# IDEAL GAS LAW Ideal Gas Law Derivation Recall

• Slides: 20

IDEAL GAS LAW

Ideal Gas Law Derivation § Recall that P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 § Also we learned that § At Oo. C, and 1 atm (101. 3 k. Pa) that 1 mole of gas occupies 22. 4 L (molar volume)

Ideal Gas Law Derivation § We can set one side of the equation under standard conditions ie. § n = 1 mole § V = 22. 4 L § P = 101. 3 k. Pa § T = 0 o. C

Ideal Gas Law Derivation P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 So…. (101. 3 k. Pa) (22. 4 L) = P 2 V 2 (1 mole) (273 K) n 2 T 2

Ideal Gas Law Derivation 8. 314 k. Pa x L = P V mol x K n. T Overall…. PV = n. RT 8. 314 k. Pa. L/mol K = R = Ideal Gas Law Constant

Ideal Gases l. Behave as described by the ideal gas equation; no real gas is actually ideal l. Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less l. In real gases, particles attract each other reducing the pressure l. Real gases behave more like ideal gases as pressure approaches zero.

PV = n. RT R is known as the universal gas constant Using other STP conditions P V R = PV n. T = (1. 00 atm)(22. 4 L) (1 mol) (273 K) n T = 0. 0821 L-atm mol-K

Additional R Values What is the value of R when the STP value for P is 760 mm. Hg? R = PV n. T = (760 mm Hg) (22. 4 L) (1 mol) (273 K) = 62. 4 L-mm Hg mol-K

Learning Check G 16 Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2. 86 mol of gas occupies a 20. 0 L tank at 23°C, what is the pressure (k. Pa) in the tank in the dentist office?

Solution G 16 Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20. 0 L T = 23°C + 273 296 K n = 2. 86 mol P = ? ?

Rearrange ideal gas law for unknown P P = n. RT V Substitute values of n, R, T and V and solve for P P = (2. 86 mol)(8. 314)(296 K) (20. 0 L) = 351. 91 k. Pa

What is the volume (in liters) occupied by 49. 8 g of HCl at STP? T = 0 0 C = 273. 15 K PV = n. RT V= P P = 101. 3 k. Pa n = 49. 8 g x 1 mol HCl = 1. 37 mol 36. 45 g HCl 1. 37 mol x 8. 314 x 273. 15 K V= V = 30. 6 L 1 atm

Learning Check G 17 A 5. 0 L cylinder contains oxygen gas at 20. 0°C and 98 k. Pa. How many grams of oxygen are in the cylinder?

Solution G 17 Solve ideal gas equation for n (moles) n = PV RT = (98 k. Pa)(5. 0 L)(mol K) (62. 4 mm. Hg L)(293 K) = 0. 20 mol O 2 x 32. 0 g O 2 = 6. 4 g O 2 1 mol O 2

Molar Mass of a gas What is the molar mass of a gas if 0. 250 g of the gas occupy 215 m. L at 0. 813 atm and 30. 0°C? n = PV = (0. 813 atm) (0. 215 L) = 0. 00703 mol RT (0. 0821 L-atm/mol. K) (303 K) Molar mass = g = 0. 250 g = 35. 6 g/mol 0. 00703 mol

Density of a Gas Calculate the density in g/L of O 2 gas at STP. From STP, we know the P and T. P = 1. 00 atm T = 273 K Rearrange the ideal gas equation for moles/L PV = n. RT P = n RTV RT V

Substitute (1. 00 atm ) mol-K = (0. 0821 L-atm) (273 K) 0. 0446 mol O 2/L Change moles/L to g/L 0. 0446 mol O 2 1 L x 32. 0 g O 2 1 mol O 2 = 1. 43 g/L Therefore the density of O 2 gas at STP is 1. 43 grams per liter

Formulas of Gases A gas has a % composition by mass of 85. 7% carbon and 14. 3% hydrogen. At STP the density of the gas is 2. 50 g/L. What is the molecular formula of the gas?

Formulas of Gases Calculate Empirical formula 85. 7 g C x 1 mol C = 7. 14 mol C/7. 14 = 1 C 12. 0 g C 14. 3 g H x 1 mol H = 14. 3 mol H/ 7. 14 = 2 H 1. 0 g H Empirical formula = CH 2 EF mass = 12. 0 + 2(1. 0) = 14. 0 g/EF

Using STP and density ( 1 L = 2. 50 g) 2. 50 g 1 L x 22. 4 L 1 mol n = EF/ mol = = 56. 0 g/mol = 4 14. 0 g/EF molecular formula CH 2 x 4 = C 4 H 8