II. The Gas Laws (p ) Ch. 10 & 11 - Gases

A. Boyle’s Law b The pressure and volume of a gas are inversely related at constant mass & temp P V PV = k

GAS LAW’S b Pressure & Volume at Constant Temperature b *Remember - pressure is caused by moving molecules hitting container. b Therefore, as Volume increases, Pressure Decreases b as Volume decreases, Pressure Increases b THUS, Pressure & Volume are Inversely Proportional.

A. Boyle’s Law P V PV = k

GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

GAS LAW’S b Temperature & Volume at Constant Pressure b If pressure is constant, gases expand when heated and contract when cooled. Therefore, Temperature & Volume are Directly Proportional.

V T B. Charles’ Law b The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

V T B. Charles’ Law

GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

GAS LAW’S b Pressure & Temperature at Constant Volume b At higher temperatures, gaseous molecules have more kinetic energy, faster speed and hit sides of container with more force increasing pressure. Therefore, Temperature & Pressure are directly proportional.

P T C. Gay-Lussac’s Law b The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

P T C. Gay-Lussac’s Law

GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(296K) T 2 = 217 K = -56°C

= kPV PTPT VTVT T D. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

GIVEN: V 1 = 7.84 cm 3 P 1 = 3 atm T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 1 atm T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (3 atm)(7.84 cm 3 )(273 K) =(1 atm) V 2 (298 K) V 2 = 21.5 cm 3 E. Gas Law Problems b A gas occupies 7.84 cm 3 at 3atm & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

Dalton’s Law of Partial Pressures b The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. P t = P 1 + P 2 + P 3 … The pressure of each gas in a mixture is called the partial pressure of that gas

Gas collected by HO displacement Gas collected by H 2 O displacement b Compare gas collected to room pressure and then P atm = P gas + P water vapor

GAS LAW’S b Pressure & Moles, as well as, Volume & Moles are directly proportional, at constant Temperature. b VOLUME OF 1 MOL AT STP IS – 22.4L