3/1/10

TEST TOMORROW!!

The Chapter 10 and 11 test is tomorrow! Be sure to study, especially calculations! Also tomorrow, an outline of chapter 11 is due. Today we worked on our outlines during class, so here's all the blog notes from this chapter compiled into one to help with your outline:

Please comment with study tips or online quizzes etc.

ch 11

the pressure formula is p=f/a such that p-pressure, f=force, and a is area

always remember that area is squared

the SI unit for area is N which means newton. it will increase the speed of one kilogram mass by one meter per second that that force is applied.

pressure is a force per unit area, therefore pressure of a 500 N on a floor with an Area of 325 cm^2 is:

500 n / 325 cm^2 = 1.5 N/Cm^2

*the greater the force -> greater pressure

*smaller the area-> greater pressure KNOW THOSE TWO THINGS

introduced by Evangelista Toricelli who was constantly picked on by his parents, the ultimate teaser being his girlish name....

water pumps can raise about 34 feet

thought that it must be dependent on weight and weight of air

reasoned that since mercury was 14 times less dense than water, it would be 1/14 of 34 feet

tested it and it in rose 30 in.

pressures can be also measured in units of atmospheres. Because the average pressure is 760 mm of Hg or one torr named after evangelista.

In pascals pressure is exerted by one N on one square meter

Daltons law

the measure of full pressure in a gas is the sum of the measure of partial pressures

- To determine the pressure of a gas inside a collection bottle, you would use this equation, which is an instance of Dalton's Law of Partial Pressure:

- P(atmosphere) = P(gas) + P(water)

- If you raise the bottle until the water levels inside and outside the bottle are the same, the total pressure outside and inside will be the same.

- Reading the atmospheric pressure on a barometer and looking up the value of P(water) at the temperature of the experiment in a table (p. 859 in our book), you can calculate the P(gas).

Sample Problem B

- Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0 degrees C respectively. What was the partial pressure of the oxygen collected?

G: Total Pressure = P(atmosphere) = 731.0 torr

P(water) = 17.5 torr (vapor pressure of water at 20.0 C from table A)

P(atmosphere) = P(oxygen) + P(water) ; solve for P(oxygen)

P(oxygen) = P(atm) - P(water)

substitute: P(oxygen) = 731.0 torr - 17.5 torr = 713.5 torr

- Boyle's Law

- Robert Boyle discovered that doubling the pressure on a sample of gas at constant temperature reduced its volume by 1/2

- explained by the Kinetic-Molecular Theory (Dr. B said to make sure and know this!)

- the pressure of a gas is caused by moving molecules hitting the container walls

- if the volume of the container is decreased, more collisions will occur and the pressure will increase

- if the volume of the container is increased, less collisions will occur and the pressure will decrease

- Boyle's Law states that the volume of a fixed mass of gas varies inversely with the pressure at constant temperature.

- Formula: PV=K (P=pressure, V=volume, K= constant)

-the inverse would be a straight line (V=K/P)

- Because of the transitive property of equality, since two different quantities are equal to the same thing (volume x pressure = K), it can be concluded that two separate sets of conditions are equal to each other (P1V1 = P2V2)

Charles Law (cont)

Charles Law: Volume-Temperature between volume and temperature was discovered by the French scientist Jacques Charles in 1787.

Charles found that the volume changes by 1/273 of the original volume for each Celsius degree, at a constant pressure and at an initial temperature of 0 degrees C.

The temperature 273 is absolute zero and is given a value of zero in the Kelvin temperature scale. The relationship between the two temperature scales is K=273.15 +degrees C.

Charles Law states that the volume of a fixed mass of gas at a constant pressure varies directly with the Kelvin temperature.

Gas volume and Kelvin temperature are directly proportional to each other at constant pressure.

Mathematically, Charles Law can be expressed as: V=KT or V/T=K where V is the volume, T is the Kelvin temperature, and K is a constant. the ratio V/T for any set of volume temperature values always equals the same K.

The equation reflects the fact that volume and temperature are directly proportional to each other at constant pressure.

The form of Charles Law that can be applied directly to most volume- temperature gas problems is: V1/T1 = V2/T2.

V1 and T1 represent initial conditions, and V2 and T2 represent another set of conditions.

Given three of the four values, V1, T1, V2, and T2, you can use this equation to calculate the 4th value for a system at constant pressure.

Gay-Lussacs Law: Pressure Temperature Relationship

At a constant volume, the pressure of a gas increases with increasing temperature.

Gas pressure is the result of collisions of molecules with container walls.

The energyu and frequency of collisions depend on the average kinetic energy of the molecues.

Pressure is directly proportional to Kelvin temperature.

Gay Lussacs Law: The pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature.

This law is named after Joseph Gay-Lussac, who discovered it in 1802.

Mathematically, Gay Lussacs Law can be expressed as P=KT or P/T=K where P is pressure, T is the Kelvin temperaure, and K is a constant. The ratio P/T for any set of volume-temperature values always equals the same K.

• Boyle's Law, Charles' Law, and Gay-Lussac's Law can be combined into a single equation that can be used for situations in which temperature, pressure, and volume all vary at the same time.
• This is the combined gas law, PV/T=k, or P1V1/T1=P2V2/T2.
• Each gas law can be derived from the combined gas law when the proper variable is kept constant.
• Sample Problem F can be found in your book.
• In the early 1800s, French chemist Joseph Gay-Lussac observed that 2L hydrogen can react with 1 L oxygen to form 2L water vapor.
• This reaction shows a simple 2:1:2 ratio in the volumes of reactants and products. This same ratio applies to any volume proportions.
• Gay-Lussas's law of combining volumes of gases (that's a mouthful) states that at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers.
• 1811: Avogadro explained Gay-Lussac's law of combining volumes of gases without violating Dalton's idea of indivisible atoms.
• Avogadro reasoned that, instead of always being in monoatomic form, when they combine to form products, gas molecules can contain more than one atom.
• Avogadro's law: equal volumes of different gases contain the same number of molecules, at given pressure and temperature. Also, gas volume is directly proportional to the amount of gas at a given temperature or pressure. V=kn.
• Dalton had guessed that the formula for water was HO, but Avogadro's reasoning established that water must contain twice as many hydrogen atoms as oxygen atoms because of the volume ratios in which the gases combine.
• Ergo, Avogadro's idea of diatomic gases was consistent with all other knowledge and laws.

• You can use the volume ratios as conversion factors in gas stoichiometry problems as you would mole ratios.
• Ideal Gas Law
• You have learned about equations describing the relationships between 2 or 3 of the 4 variables - Pressure, Volume, Temperature and number of moles - needed to describe a sample at a time.
• All of the laws you have learned thus far can be combined into a single equation, the IDEAL GAS LAW: the mathematical relationship among pressure, volume, temperature, and number of moles of a gas.
• R is a constant
• PV=nRT
• In the equation representing the Ideal gas law, R = idea gas constant
• Its value depends on the units chosen for pressure, volume, and temperature in the rest of the equation.
• Measured values of P, V, Temp., and n for a gas at near-ideal conditions can be used to calculate R
• R = 0.082058

At the begining of the lab Dr. B will give us the barometric pressure, but not the units we need, we will have to convert it as part of the lab.
During the lab there will be no chewing on any substances. HCl will be used during the lab and it will burn your skin.
......................................................Missed some info...................................................
Add HCl the the tube then hold the tube with HCl at an angle when squeezing the water slowly into the tube, try not to let the water and HCl mix because it will cause the reaction to go slower. When putting the capper and magnesium in the tube put it in close to the top and not far in the tube. Then you put your finger over the top of the tube and flip it over and into the beaker with water. It should start to bubble and gas will start to form at the what used to be the bottom of the tube, now the top.
NOTES - CH11 sec 3 cont.
The Ideal Gas Law cont.
The Ideal Gas Constant cont.

• The calculated value of R is usually rounded to 0.0821(L x atm)(md x k)
• Dr. B wants Rto equal 0.08206
• use this value in ideal gas law calculations when the volume is in liters, the preasure is in atmospheres, and the temp is in kelvins
• The ideal gas law canbe applied to determine the existing conditions of a gas sample when three of the four values; P,T,V, and n; are known
• be sure to match the units of the known quantities and the units of R
• Sample problem 1: what is the pressure in atmospheres exerted by a 0.500mol smple of nitrogen gas in a 10.0 L container at 298K?........P=nRT/V..........P=(0.500mol)(0.08206L x atm)(298K)/10.0L 122atm

Graham's Law of Efusion

• Rates of effusion and diffusion depend on the relative velocities of gas molecules. the velocity of a gas varies inversely with the square root of its molar mass.
• recall that the average kinetic energy of the molecules in any gas depends on the temperature and equalys(1/2)mv^2
• for 2 different gases, A and B, at the same temperature, the following relationship is true 1/2 MaVa^2= 1/2 MbVb^2
• from the equation relationg the kinetic energy of two different gases at the same conditions, one can derive an equation relating the rates of effusion of two gases with their molecular mass. Rate of effusion of A/rate of effusion of B= square root of Mb/ square root of Ma
• this equation is known as Graham's law of effusion which states that the rates of effusion of gasses at the same temperature and pressure are inversely proportional to the square roots of their molar mass