Loading in 5 sec....

Chapter 11 – Molecular Composition of GasesPowerPoint Presentation

Chapter 11 – Molecular Composition of Gases

- 152 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Chapter 11 – Molecular Composition of Gases' - verdad

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

11-1 Volume-Mass Relationships of Gases

- Joseph Gay-Lussac, French chemist in the 1800s, found that at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers.
- This is called Gay-Lussac’s law of combining volumes of gases.
- Today we know that these volume relationships are given by the coefficients of a balanced chemical equation and are equivalent to the mole ratios of gaseous reactant and products.

1-1 Gay-Lussac’s Law of Combining Volumes

- hydrogen + oxygen water vapor
- hydrogen + chlorine hydrogen chloride

11-1 Volume-Mass Relationships of Gases

- In 1811, Amedeo Avogadro proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
- This is known as Avogadro’s law.

11-1 Avogadro’s Law

- H2(g) + Cl2(g) 2HCl(g)
- One volume of hydrogen combines with one volume of chlorine to form two volumes of hydrogen chloride.
- One molecule of hydrogen combines with one molecule of chlorine to form two molecules of hydrogen chloride.

11-1 Avogadro’s Law

- Gas volume (V) is directly proportional to the number of particles (n) at constant temperature and pressure.
V = kn

or

k = V/n

11-1 Molar Volume of Gases

- One mole of any gas will occupy the same volume as any other gas at the same temperature and pressure, regardless of the mass of the particle.
- The volume occupied by one mole of any gas at STP is called the standard molar volume of a gas and is equal to 22.4 L/mol.

11-1 Molar Volume of Gases

- A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present?
- At STP, a sample of neon gas occupies 0.55 L. How many moles of neon gas does this represent?

11-2 The Ideal Gas Law

- To describe a gas sample, four quantities are needed: pressure, volume, temperature, and number of moles
- V α 1/P (Boyle’s Law)
- V α T (Charles’ Law)
- V α n (Avogadro’s Law)
- V α 1/P x T x n
- V = R x 1/P x T x n
- PV = nRT
- R is the gas constant.

P is pressure in atm or kPa

V is volume in L (dm3)

N is moles

T is temp in Kelvin

11-2 The Value of R

- Remember, one mole of any gas at STP has a volume of 22.4 L. We can use this definition to determine the value of R.
- If pressure is in atm…
- If pressure is in kPa…
- Remember, 1 L = 1 dm3

11-2 The Ideal Gas Law

- The ideal gas law is the mathematical relationship among pressure, volume, temperature and the number of moles of a gas.
- PV = nRT

P – pressure (atm or kPa)

V – volume (L)

n – moles

R – gas constant (0.0821 L.atm/mol.K or 8.31 L.kPa/mol.K)

T – temperature (K)

11-2 Ideal Gas Law

- What volume will be occupied by 0.21 moles of oxygen gas at 25°C and 1.05 atm of pressure?

11-2 Ideal Gas Law

- A sample of carbon dioxide gas has a mass of 1.20 g at 25°C and 1.05 atm. What volume does this gas occupy?

11-2 The Ideal Gas Law

- Variations:
n = m/MM

so PV = mRT/MM

and MM = mRT/PV

D = m/V

so D = MMP/RT

11-2 The Ideal Gas Law

- What is the molar mass of a gas if 0.427 g of the gas occupies a volume of 125 mL at 20.0°C and 0.980 atm?

11-2 The Ideal Gas Law

- What is the density of argon gas, Ar, at a pressure of 551 torr and a temperature of 25°C?

11-3 Stoichiometry of Gases

- Volume-Volume calculations – just use the mole ratio!
C3H8 + 5O2 3CO2 + 4H2O

- What volume, in L, of oxygen, is required for the complete combustion of 0.350 L of propane?

11-4 Effusion and Diffusion

- Diffusion – the gradual mixing of two gases due to their spontaneous, random motion.
- Effusion – the process by which the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

11-4 Effusion and Diffusion

The rates of effusion and diffusion depend on the relative velocities of the gas molecules.

- The velocity of a gas varies inversely with its mass. Lighter molecules move faster than heavier molecules at equivalent temperatures.

Remember, temperature is a measure of average kinetic energy.

Particles of two gas samples (A and B) at the same temperature have the same average kinetic energy.

KE =1/2 mv2

Graham’s Law of Effusion compares rates of effusion and diffusion for gases.

The relationship can be derived easily.

11-4 Effusion and Diffusion11-4 Effusion and Diffusion energy.

- The rate of effusion or diffusion for gases depends on the average velocities of the particles.
- Graham’s Law of Effusion – For two gases, A and B, at the same temperature, the following relationship exists…

11-4 Effusion and Diffusion energy.

- Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.

11-4 Effusion and Diffusion energy.

- If a molecule of neon gas travels at an average of 400 m/s at a given temperature, estimate the average speed of a molecule of butane gas, C4H10, at the same temperature.

Download Presentation

Connecting to Server..