Section 13.1

Pressure One of the most obvious properties of a gas is that it exerts pressure on its surroundings. The gases most familiar to us form the earth’s atmosphere. A device that measures atmospheric pressure is the barometer. Atmospheric pressure results from the mass of air being pulled toward the center of the earth by gravity Low pressure system- stormy weather High pressure system- fair weather

Pressure Pressure varies with altitude- higher altitude has less air pushing down on the earth’s surface than at sea level Units of pressure- atm, mm Hg (torr), Pa 1 standard atmosphere= atm = mm Hg = torr = 101,325 Pa StandardTemperaturePressure= 0 0 C and 1.00 atm (STP)

Examples of conversions Convert 732 mm Hg into torr and atm 732mm Hg = 732 torr 732 mmHg x 1 atm =.963 atm 760 mmHg Convert 2.79 x 10 5 Pa into torr and atm 2.79 x 10 5 Pa x 1 atm = 2.75 atm 101,325 Pa 2.79 x 10 5 Pa x 76o torr = torr 101,325 Pa

Pressure and volume- Boyle’s law Boyle’s law – the volume of a gas and the pressure of a gas are inversely proportional ( when one increases the other decreases ) at a constant temperature P1 V1 = P2 V2 Old pressure old volume = new pressure new volume Pressure must be in the same units and volume must be in the same units

Example #1 of Boyle’s law A sample of neon to be used in a neon sign has a volume of 1.51 L at a pressure of 635 torr. Calculate the volume of the gas after it is pumped into the glass tubes of the sign, where is shows a pressure of 785 torr. Initial After P1V1 = P2V2 P 635 torr 785 torr V2= P1V1 V 1.51 Liters ?? P2 V2 = (635 torr)(1.51L) (785 torr) V2= 1.22 L

Example #2 of Boyle’s law A sample of carbon dioxide occupies a volume of 3.50 liters at 125 kPa pressure. What pressure would the gas exert if the volume was decreased to 2.00 liters? Initial After P1V1=P2V2 P 125 kPa?? P2 = P1V1 V 3.50 L 2.00L V2 P2= (125kPa) (3.50L) 2.00 L P2= 219 kPa

Example #3 of Boyle’s law A 2.00 L container of nitrogen is expanded to 6.40 L. If the gas is now at standard pressure, what was the pressure in the beginning? Initial After P1V1= P2V2 P ?? 1.00 atm P1= P2V2 V 2.00L 6.40 L V1 P1= (1.oo atm) (6.40 L) 2.00L P1= 3.20 atm

Volume and temperature- Charles’s law Charles’s law- the volume and temperature ( in Kelvin) of a gas are directly proportional (at a constant pressure ) Celsius temperature = Kelvin temperature V 1 = V 2 T 1 T 2 Temperature must be in Kelvin because you can’t have a negative volume!! Volume must be in the same units

Example of Charles’s Law A child blows a soap bubble that contains air at 28 o C and has a volume of 23cm 3 at 1 atm. As the bubble rises, it encounters a pocket of cold air (temperature 18 o C). If there is no change in pressure, will the bubble get larger or smaller as the air inside cools to 18 o C? Calculate the new volume of the bubble. V1 = V2 T1 T2

answer V2 = V1 T2 V2= (23 cm 3 ) (291K) T1 (301K) = 301K = 291 K V2= 22 cm 3

Volume and Moles- Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas is Avogadro’s Law V 1 = V 2 n 1 n 2

Example for Avogadro’s law Consider two samples of nitrogen gas. Sample 1 contains 1.5 moles of N2 and has a volume of 36.7L while sample 2 has a volume of 16.5 L. Both are at 25 o C and 1 atm. Calculate the number of moles of N2 in sample 2. V1 = V2 n1 n2

answer n2= (n1 ) (V2) n2= ( 1.5 mols) (16.5L) (V1) (36.7L) n2=.67 moles

Gay-Lussac’s Law P1 = P2 T1 T2 Pressure and Temperature have a directly proportional relationship