1 Gas Laws Gas temperatures must be in Kelvins for Boyle Boyle s - - PDF document

SLIDE 1

1

Pressure

• Caused by particles colliding with surface of container
• P = force of collision/ unit area

Units of Pressure

1 atm = 760 torr = 760 mmHg 101.3 kPa =

Convert the following pressures to atmospheres:

891 mm Hg 93.0 kPa 501 torr mm Hg 1 atm 760 mm Hg

• = 1.17 atm

93.0 kPa 1 atm 101.3 kPa

• = 0.918 atm
• 501 torr

1 atm 760 torr

• = 0.659 atm

Atmospheric Pressure- the pressure of all the molecules of air above and around us. Higher you go in atmosphere: atmospheric P ↓ Deeper you go in ocean: water P ↑

• Smaller column of air above pushing down
• larger column of water above pushing down

Column of mercury

atmospheric pressure measured with a barometer:

• The height of the Hg = P in mm Hg
• r torr
• At sea-level, P = 760 mm Hg

With a barometer gas pressure measured with a manometer:

760 mm Hg 186 mm Hg 93 mm Hg

Gas P = 760 +186 = 946 mm Hg Gas P = 760 + 93 = 853 mm Hg 50 mm What is the pressure of the gas in the flask if the atmospheric P = 1.02 atm? Hg atmosphere gas

1.02 atm atm 760 mm Hg 1 atm

• = 775 mm Hg

775 mm Hg 775 mm Hg - 50 mm Hg = 50 mm Hg = 725 mm Hg

SLIDE 2

2

Kelvins =

Gas temperatures must be in Kelvins for all calculations:

Kelvins = °C + 273

Gas Laws Charles Charles’ ’s Law s Law Boyle Boyle’ ’s Law s Law

If T held constant PV = constant How are P and V of a gas related? Inversely related P1V1 = P2V2 Initial conditions = final conditions

V T = constant

• Only holds true if T in K

and P held constant How are T and V of a gas related? Directly related

V

1

T

1

= V

2

T

2

Avogadro Avogadro’ ’s Law s Law

• equal volumes of gas at same T and P contain the same

number of moles

• V and # moles of gas are directly proportional
• Molar volume of any gas at STP= 22.4 L

V

1

n1 = constant

V

1

n1 = V

2

n2

STP = 0°C 1 atm

Combined Gas Law Combined Gas Law

PV T = constant

• Use for any problems involving changes in P, T or V

Initial conditions final conditions P1 = 0.454 atm V1 = 3.48 L T1 = -15°C P2 = 0.616 atm V2 = ? T2 = 36°C

P

1V 1

T

1

= P

2V 2

T

2

initial final A gas has a pressure of 0.454 atm at -15°C and a volume

• f 3.48 L. If the conditions are changed so that the

temperature is 36°C and pressure is 0.616 atm, what will the new volume be? problem 1 + 273 = 258K + 273 = 309K

P

1V 1

T

1

= P

2V 2

T

2

V

2 = 3.07L

V

2 = V 1 T 2 P 1

T

1 P 2

V

2 =

V

2 = 3.48L

3.48L 309K 309K 0.454 atm 258K 258K 0.616 atm

Initial conditions final conditions P1 = 1.0 atm V1 = 0.675 L T1 = 35°C P2 = 1.0 atm V2 = 535 ml T2 = ? A gas at 35°C and 1 atm has a volume of 0.675 L. With pressure constant, the temperature is lowered and the volume

• f the gas reduces to 535 ml. Calculate the final T.

P

1V 1

T

1

= P

2V 2

T

2

Eliminate P

V

1

T

1

= V

2

T

2

T

2 =

problem 2 + 273 = 308K

T

2 = V 2 T 1

V

1

• K = 244 K

0.535 L 0.535 L 308K 0.675 L

= 0.535 L

SLIDE 3

3

v P

The Ideal Gas Law The Ideal Gas Law

• Summarizes all the Gas Laws:

Boyle: V inversely proportional P

V T

Absolute zero

Charles: Avogadro: V directly proportional T V directly proportional n