volume Temp o C All gases extrapolate to zero volume at a temperature corresponding to – o C (absolute zero). He CH 4 H2OH2O H2H2

Kelvin Temperature Scale Kelvin temperature (K) is given by K = o C where K is the temperature in Kelvins, o C is temperature in Celcius Using the ABSOLUTE scale, it is now possible to write Charles’ Law as V / T = constant Charles’ Law Gay-Lussac also showed that at fixed volume P / T = constant Combining Boyle’s law, Charles’ law, and Gay-Lussac’s law, we have P V / T = constant Gay-Lussac Charles

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

Gas Law Calculations Boyle’s Law PV = k Boyle’s Law PV = k Charles’s Law V T Charles’s Law V T Combined Gas Law PV T Combined Gas Law PV T Ideal Gas Law PV = nRT Ideal Gas Law PV = nRT = k T and V change P, n, R are constant P, V, and T change n and R are constant P and V change n, R, T are constant

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 = 0.8 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

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: PV = nRT 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?

27dFields-Jeff/GasLaw1.htm Ideal Gas Law

The production of oxygen by thermal decomposition Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 423 Oxygen plus water vapor KClO 3 (with a small amount of MnO 2 )

27dFields-Jeff/GasLaw1.htm Density of Gases Table Density of Gases Table