WHAT FORMULA IS USED FOR SOLVING

BOYLE’S LAW? FOR CHARLES’ LAW?Quick Write p128:

Gay-Lussac's Law

Joseph-Louis Gay-Lussac In the early 1800’s, Gay-Lussacascended to a height ofapproximately 23,000 ft in a hotair balloon to study variations inthe Earth's electro-magneticintensity relative to altitude. Onthis flight, he experienced theeffects of oxygen deprivation butstill managed to collect airsamples at over 20,000 ft, studythe variation of pressure andtemperature, and continue hisobservations on electro-magnetism.

What is Gay-Lussac’s Law?

Gay-Lussac's Law states

that the pressure of a

sample of gas at constant

volume, is directly

proportional to its

temperature in Kelvin.

Mathematically Gay-Lussac’s Law can beexpressed as P1 = P2

T1 T2

P1 T2 = P2 T1

• P1 is the initial pressure

• T1 is the initial temperature (in Kelvin)

• P2 is the final pressure

• T2 is the final temperature (in Kelvin)

Gay-Lussacs’s Law Practice Problems

1. A tank of gas has a pressure of 2.75 atm at 20°C. What will be the new pressure in the tank if its temperature is increased to 100°C?

P1 T2 = P2 T1

P1:

T1:

P2:

T2:

2. A gas system has initial pressure of 793

kPa and temperature of 34.5°C If the

pressure changes to 446 kPa, what will the

resultant temperature be in K?

P1 T2 = P2 T1

P1:

T1:

P2:

T2:

COMBINED GAS LAW

The combined gas law is a combination of

Boyle's Law and Charles's Law; hence its

name the combined gas law. In the

combined gas law, the volume of gas is

directly proportional to the absolute

temperature and inversely proportional to

the pressure.

Mathematically, Combined Gas Law can be expressed as P1V1 = P2V2

T1 T2

P1V1 T2 =P2V2 T1 • P1 is the initial pressure

• V1 is the initial volume

• T1 is the initial temperature (in Kelvin)

• P2 is the final pressure

• V2 is the final volume

• T2 is the final temperature (in Kelvin)

Combined Gas Law Practice Problems

1. A sample of gas occupies a volume of 75 mL at a pressure of 725 mmHg and a temperature of 18°C. What volume would this sample of gas occupy at 800 mmHg and 298K?

P1V1 T2 =P2V2 T1

P1:

V1:

T1:

P2:

V2:

T2:

2. A sample of gas occupies a volume of 3.75 L at a pressure of 1.15 atm and a temperature of 25°C. At what

temperature will the sample of gas occupy 4 L at a pressure of 1 atm?

P1V1 T2 =P2V2 T1

P1:

V1:

T1:

P2:

V2:

T2:

Abbreviationsatm – atmospheremm Hg - millimeters of mercurytorr - another name for mm HgPa - Pascal (kPa = kilo Pascal)K - Kelvin°C - degrees Celsius

Problems:

1. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and

600.0 torr of pressure?

2.500.0 liters of a gas are prepared at 700.0 mm Hg and 200.0 °C. The gas is placed into a tank under high pressure. When the

tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the volume of the gas?

3. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0 °C to 30.0 °C and a

pressure change from 760.0 mm Hg to 360.0 mm Hg?

4.What is the volume of gas at 2.00 atm and 200.0 K if its original volume was 300.0 L at 0.250 atm and 400.0 K.

5.At conditions of 785.0 torr of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume

of the same gas at 745.0 torr and 30.0 °C?

6.A gas occupies a volume of 34.2 mL at a temperature of 15.0 °C and a pressure of 800.0 torr. What will be the volume of this

gas at standard conditions?

7.The volume of a gas originally at standard temperature and pressure was recorded as 488.8 mL. What volume would the same

gas occupy when subjected to a pressure of 100.0 atm and temperature of minus 245.0 °C?

8.At a pressure of 780.0 mm Hg and 24.2 °C, a certain gas has a volume of 350.0 mL. What will be the volume of this gas under

STP

ConversionsK = °C + 2731 cm3 (cubic centimeter) = 1 mL (milliliter)1 dm3 (cubic decimeter) = 1 L (liter) = 1000 mLStandard Conditions0.00 °C = 273 K1.00 atm = 760.0 mm Hg = 101.325 kPa = 101,325 Pa

Combined Gas Law Worksheet: ISN p 127