gas laws and gas stoichiometry. kinetic –molecular theory particles of matter (solid, liquid, or...

Gas Laws and Gas Stoichiometry

Kinetic –Molecular Theory• Particles of matter (solid, liquid, or gas) are always in motion.

This motion has consequences .

• Solid (defined shape and definite volume)• Liquid (undefined shape and definite volume)• Gases (undefined shape and volu)me

Kinetic-Molecular Theory of Gases• Gases are large numbers of tiny particles

• Particles always moving in straight lines in all directions. Therefore, they have “kinetic energy”

• These particles have elastic collisions between other particles and the container• Elastic – no loss of KE after collision

• There is no attraction or repulsion between the particles

• Kinetic energy of the particles is proportional to the temperature of the matter

Ideal Gas vs. Real gas

• Ideal Gas:

• Completely follows kinetic molecular theory (K-M theory) of gases• Hydrogen, maybe• Helium, sometimes

• Real Gas:

• Does not behave by K-M theory

K-M theory and nature of gases• Expansion

• Fluidity (liquids and gases)

• Low Density

• Compressibility

• Diffusion

Characteristics

1. Expansion

2. Fluidity

Characteristics

3. Low density

Characteristics

4. Compressibility

Characteristics• Diffusion

4 quantities that can be measured• Volume – How much space it takes up

• Pressure – Amount of collisions particles have with container

• Temperature – Average kinetic energy of the gas particles

• Quantity or number of molecules – moles

• We use these quantities to “work” with gases

Pressure

• Pressure = Force / Area

• Units of Pressure

• 1 atm = 760 mm Hg = 760 torr = 1.01 x 105 pascals

Convert a pressure of 0.830 atm to mmHg

Temperature Scales• Absolute zero = -273.15oC = 0 K (not 0oK)

• Therefore K = 273 + oC• 0 oC = 273 K

• As temperature increases, the number of gas collisions increases.

Temp. conversions

K = oC + 273

25.0 oC = ? K

STP• “Slap The Pupils” -- No• “Such Total Pigs” -- No• Standard Temperature and Pressure

• The volume of a gas depends on temperature and pressure. In order to compare volumes of gases, need a standard.

• 0 degrees Celsius, 273 K• 1 atm of pressure

Boyle’s Law• Volume off a fixed mass of gas

varies inversely with pressure at a constant temperature

Boyle’s Law

• P1V1 = k and P2V2 = k

• Therefore P1V1 = P2V2

A sample of oxygen gas occupies a vol. of 150 mL at a pressure of 720 mmHg. What would the volume be at 750 mmHg press.?

France, early 1800’s• Hot air balloons were extremely popular• Scientists were eager to improve the performance of their

balloons. Two of the prominent French scientists were Jacques Charles and Joseph-Louis Gay-Lussac,

Charles’ Law

• The volume of a fixed mass of gas varies directly with the Kelvin temperature at constant pressure

Charles’ Law

• V1/ T1 = V2/ T2

A sample of Ne gas has a vol. of 752 mL at 25.0 deg. C. What is the vol. at 50.0 deg. C?

T1 = 25.0 deg. C = 298 K

T2 = 50.0 deg C = 323 K

V1 = 752 mL

Gay-Lussac’s Law• The pressure of a fixed mass of

gas varies directly with the Kelvin temp. at constant volume

• P1/ T1 = P2/ T2

P1/ T1 = P2/ T2

A sample of N gas is at 3.00 atm of pressure at 25 deg C what would the pressure be at 52 deg C?P1 = 3.00 atm

T1 = 25 deg C = 298 K

T2 = 52 deg C = 325 K

P2 = ?

P1/ T1 = P2/ T2

or P2 = P1T2/ T1

Combined Gas Law

• Relates Pressure, Volume, and Temperature• Notice it is a combination of Boyles, Charles, and Gay-Lussac’s

Laws

• P1V1/ T1 = P2V2/ T2

A helium filled balloon has a vol. of 50.0 L at 25 deg C and 820. mmHg of pressure. What would the vol. be at 650 mmHg pressure and 10. deg C?

• P1V1/ T1 = P2V2/ T2

• V1 = 50.0 L

• T1 = 25 deg C = 298 K

• P1 = 820. mmHg

• T2 = 10. deg C = 283 K

• P2 = 650 mmHg

• V2 = ?

Molar Volume of a Gas• One mole of a gas has the same volume at STP as any other gas• 22.4 L / mole at STP

What volume would 0.0680 mol of oxygen gas occupy at STP?

What about 0.0680 mol of nitrogen gas at STP?

Practice Problem

Ideal Gas Law

PV = nRTn = Number of MolesR = Ideal Gas Constant

0.0821 atm*L / mol*K V = Volume (must be in liters)P = Pressure (must be in atmospheres)T = Temperature (must be in Kelvin)

Practice Problem

What is the P (in atm) exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?

Pressure = 1.22 atm

Gas Stoichiometry Uses volume-volume calculations (e.g. L L) Liters can be used just like mole to mole ratios in a factor label

problem

Practice Problem How many L of oxygen are required for the complete combustion of

0.250 L of propane?

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

5 L O2 needs 1 L C3H8

Practice Problem How many grams of calcium carbonate must be decomposed to

produce 2.00 L of CO2 at STP?

CaCO3(s) CaO(s) + CO2(g)

2.00 L CO2 x 1 mol CO2 x 1 mol CaCO3 X 100.086 g CaCO3

22.4 L CO2 1 mol CO2 1 mol CaCO3

= 8.94 g CaCO3