- 141 Views
- Uploaded on
- Presentation posted in: General

CHAPTER 9

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

- GASES: Their Properties and Behavior

- force per unit area
- Standard Pressure
- 760 mm Hg
- 760 torr
- 76 cm Hg
- 1 atmosphere
- 29.925 inches Hg
- 101.3 kPa
- 1.013 bar

density Hg = 13.6 g/mL

Evangelista Torricelli observed the first vacuum in 1643

At a given temperature, the product of pressure and volume of a definite mass of gas is constant.

- V = volume P = pressure k = constant

k1 = k2 (for same sample of gas at same T)

- Example 1: At 25 °C a sample of He has a volume of 400 mL under a pressure of 760 torr. What volume would it occupy under a pressure of 2.00 atm at the same T?

You must use the same units of pressure!

- Boyle recognized that temperature had a predictable effect on volume. Boyle’s Law
- Boyle was unable to define the relationship between temperature and volume.
- The relationship was solved by the balloonist, Charles. Charles’ Law
- Charles defined the relationship by using the temperature scale developed by Lord Kelvin.

absolute zero = –273.15 °C

- The volume of a gas is directly proportional to the absolute temperature at constant pressure.

K = °C + 273 V = volume T = temperature in K

k1 = k2 (for same sample of gas at same P)

- Example : A sample of hydrogen, H2, occupies 100 mL at 25 °C and 1.00 atm. What volume would it occupy at 50 °C under the same pressure?
- T1 = 25 °C + 273 = 298 K
- T2 = 50 °C + 273 = 323 K
- 273.15 is seldom used because the fraction is usually lost to significant figures

- STP
- P = 1.00000 atm or 101.3 kPa
- T = 273 K or 0 °C
- STP is a defined value. You will see this frequently in problems, quizzes and on exams and will be expected to know its values by memory.

- Boyle’s and Charles’ Laws can be combined into one statement called the Combined Gas Laws

Combined Gas Law

This is one of two equations that must be memorized.

- This general formula of the combined gas law can be rearranged to solve for any new T, P, or V

- Example : A sample of nitrogen gas, N2, occupies 750 mL at 75 °C under a pressure of 810 torr. What volume would it occupy at standard conditions?

V1 = 750 mL T1 = 348 K T2 = 273 K

P1 = 810 torr P2 = 760 torr

- Example : A sample of methane, CH4, occupies 260 mL at 32 °C under a pressure of 0.500 atm. At what temperature would it occupy 500 mL under a pressure of 1200 torr?

T1 = 305 K V1 = 260 mL V2 = 500 mL

P1 = 0.5 atm = 380 torr P2 = 1200 torr

This could also be worked using 1200 torr = 1.58 atm

- Avogadro’s Law
- At the same T and P, equal volumes of two gases contain the same number of molecules of gas.
- One mole of any gas has the same V at STP.
- standard molar volume = 22.4 L
- You are expected to know this value from memory!

- Example : One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given T & P.
- What is its molar mass?

What is its density at STP?

Remember at STP 1 mol occupies 22.4 L

The combination of these three laws yields

This rearranges to give the Ideal Gas Law

R is a proportionality constant known as the Universal Gas Constant.

At STP and 1 mol of gas

The various R values will be given to you on the exams.

- R has several values depending upon the choice of units
- R = 0.0821 L•atm/mol•K
- R = 8.314 J/mol•K
- R = 8.314 kg m2/s2•mol•K
- R = 8.314 dm3 kPa/mol•K
- R = 1.987 cal/mol•K
- Warning! To use the ideal gas law:
- P (pressure) must be in atmospheres (atm); and V (volume) must be in liters (L).

- You should memorize both the Ideal Gas Law

and the Combined Gas Law formulas

These two formulas will solve any problem involving gases.

- Example : What volume would 50.0 g of ethane, C2H6, occupy at 140 °C under a pressure of 1820 torr? (mol.wt. C2H6 = 30 g)

T = 413 K P = 1820 torr = 2.39 atm

- Example : Calculate the number of moles in, and the mass of, an 8.96 L sample of methane, CH4, measured at standard conditions. (mol.wt. CH4 = 16)

mass = 0.400 mol 16 (f.wt. CH4) = 6.40 g

- Example : A compound that contains only carbon & hydrogen is 80.0% C and 20.0% H by mass. At STP 546 mL of the gas has a mass of 0.732 g . What is the molecular (true) formula for the compound?

- Example : A compound that contains only carbon & hydrogen is 80.0% C and 20.0% H by mass. What is the molecular (true) formula for the compound?

- The Dumas method is an antiquated method of determination of molecular weight based upon the accurate weight of and exactly known volume of gas.
- Analysis by measuring the weight of a material is known as gravimetric analysis.
- Mass spectroscopy is currently the preferred method of molecular weight determination. However if you don’t have a spare $1.5 million (or more) laying around for the instrument, the Dumas method works fine.

- Example : A 1.74 g sample of a compound that contains only carbon & hydrogen contains 1.44 g of C and 0.300 g of H. At STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular (true) formula?

- Example : A 1.74 g sample of a compound that contains only carbon & hydrogen contains 1.44 g of C and 0.300 g of H. What is its molecular (true) formula?

- Example : A 250 mL flask contains 0.423 g of vapor at 100 °C and 746 torr. What is molar mass of compound?

- If you have a mixture of gases, the total number of moles of gas present it equal to the sum of the moles of each kind of gas present.

Dalton’s Law of Partial Pressures

The total pressure exerted by a mixture of ideal gases is the sum of the partial pressures of the individual gases. (at constant T and V)

- Example : If 100 mL of hydrogen, measured at 25 °C and 3.00 atm pressure, and 100 mL of oxygen, measured at 25 °C and 2.00 atm pressure, were forced into one of the containers at 25 °C, what would be the pressure of the mixture of gases?

T (25 °C) and V (100 mL) are constant.

Since the pressure of each gas was measured in 100 mL, their pressures could be added. If the volume of either or both gases had changed, the pressure in that volume for a particular gas would need to be calculated before the values could be added together.

- Vapor pressure is defined as the pressure exerted by gas (vapor) over the liquid (and some solids) at equilibrium.
- Important! The gas must be in physical contact with the liquid (or solid) the vapor came from or there is no vapor pressure. Vapor pressure depends ONLY on the temperature. Adding more liquid to the container WILL NOT CHANGE the vapor pressure

- Example : A sample of hydrogen was collected by displacement of water at 25 °C. The atmospheric pressure was 748 torr. What pressure would the dry hydrogen exert in the same container?
- Vapor pressure of water at 25 °C = 24 torr

- Example: A sample of oxygen was collected by displacement of water. The oxygen occupied 742 mL at 27 °C. The barometric pressure was 753 torr. What volume would the dry oxygen occupy at STP?
- Vapor pressure H2O at 27 °C from table = 27 torr

- Postulate 1
- gases are discrete molecules that are far apart
- have few intermolecular attractions
- molecular volume small compared to gas’s volume

- Proof - Gases are easily compressible.

- Postulate 2
- molecules are in constant straight line motion
- varying velocities
- moleculeshave elastic collisions
- Proof - Brownian motion displays molecular motion.
- Proof - A sealed, confined gas exhibits no pressure drop over time.

- Postulate 3
- kinetic energy is proportional to absolute T
- average kinetic energies of molecules of different gases are equal at a given T

- Proof - Brownian motion increases as temperature increases.

- Boyle’s & Dalton’s Law
- P 1/V as V increases collisions decrease so P drops
- Ptotal = PA + PB + PC + .....because few intermolecular attractions
- Charles’ Law
- VT increased T raises molecular velocities, V increases to keep P constant

- diffusion– intermingling of gases
- effusion– gases pass through porous containers
- rates are inversely proportional to square roots of molecular weights or densities

- Example : Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SO2, at the same T and P.
- Mol.wt. He = 4 mol.wt. SO2 = 64.1

Therefore He effuses 4 faster than SO2.

- Example : A sample of hydrogen, H2, was found to effuse through a pinhole 5.2 times as rapidly as the same volume of unknown gas ( at the same T & P). What is the molecular weight of the unknown gas? mol.wt. H2 = 2.0

2

The kinetic energy of a molecule is dependent upon its velocity (u).

- R = 8.314 kg·m2/s2·K·mol
- M must be in kg/mol because of R

The root-mean-squarevelocity of a molecule, urms, is calculated with the following formula

- Exercise : What is velocity of N2 molecules at room T, 25 °C? mol.wt. N2 = 0.028 kg/mol

- Example : What is the velocity of He at room T, 25 °C? (mol.wt. He = 0.004 kg/mol)

- Real gases behave ideally at ordinary T’s and P’s
- the gases have high velocity, and are fairly dilute so there is little interaction

- However, at low T and/or high P– large deviations are noted in their behavior
- the molecules are forced close together so there is interaction (dipole moment)
- the molecules occupy volume and have mass

- These interactions strongly affect their properties

- The van der Waals’s equation is a method of correcting for deviations from ideal gas behavior due to the intermolecular interaction between the molecules and the volume they occupy.

- a and b are van der Waal’s constants
- a is the intermolecular attractive force
- b is the measured molecular volume of the gas
- These values are determined experimentally for each gas. At high T or low P, use the ideal gas law.

- London forces are attractive forces found in nonpolar gases.
- Forcing molecules together, even nonpolar gases, will induce dipole moments in the molecules.

- Dipole-dipole interactions are attractive forces found in polar gases
- These are weak electrostatic attractions between the partially positive side of one polar molecule and the partial negative side of another polar molecule.

- Hydrogen bonding is a relatively strong interaction between a H (attached to N, O, or F) and another N, O, or F
- Hydrogen bonding is much stronger than the other intermolecular forces but only occurs in special cases.

- Example : Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L at 200 °C using the ideal gas law. (mol.wt. NH3 = 17.0 g)

Example : Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L at 200 °C using the van der Waal equation.

This is a 7.6% difference from the 38.4 atm calculated for this same problem using the ideal gas law.

The Chemistry of Explosions

- 2 (122.6g) 2 (74.6g) 3(32g)
- ~ 80 mL (vol)~80 mL67.2 L at STP

2 (79 g) 2 (28 g) 4 (18 g) 32 g

~80 mL 44.8 L 89.6 L 22.4 L

The rapidly expanding gases are responsible for the explosive power. Consider that the average temperature of one of these explosions is >2500 °C (or 10 STP) and you glimpse the destructive power released.

- Example : What volume of oxygen measured at STP, can be produced by the thermal decomposition of 120 g of KClO3?

- Gases are the simplest state of matter.
- Liquids and solids are more complex.