BEHAVIOR OF GASES Chapter 12 1

THREE STATES OF MATTER 2

3 Importance of Gases Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 ---> 2 Na + 3 N 2

General Properties of Gases There is a lot of “free” space in a gas. Gases can be expanded infinitely. Gases occupy containers uniformly and completely. Gases diffuse and mix rapidly. 4

KINETIC MOLECULAR THEORY (KMT) Theory used to explain gas laws Gases consist of molecules in constant, random motion. P arises from collisions with container walls. Collisions elastic. No attractive / repulsive forces between molecules. Volume of molecules is negligible. 5

Properties of Gases Gas properties can be modeled using math. Model depends on— V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atm) 6

Pressure Air Pressure is measured with a BAROMETER (developed by Torricelli in 1643) 7

Pressure Hg rises in tube until force of Hg (down) balances the force of air (pushing up). P of Hg pushing down related to Hg density column height 8

Pressure Column height measures P 1 atm= 760 mm Hg or 29.9 inches Hg = 34 feet of water SI unit is PASCAL, Pa, 1 atm = kPa 9

Avogadro’sAvogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = n k V and n are directly related. 10 twice as many molecules

Avogadro’s Hypothesis and Standard Molar Volume 11 1 mol of ANY gas occupies 22.4 L at STP mol1mol1mol1mol1mol1mol = ? L V / n = 22.4 L/mol at STP

Boyle’s Law If n and T are constant, then PV = (nRT) = k This means, P and V are inversely related 12 Robert Boyle ( ).

BOYLE’S LAB P inversely proportional to V

Practice Problem If you had a gas that exerted 202 kPa of pressure and took up a space of 3.00 liters. If you decide to expand the tank to 7.00 liters, what would be the new pressure? (Assume constant T) P 1 V 1 =P 2 V kPa x 3.00 liters = P 2 x 7.00 liters 606 = P 2 x 7.00 liters P 2 = 86.8 kPa

15 torr atm mmHg = torr V = ft 3 n T = n 1 +n 2 (PV) T = (pv) 1 +(pv) 2

Charles’s Law If n and P are constant, then V = (nR/P)T = kT V and T are directly related. 16 Jacques Charles ( ).

17 Charles’s original balloon Modern long-distance balloon

V is directly proportional to T

Practice Problem If you took a balloon outside C that was originally inside at C at 2.0 liters, what volume would the balloon occupy when cold? (constant P) T 1 /V 1 =T 2 /V 2 (20+273) = (5+273) 2.00 L X liters V 2 = 278*2/293 V 2 = 1.9L

NO F 20 K = 273 +C 13 C 27 C 21 C

Gas Volume, Temperature, and Pressure 21 COMBINED GAS LAW: Combines Charles and Boyle’s Law P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 = P 2 V 2 T 1 T 2

22 STP STP = S.C. (STANDARD CONDITIONS) 3 atm 35 C 5 C

23 27C C 1 atm = 760 torr = 101 KPa 27C