Gases

Gases:

• Gases are composed of XX that are moving

around very XX in their container(s).

- Examples: H2 (g), HF (g), He (g)

• All gases have the following characteristics:

• They take on the XX and of their containers.

• They are the most XX of the states of matter.

• They will mix XX and completely when confined to the

same container.

• Gases have much lower XX than liquids and solids

Gases are composed of particles that are moving

around very fast in their container(s).

- Examples: H2 (g), HF (g), He (g)

• All gases have the following characteristics:

• They take on the volume and shape of their containers.

• They are the most compressible of the states of matter.

• They will mix evenly and completely when confined to the

same container.

• Gases have much lower densities than liquids and solids

The four variables we use to describe a gas are:P,V,T,R,N

P=pressure V= volume T=temperature n=moles of gas

The four variables we use to describe a gas are:P,V,T,R,N

P=pressure V= volume T=temperature n=moles of gas

Kinetic Molecular Theory (KMT)

Kinetic Molecular Theory explains the behavior of XX XX

1. Gas molecules move randomly at XX speeds and in all possible XX.

2. The average kinetic energy of gas molecules is proportional to the temperature in Kelvin. As temperature XX average kinetic energy XX.

3. The volume occupied by the actual gas particles is XX compared with the distance between the particles. This means gases contain a large amount of XX XX

4. Ideal gas particles do not exert attractive or repulsive forces on one another.

5. Collisions between gas particles are elastic. Energy can be transferred between particles during a collision, but the total energy is not lost.

Kinetic Molecular Theory (KMT)

Kinetic Molecular Theory explains the behavior of gas particles.

1. Gas molecules move randomly at different speeds and in all possible directions.

2. The average kinetic energy of gas molecules is proportional to the temperature in Kelvin. As temperature increases, average kinetic energy increases.

3. The volume occupied by the actual gas particles is negligible compared with the distance between the particles. This means gases contain a large amount of empty space.

4. Ideal gas particles do not exert attractive or repulsive forces on one another.

5. Collisions between gas particles are elastic. Energy can be transferred between particles during a collision, but the total energy is n

Pressure and Temperature at the Molecular Level

Pressure is related to how often gas molecules XX with the XX of their container.

More collisions with the walls of the container result in greater XX

Temperature is related to XX XX.

Higher temperature means the gas particles move XX on average

Pressure and Temperature at the Molecular Level

Pressure is related to how often gas molecules collide with the walls of their container.

More collisions with the walls of the container result in greater pressure.

Temperature is related to molecular motion.

Higher temperature means the gas particles move faster on average

Gas Pressure

Pressure is force per unit area.

Common pressure units include:

• xx = atmosphere

• xx

• xxxx = millimeters of mercury

• XX = pascal

• xx = pounds per square inch

Important pressure conversions:

1 atm = 760 torr

1 atm = 760 mmHg

1 atm = 1.01325 × 10⁵ Pa

Standard pressure = 1 atm.

The density of mercury given in the lecture is 13.6 g/mL. Module 5_Gases

Example of converting torr to atm:

810 torr × (1 atm / 760 torr) = 1.07 atm

Gas Pressure

Pressure is force per unit area.

Common pressure units include:

• atm = atmosphere

• torr

• mmHg = millimeters of mercury

• Pa = pascal

• psi = pounds per square inch

Important pressure conversions:

1 atm = 760 torr

1 atm = 760 mmHg

1 atm = 1.01325 × 10⁵ Pa

Standard pressure = 1 atm.

The density of mercury given in the lecture is 13.6 g/mL. Module 5_Gases

Example of converting torr to atm:

810 torr × (1 atm / 760 torr) = 1.07 atm

Boyle's Law

Boyle's Law describes the relationship between XX and XX

Temperature and the number of moles remain XX.

The equation is:

XX=XX

Pressure and volume are XX XX.

As pressure XX, volume XX

As pressure XX, volume XX.

P X → V X

P X → V X

For example, if a gas occupies 1.0 L at 1 atm and the pressure increases to 2 atm, the volume decreases to 0.50 L.

This happens because increasing the pressure compresses the gas into a smaller volume.

Boyle's Law

Boyle's Law describes the relationship between pressure and volume.

Temperature and the number of moles remain constant.

The equation is:

P₁V₁ = P₂V₂

Pressure and volume are inversely proportional.

As pressure increases, volume decreases.

As pressure decreases, volume increases.

P ↑ → V ↓

P ↓ → V ↑

For example, if a gas occupies 1.0 L at 1 atm and the pressure increases to 2 atm, the volume decreases to 0.50 L.

This happens because increasing the pressure compresses the gas into a smaller volume.

Combined Gas Law

The combined gas law is used when XX, XX, and XX change for the same sample of gas.

The equation is:

XX = XX

Temperature must be in XX

It is helpful to organize these problems into "before" and "after."

Before:

P₁
V₁
T₁

After:

P₂
V₂
T₂

Example from the lecture:

A sample of N₂ occupies 750 mL at 75°C and 810 torr. Find the volume at STP.

Before:

V₁ = 750 mL
P₁ = 810 torr
T₁ = 75°C + 273 = 348 K

After:

V₂ = ?
P₂ = 760 torr
T₂ = 273 K

Set up:

(810 × 750) / 348 = (760 × V₂) / 273

Solving gives approximately:

V₂ = 627 m

Combined Gas Law

The combined gas law is used when pressure, volume, and temperature change for the same sample of gas.

The equation is:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Temperature must be in Kelvin.

It is helpful to organize these problems into "before" and "after."

Before:

P₁
V₁
T₁

After:

P₂
V₂
T₂

Example from the lecture:

A sample of N₂ occupies 750 mL at 75°C and 810 torr. Find the volume at STP.

Before:

V₁ = 750 mL
P₁ = 810 torr
T₁ = 75°C + 273 = 348 K

After:

V₂ = ?
P₂ = 760 torr
T₂ = 273 K

Set up:

(810 × 750) / 348 = (760 × V₂) / 273

Solving gives approximately:

V₂ = 627 m

Avogadro's Law

Avogadro's Law describes the relationship between XX and number of XX

XX and XX remain constant.

The equation is:

XX=XX

Volume and moles are directlyXX

As the number of moles XX, volume XX

As the number of moles XX, volume XX

n X → V X

n X → V X

Equal volumes of gases under the same conditions contain equal numbers of moles.

The identity of the gas does not matter for this relationship

Avogadro's Law

Avogadro's Law describes the relationship between volume and number of moles.

Pressure and temperature remain constant.

The equation is:

V₁/n₁ = V₂/n₂

Volume and moles are directly proportional.

As the number of moles increases, volume increases.

As the number of moles decreases, volume decreases.

n ↑ → V ↑

n ↓ → V ↓

Equal volumes of gases under the same conditions contain equal numbers of moles.

The identity of the gas does not matter for this relationship

How the Gas Laws Connect

Boyle's Law:

X∝ X

Charles's Law:

X ∝ X

Avogadro's Law:

X ∝ X

Combining these relationships gives:

X ∝XX

This relationship leads to the Ideal Gas Law:

XX = XX

How the Gas Laws Connect

Boyle's Law:

V ∝ 1/P

Charles's Law:

V ∝ T

Avogadro's Law:

V ∝ n

Combining these relationships gives:

V ∝ nT/P

This relationship leads to the Ideal Gas Law:

PV = nRT

Standard Temperature and Pressure (STP)

STP stands for Standard Temperature and Pressure.

At STP:

Temperature = XX = XX

Pressure = XX atm

At STP, one mole of an ideal gas occupies approximately:

1 mol gas = XX XX

Therefore:

1 mol = 22.4 L

2 mol = 44.8 L

0.5 mol = 11.2 L

The 22.4 L/mol conversion applies to the STP conditions used in this module. Module 5_Gases

Standard Temperature and Pressure (STP)

STP stands for Standard Temperature and Pressure.

At STP:

Temperature = 0°C = 273.15 K

Pressure = 1.00 atm

At STP, one mole of an ideal gas occupies approximately:

1 mol gas = 22.4 L

Therefore:

1 mol = 22.4 L

2 mol = 44.8 L

0.5 mol = 11.2 L

The 22.4 L/mol conversion applies to the STP conditions used in this module. Module 5_Gases

Rearranging the Ideal Gas Law

Original equation:

PV = nRT

To solve for pressure:

XX

To solve for volume:

XX

To solve for moles:

XX

To solve for temperature:

XX

Rearranging the Ideal Gas Law

Original equation:

PV = nRT

To solve for pressure:

P = nRT/V

To solve for volume:

V = nRT/P

To solve for moles:

n = PV/RT

To solve for temperature:

T = PV/nR

Choosing Which Gas Law to Use

First determine whether the problem describes a change from one condition to another.

Words such as "increases from," "decreases from," "changes to," or "what volume would it occupy at" usually indicate a before-and-after problem.

If pressure and volume change while T and n stay constant:

XX

If volume and temperature change while P and n stay constant:

XX

If volume and moles change while P and T stay constant:

XX

If pressure, volume, and temperature change:

XX

If there is only one set of conditions and the problem gives some combination of P, V, n, and T:

XX

Choosing Which Gas Law to Use

First determine whether the problem describes a change from one condition to another.

Words such as "increases from," "decreases from," "changes to," or "what volume would it occupy at" usually indicate a before-and-after problem.

If pressure and volume change while T and n stay constant:

P₁V₁ = P₂V₂

If volume and temperature change while P and n stay constant:

V₁/T₁ = V₂/T₂

If volume and moles change while P and T stay constant:

V₁/n₁ = V₂/n₂

If pressure, volume, and temperature change:

(P₁V₁)/T₁ = (P₂V₂)/T₂

If there is only one set of conditions and the problem gives some combination of P, V, n, and T:

PV = nRT Module 5_Gases

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Using the Ideal Gas Law to Find Mass

If P, V, and T are given and the problem asks for grams, first use the Ideal Gas Law to determine moles.

n = PV/RT

Then convert moles to grams using molar mass:

mol × (g/mol) = grams

The overall pathway is:

P, V, T → moles → grams

The lecture includes an example using CO₂ collected in a flask and gives a final mass of 0.572 g CO₂.

Using the Ideal Gas Law to Find Mass

If P, V, and T are given and the problem asks for grams, first use the Ideal Gas Law to determine moles.

n = PV/RT

Then convert moles to grams using molar mass:

mol × (g/mol) = grams

The overall pathway is:

P, V, T → moles → grams

The lecture includes an example using CO₂ collected in a flask and gives a final mass of 0.572 g CO₂.

Finding Molar Mass Using Gas Information

Molar mass is:

MM = mass/moles

If P, V, T, and mass are given, first determine moles:

n = PV/RT

Then calculate:

MM = mass/n

The pathway is:

P, V, T → moles → molar mass

Finding Molar Mass Using Gas Information

Molar mass is:

MM = mass/moles

If P, V, T, and mass are given, first determine moles:

n = PV/RT

Then calculate:

MM = mass/n

The pathway is:

P, V, T → moles → molar mass

Molecular Formula Using Gas Information

Gas information can be used to determine the molar mass of a compound.

The general process is:

1. Determine theXX XX

2. Calculate the empirical formula XX

3. Use the gas information to determine the actual molar mass.

4. Divide the actual molar mass by the empirical formula mass.

5. The result should be a whole-number multiplier.

6. Multiply every subscript in the empirical formula by that number.

The lecture example gives:

80.0% C

20.0% H

Empirical formula = CH₃

Empirical formula mass = 15 g/mol. Module 5_Gases

Molecular Formula Using Gas Information

Gas information can be used to determine the molar mass of a compound.

The general process is:

1. Determine the empirical formula.

2. Calculate the empirical formula mass.

3. Use the gas information to determine the actual molar mass.

4. Divide the actual molar mass by the empirical formula mass.

5. The result should be a whole-number multiplier.

6. Multiply every subscript in the empirical formula by that number.

The lecture example gives:

80.0% C

20.0% H

Empirical formula = CH₃

Empirical formula mass = 15 g/mol. Module 5_Gases

Gas Density

The normal density equation is:

d = m/V

For gases, combining density with the Ideal Gas Law gives:

XX

d = density
P = pressure
MM = molar mass
R = gas constant
T = temperature

The equation can also be rearranged to determine molar mass:

XX

Gas density is directly proportional to molar mass when the other conditions are the same.

Therefore:

MM ↑ → density ↑

A gas with a larger molar mass will be denser than a gas with a smaller molar mass under the same conditions. Module 5_Gases

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Gas Density

The normal density equation is:

d = m/V

For gases, combining density with the Ideal Gas Law gives:

d = PMM/RT

d = density
P = pressure
MM = molar mass
R = gas constant
T = temperature

The equation can also be rearranged to determine molar mass:

MM = dRT/P

Gas density is directly proportional to molar mass when the other conditions are the same.

Therefore:

MM ↑ → density ↑

A gas with a larger molar mass will be denser than a gas with a smaller molar mass under the same conditions. Module 5_Gases

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Gas Stoichiometry

Gas stoichiometry follows the same basic rules as regular stoichiometry.

The most important idea is that moles are the bridge between substances.

The general pathway is:

Gas information A → mol A → mole ratio → mol B → requested information B

If P, V, and T are given, use:

n = PV/RT

to determine moles.

Then use the coefficients in the balanced chemical equation as the mole ratio.

If the final answer needs to be grams, convert the final moles to grams using molar mass.

Gas Stoichiometry

Gas stoichiometry follows the same basic rules as regular stoichiometry.

The most important idea is that moles are the bridge between substances.

The general pathway is:

Gas information A → mol A → mole ratio → mol B → requested information B

If P, V, and T are given, use:

n = PV/RT

to determine moles.

Then use the coefficients in the balanced chemical equation as the mole ratio.

If the final answer needs to be grams, convert the final moles to grams using molar mass.