Ch.14: Gas Laws Part 1

Pressure of the Atmosphere Called “atmospheric pressure,” or the force exerted upon us by the atmosphere above. A measure of the weight of the atmosphere pressing down upon us. Measured using a barometer - A device that can weigh the atmosphere above us.

A Mercury Barometer

Common Units of Pressure Unit Atmospheric Pressure Scientific Field pascal (Pa); 1.01325 x 105 Pa SI unit; physics, kilopascal(kPa) 101.325 kPa Chemistry atmosphere (atm) 1 atm* Chemistry millimeters of mercury 760 mmHg* Chemistry, medicine, ( mm Hg ) biology torr 760 torr* Chemistry pounds per square inch 14.7 lb/in2 Engineering ( psi or lb/in2 ) bar 1.01325 bar Meteorology, chemistry, physics

Problem 1: Converting Units of Pressure Problem: A chemist collects a sample of Carbon dioxide from the decomposition of limestone (CaCO3), the height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure in torr, atmospheres, and kilopascals. Plan: The pressure is in mmHg, so we use the conversion factors from The preceding table to find the pressure in the other units. Solution: converting from mm Hg to torr: (a) PCO2 (torr) = converting from torr to atm: (b) PCO2( atm) = converting from atm to kPa: (c) PCO2(kPa) =

Boyle’s Law : P - V Relationship Pressure is inversely proportional to volume at fixed temperature and fixed amount: P1V1 = P2V2 Where: P is the pressure exerted by the sample on the surroundings. V is the volume occupied by the sample in Liters

Problem 2: Applying Boyles Law Problem: A gas sample at a pressure of 1.23 atm has a volume of 15.8 cm3, what will be the volume if the pressure is increased to 3.16 atm? Plan: We begin by converting the volume that is in cm3 to mL and then to liters, then we do the pressure change to obtain the final volume. Solution: P1 = 1.23 atm P2 = 3.16 atm V1 = 15.8 cm3 V2 = unknown T and n remain constant V1 = V2 =

Charles’s Law : V - T Relationship Volume is directly proportional to temperature at fixed pressure and fixed amount: V1 = V2 T1 T2 Where: V is the volume occupied by the sample in Liters. T is the temperature in degrees Kelvin. Kelvin is the SI unit for temperature. To convert celsius to Kelvin, simply add 273. K = Co + 273

Problem 3: Charles Law A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant. V1 = T1 = V2 = T2 = T2 = T2 = oC =

Gay-Lussac’s Law How are temperature and pressure of a gas related? Pressure of a gas varies directly with the temperature. P1 = P2 T1 T2 Where P = pressure, T = temperature in Kelvin

Problem 4: Gay-Lussac’s Law Problem: A copper tank is compressed to a pressure of 4.28 atm at a temperature of 350 K. What will be the pressure if the temperature is raised to 95.6 oC? Solution: T1 = T2 = P1 =

Combined Gas Law Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law can all be combined together to show the relationship of pressure, volume, and temperature of a specific amount of a gas. P1V1 = P2V2 T1 T2

Avogadro’s Law : V - n Relationship Equal volumes of gases at the same temperature and pressure contain the same number of particles. Standard Temperature and Pressure (STP) A set of Standard conditions have been chosen to make it easier to understand the gas laws, and gas behavior. Standard Temperature = 00 C = 273.15 K Standard Pressure = 1 atmosphere = 760 mm Mercury At these “standard” conditions 1.0 mole of a gas will occupy a “standard molar volume”. Standard Molar Volume = 22.414 Liters = 22.4 L