# Gas Behavior Notes

Kinetic Molecular Theory

- Kinetic molecular theory - used to describe the behaviors of an ideal gas
- A gas is a collection of particles (molecules or atoms) in constant, straight0line motion.
- Gas particles do not attract or repel each other - they do not interact. All collisions are perfectly elastic (the energy before the collision equals the *energy after the collision - as you increase temperature, this becomes more likely). This is why gases behave better at higher temperatures
- There is a lot of space between gas particles compared with the size of the particle themselves - assume the volume is zero. Hydrogen gas behaves more *ideally than carbon dioxide gas because it is smaller
- The average kinetic energy of gas particles is proportional to the temperature of the gas in Kelvins
- Factors that make a gas more ideal:
- Increase the temperature of the gas
- Decrease the pressure
- Properties of gasses

- Gasses are highly compressible - this is because there is a lot of space between particles
- Take the shape of their container
- Low interactions
- Less dense than solids and liquids
- Gas is not a condensed state of matter - solids and liquids are

Pressure

- Pressure
- Caused by the collision of gas molecules
- If you increase the number of the collisions, the pressure increases
- Factors Depending on Pressure
- The amount of gas - as amount of gas increases in a fixed space, the amount of collisions increase and so the pressure increases as well
- Altitude - as altitude increases, pressure decreases
- Temperature - as temperature increases, pressure increases because the amount of collisions increases and those collisions have a higher energy
- Volume - as volume increases, pressure decreases and vice versa

STP

- Standard temperature and pressure
- Standard temperature - 0 degrees C or 273 K
- Units for Pressure
- Barometer measures pressure
- Units
- Atmosphere (atm)
- Pascal (Pa)
- Usually kPa because pascals are so small
- Millimeters of mercury (mm Hg)
- Torr - 1 torr equals one mm Hg
- Pounds per square inch (psi)
- Standard Pressure Units
- 1 atm
- 101.3 kPa
- 760 mm Hg or torr
- 14.7 psi
- Problems
- Convert 159.2 kPa to atm and mm Hg

- 159.2 kPa/1 * 1atm/101.3 kPa = 1.571 atm
- 159.2 kPa/1 * 760mm Hg/101.3 kPa = 1194 mm Hg
- Boyle’s Law
- Boyle’s law
- When the temperature of a gas remains constant, the pressure of the gas is inversely proportional to the volume of the gas
- PV = k
- K stays the same so long as the temperature stays the same
- P1V1 = P2V2

- Problems
- The volume of a gas is 4.23 L at 99.6 kPa and 24 degrees C. What volume will the gas occupy at the 93.3 kPa and 24 degrees C?
- 4.23L * 99.6 kPa = xL * 93.3 kPa
- X = 4.52 L
- A gas with a volume of 525 mL at a pressure of 625mm Hg expands to a volume of 725 mL. What is the pressure if the temperature remains constant?
- 525mL * 625 torr = 725mL * x torr
- X = 453 torr

Charles’s law

- Charles’s law
- When the pressure of a gas remains constant the volume of the gas is directly proportional to the Kelvin temperature
- Absolute zero (0 K - no molecular motion)
- V/T = K
- V1/T1 = V2/T2
- Problems
- The volume of a gas is 0.8 dm^3 at 0.998 atm and 0 degrees C. What volume will it occupy at 0.998 atm and 24 degrees C?
- 0.8 dm^2/273.15K = x dm^3/297.15 K
- 0.87 dm3
- 2.25L of air at -25 degrees C is warmed to 80 degrees C. What is the new volume if the pressure remains constant?
- 2.25L/248.15K = xL/353.15K
- X = 3.20 L

Combined Gas Law

- Combined gas law
- P1V1/T1 = P2V2/T2
- Problems
- The volume of a gas at 26 degrees C and 75 kPa is 10.5 L. What would be the final temperature if the volume was 9.25L and the pressure was 116 kPa
- 75kPa * 10.5L/299 degrees C = 9.25L * 116 kPa/x degrees K
- X = 407 K
- The volume of a gas is 26.0 mL at 35.0 degrees C and 725 mm Hg. What will the volume be at STP?
- 725 torr * 26.0 mL/308 K = 760 torr * x mL/273.15 K
- X = 22.0 mL
- Avogadro’s Law
- Avogadro’s Law

- When the pressure and temperature of a gas remain constant, the volume of the gas is directly proportional to the number of moles
- V/n = k
- V1/n1 = V2/n2
- Problem
- A 4.75 L sample of helium contains 12.3g of helium. How many additional moles of helium gas must be added to the sample to obtain a volume of 6.25L? Assume
**constant temperature and pressure.**12.3g He/1 *1 mol He/4.00g He = 3.08 mol He - 4.75 L/3.08 mol = 6.25L/x mol
- X = 4.05 mol
- 4.05 - 3.08 = 0.97 mol

The Ideal Gas Law

- Ideal Gas Law
- PV = nRT
- R is the universal gas constant
- R =0.08206 atm *L/k*mol
- Problems
- A balloon contains 7.28g of He at 104.2 kPa and a temperature of 21.0C. What is the volume of the balloon?
- 1.029 atm * x L = 1.82 mol He * 0.08206 atm*L/k*mol * 294.15 K
- X = 42.6 L
- A container has a volume of 725 mL with nitrogen gas has pressure of 97.8 kPa at 25.0 degrees C. How many grams of nitrogen gas does the container hold?
- 0.965 atm * 0.725 L = x mol * 0.08206 atm*L/K*mol * 298.15 K
- X = 0.0286 mol
- 0.0286 mol N2/1 * 28.02 g N2/1 mol N2 = 0.801 g N2

- Molar Mass of a Gass
- PV = nRT
- R = 0.08206 atm*L/mol*K
- MM = g/mol
- A sample of gas has a mass of 0.435g. Its volume is 225 mL at a temperature of 45.2 degrees C and a pressure of 883 mmHg. Find its molar mass
- 883 torr = 1 atm/760 torr = 1.16 atm
- 1.16 atm * 0.225 L = x mol * 0.08206 atm*L/mol*K * 318.35 K
- X = 0.9996 mol
- 0.4535g/0.9996 mol = 43.5 g/mol
- A sample of gas has a mass of 925 mg. Its volume is 275mL at a temperature of 76.2 degrees C and a pressure of 928 mmHg. Find its molar mass.
- 928mmHg/1 * 1atm/760 torr = 1.2211 atm
- 1.211 atm * 0.275 L = x mol * 0.08206 atm*L/mol*K * 349.35K
- X = 0.01162 mol
- MM = g/mol = 0.925g/0.01162 mol = 79.6 g/mol
- Mixtures of Gases; Dalton’s Law
- Partial pressure - the pressure exerted by an individual gas
- Dalton’s law of partial pressures - the total pressure of a mixture of gases is equal to the sum of partial pressures of the gases.

- P (total) = p1 + p2 + p3
- Problems
- The pressure in an automobile tire filled with air is 245.0 kPa. The Po2 is 51.3 kPa; pCo2 is 23.5kPa. What is the partial pressure of nitrogen gas?
- 245.0 kPa + 51.3 kPa + 23.5 kPa + xkPa = 170.2 kPa (n2)
- An air sample contains 1.38% Co2. If the total pressure is 748 mmHg, what is the partial pressure of CO2?
- X Pco2/478 mmHg = 0.0138 Co2/1 = 10.3

Graham’s Law of Effusion

- Diffusion - the movement of particles from a higher to a lower concentration until they reach equilibrium
- Effusion - when a gas escapes through a tiny hole (from a higher to a lower concentration)
- When you blow up a balloon, it is a bit smaller the next day because the gas is going to escape
- Hydrogen gas travels faster because it is smaller and it has a lower molar mass
- Graham’s Law of effusion
- The effusion rate is inversely proportional to the square root of the gases’ molar mass or density
- Rate1/Rate2 = sqrt(M2/M1)
- Rate1 is rate of the lower mass because is is going to travel faster and rate2 is going to the rate of the higher molar mass because it is going to travel *slower
- M2 is going to be the higher molar mass and M1 is the lower molar mass
- Problems
- Compare the effusion rates of helium and nitrogen gas.
- Rate1/rate2 = sqrt(28.02/4.00) = 2.65
- Helium travels 2.65 times faster than nitrogen gas
- Compare the effusion rates of oxygen gas and sulfur dioxide
- Rate2/rate2 = sqrt(64.02/32) = 1.41
- Oxygen gas travels 1.41 times faster than sulfur dioxide
- Gases in Chemical Reactions

- Problems
- In this reaction 4.58 L of O2 were formed at 742 mm Hg and 25 degrees C. How many grams of Ag2O decomposed?
- 2Ag2O(s) → 4Ag(s) + O2(g)

- PV = nRT
- 742 mm Hg * 1 atm/760 mm Hg = 0.976 atm
- (0.976 atm)(4.58L) = x(0.08206 atm*L/mol * K)(298K)
- X = 0.183 mol O2
- 0.183 mol O2/1 * 2 mol Ag2O/1 mol O2 * 231.74 g Ag2O/1 mol Ag2O = 84.8 g
- Oxygen gas reacts with powdered aluminum. How many liters of O2 gas, measured at 782 mm Hg and 23.0 degrees C are required to compeltely react with 1.25 g of Al?
- 3O2(g) + 4Al(s) = 2Al2O3(s)
- 1.25 g Al/1 * 1 mol Al/26.98 g Al * 3 mol O2/4 mol Al = 0.0347 mol O2*
- PV = nRT
- 1.0289 atm * x L = 0.0347 mol O2 * 0.08206 atm*L/mol*K * 296 K
- X = 0.818 L O2

- 1 mol of gas = 22.4 L at STP
- Problems
- How many liters of oxygen (at STP) are required to form 7.25 x 10^23 molecules of H2O?
- 2H2(g) + O2(g) → 2H2O
- 1.25 x 10^23 molec H2O * 1 mol H2O/6.022 x 10^23 molec H2O * 1 mol O2/2 mol H2O = 13.5 L O2
- The following reaction consumes 2.45 g of CO(g): CO(g) + H2O(g) → CO2(g) + H2(g). How many total liters of gas are formed if the products are collected at STP?

- 2.45 g CO/1 * 1 mol CO/28.01 g CO * 2 mol product/1 mol CO * 22.4 L/1 mol product = 3.92 L of products