GAS LAWS PROBLEMS

gas laws problems
http://misterguch.brinkster.net/gaslawworksheets.html

INTRODUCTION TO GASES

Properties of gases
http://www.chem1.com/acad/webtext/gas/


Graham's law of effusion

Gases : Graham's Laws of Diffusion and Effusion

Only a few physical properties of gases depends on the identity of the gas.
Diffusion - The rate at which two gases mix.
Effusion - The rate at which a gas escapes through a pinhole into a vacuum. Thomas Graham

Graham's Law of Diffusion

The rate at which gases diffuse is inversely proportional to the square root of their densities.
Since volumes of different gases contain the same number of particles (see Avogadro's Hypothesis), the number of moles per liter at a given T and P is constant. Therefore, the density of a gas is directly proportional to its molar mass (MM).


Combined gas law


The combined gas law is a gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law. These laws each relate one thermodynamic variable to another mathematically while holding everything else constant. Charles's law states that volume and temperature are directly proportional to each other while pressure is held constant. Boyle's law asserts that pressure and volume are inversely proportional to each other at fixed temperature. Finally Gay-Lussac's law introduces a direct proportionality between temperature and pressure at constant volume. The inter-dependence of these variables is shown in the combined gas law, which states that:
“
The ratio between the pressure-volume constant and the temperature of a system remains constant.
”
This can be stated mathematically as
where:
P is the pressure.
V is the volume.
T is the temperature (measured in kelvin).
k is a constant with units of energy divided by temperature.
For comparing the same substance under two different sets of conditions, the law can be written as:
The addition of Avogadro's law to the combined gas law yields the ideal gas law.
Retrieved from http://en.wikipedia.org/wiki/Combined_gas_law

charles' law

Background:
Charles' Law:


Assuming that pressure remains constant, the volume and absolute temperature of a certain quantity of a gas are directly proportional.
Mathematically, this can be represented as:


Temperature = Constant x Volume
or
Volume = Constant x Temperature
or
Volume/Temperature = Constant
Substituting in variables, the formula is:


V/T=K
Because the formula is equal to a constant, it is possible to solve for a change in volume or temperature using a proportion:


V/T = V1/T1
Explanation and Discussion:

Charles' Law describes the direct relationship of temperature and volume of a gas. Assuming that pressure does not change, a doubling in absolute temperature of a gas causes a doubling of the volume of that gas. A drop of absolute temperature sees a proportional drop in volume. The volume of a gas increases by 1/273 of its volume at 0°C for every degree Celsius that the temperature rises.

To explain why this happens, let's explore temperature and volume in terms of gases. Temperature is an average of molecular motion. This means that, while all of the gas molecules are moving around their container in different directions at different speeds, they will have an average amount of energy that is the temperature of the gas. The volume of the gas is the size of its container because the molecules will move in a straight line until they impact something (another molecule or the container). However, to move as they do, the molecules require kinetic energy, which is measured by temperature.

So, the volume and temperature are very closely related. If the temperature was not sufficient, the molecules would not be able to overcome the weak forces of attraction among them and would not be able to fill the container.

Charles' Law must be used with the Kelvin temperature scale. This scale is an absolute temperature scale. At 0 K, there is no kinetic energy (Absolute Zero). According to Charles' Law, there would also be no volume at that temperature. This condition cannot be fulfilled because all known gases will liquify or solidify before reaching 0 K. The Kelvin temperature scale is Celcius minus 273.15 °. Therefore, zero Kelvin would be -273.15 ° and any Celcius temperature can be converted by to Kelvin by adding 273.15 (273 is often used).

Any unit of volume will work with Charles' Law, but the most common are liters (dm3) and milliliters (cm3).




The formula for the law is:


where:

V is the volume of the gas
T is the temperature of the gas (measured in Kelvin)
k is a constant.
In other more thermodynamics-based definitions, the relationship between the fixed mass of a gas at constant pressure is inversely proportional to the temperature applied to the system, which can be further used by stipulating a system where α represents cubic expansivity of a gas, with θ representing the temperature measured of the system in Kelvins:



V = Vo(1 + αθ)
To maintain the constant, k, during heating of a gas at fixed pressure, the volume must increase. Conversely, cooling the gas decreases the volume. The exact value of the constant need not be known to make use of the law in comparison between two volumes of gas at equal pressure:

.
Therefore, as temperature increases, the volume of the gas increases.

GAS LAWS PROBLEMS

Gas Laws with One Term Constant

1. A sample of gas has an initial volume of 25 L and an initial pressure of 3.5 atm. If the pressure changes to 1.3 atm, find the new volume, assuming that the temperature remains constant.

2. A sample of neon is at 89oC and 123 kPa. If the pressure changes to 145 kPa and the volume remains constant, find the new temperature, in oC.


Boyle's Law Problems

1. The volume of the lungs is measured by the volume of air inhaled or exhaled. If the
volume of the lungs is 2.400 L during exhalation and the pressure is 101.70 KPa, and the
pressure during inhalation is 101.01 KPa, what is the volume of the lungs during inhalation?



2. The total volume of a soda can is 415 mL. Of this 415 mL, there is 60.0 mL of headspace
for the CO2 gas put in to carbonate the beverage. If a volume of 100.0 mL of gas at
standard pressure is added to the can, what is the pressure in the can when it has been
sealed?



3. It is hard to begin inflating a balloon. A pressure of 800.0 Kpa is required to initially inflate
the balloon 225.0 mL. What is the final pressure when the balloon has reached it's capacity
of 1.2 L?



4. If a piston compresses the air in the cylinder to 1/8 it's total volume and the volume is
930 cm3 at STP, what is the pressure after the gas is compressed?



5. If a scuba tank that has a capacity of 10.0 dm3 is filled with air to 500.0 KPa, what will be
the volume of the air at 702.6 KPa?

BOYLE'S LAW

The Kinetic Theory of Matter

In the second half of the nineteenth century scientists developed the kinetic theory of matter. This was an attempt to explain the properties of matter- regardless of the state the matter is in- gas, liquid or solid. The word kinetic comes from the Greek “kinetikos” meaning moving. So the kinetic theory is based on the idea that particles of matter are always in motion and that this motion has consequences. Here we discuss gases.

But first, let’s look at the assumptions of the kinetic theory.

Matter consists of a large number of tiny particles. In gases, the particles are about 1000 times farther apart than they are either in liquids or solids. This accounts for their lower density and better compressibility.
The particles are in constant motion moving in straight lines and thus have kinetic energy.
all collisions between the gas particles and the sides of the container are elastic.
There are no forces of attraction or repulsion between the particles of a gas.
The average kinetic energy of a particle is directly proportional to the Kelvin Temperature of the gas.




Objectives
In this lesson, you will:
1. Be able to state Boyle’s Law either verbally or mathematically.
2. Be able to define pressure, explain how it is measured and state the standard conditions of Temperature and Pressure.
3. Be able to use Boyle’s Law to calculate the volume-pressure change at a fixed temperature.
4. Perform and analyze a simple experiment demonstrating Boyle’s Law

Intro to Boyle's Law

Pressure Volume Relationships

We are used to living at 1 ATM of pressure, so we rarely even take notice of it. We normally don't feel the pressure on us because the human body is primarily made up of liquid, and liquids are basically non compressible. At times, however, we do notice changes of pressure, primarily in our ears. You may have noticed your ears "popping" when flying, driving in the mountains, or even going up and down in elevators. This is because our ears have an air space in them, and air, like all other gases, is compressible.

A gas will compress proportionately to the amount of pressure exerted on it. For example, if you have a 1 cubic foot balloon and double the pressure on it, it will be compressed to 1/2 cubic foot. Increase the pressure by 4, and the volume will drop to 1/4 the size etc. This theory was discovered by Sir Robert Boyle, a 17th century scientist. The theory known as Boyle's Law states: If the temperature remains constant, the volume of a given mass of gas is inversely proportional to the absolute pressure.

Let's follow an example...

Suppose you had a balloon measuring one cubic foot at the surface of the water. This balloon is under 1 ATM (14.7 psi) of pressure. If we push the balloon underwater, and take it to a depth of 33 feet, it is now under 2 ATM of pressure (29.4 lbs) - 1 ATM of pressure from the air, 1 ATM of pressure from the water. Boyle's Law then tells us that since we have twice the absolute pressure, the volume of the balloon will be decreased to one half. It follows then, that taking the balloon to 66 feet, the pressure would compress the balloon to one third its original size, 99 feet would make it 1/4 etc.

If we bring the balloon in the previous example back up to the surface, it would increase in size due to the lessening pressure until it reached the surface and returned to its one cubic foot size. This is because the air in the balloon is compressed from the pressure when submerged, but returns to its normal size and pressure when it returns to the surface.

We will achieve the same result with an open container, such as an inverted bottle, as we do with a balloon. By inverting a bottle at the surface and descending with it, the pressure from the surrounding water will compress the air and the bottle will start to fill with water. Even with no air escaping, the container will be half full of water at a depth of 33 feet due to the pressure compressing the air to half its original volume.

Along with the volume of air in the balloon or container, the surrounding pressure will affect the density of the air as well. Density, simply stated, is how close the air molecules are packed together. The air in the balloon or container at the surface is at its standard density, but when we descend to the 33-foot level where its volume is reduced to one half, the density has doubled. At 66 feet, the density has tripled. This is because the pressure has pushed the air molecules closer together.

SOLUTIONS

A solution is a special type of mixture that is homogeneous throughout. This means that the molecules or ions involved are so well mixed that the composition is uniform throughout the mixture. Think of salt-water. You cannot see salt within the water when it is fully dissolved, not even with the aid of a microscope. (This contrasts with a heterogeneous mixture in which you can identify the separate components. For example, a mixture of salt and sand is heterogeneous.)
A solvent is the component in a solution that is present in the largest amount. In a NaCl solution (salt-water), the solvent is water.A solute is the component in a solution in the lesser amount. In a NaCl solution, the salt is the solute. A solution may contain more than one solute.
There are different types of solutions. The one you are probably most familiar with is the aqueous solution.An aqueous solution is a solution in which water is the solvent. A NaCl solution is an aqueous solution.A non-aqueous solution is a solution in which water is not the solvent. Examples of non-aqueous solutions are solutions used in dry cleaning (a solution of ethene in the solvent dichloromethane).A solid solution is a solution in which a solid is the solvent. An example is a brass solution that is formed by dissolving copper in zinc.
So what happens when you drop salt into a glass of water? The water before and after does not look different (assuming that all of the salt is dissolved). However, if you took a drink of it, it certainly tastes different. It boils at a higher temperature than pure water and it conducts electricity. What happened?

Your everyday table salt consists of NaCl. Water is made of H2O molecules. When these two are combined together in a solution, NaCl actually separates into ions. [In solid NaCl, Na+ and Cl- ions are arranged in an ordered three dimensional array called a crystal lattice, as depicted in the figure below.] Thus NaCl (s) becomes Na+ and Cl- ions in solution (i.e., NaCl dissolves in water). Why does NaCl dissolve?
Even though NaCl dissolves to become ions in a solvent, the overall charge remains neutral. Remember that it is NaCl, a neutral compound, that forms the Na+ and Cl- ions. There will be an equal number of positive and negative charges; therefore, the solution will be neutral.

Concentrations of Solutions

There are a number of ways to express the relative amounts of solute and solvent in a solution.
Percent Composition (by mass)

We can consider percent by mass (or weight percent, as it is sometimes called) in two ways:
The parts of solute per 100 parts of solution.
The fraction of a solute in a solution multiplied by 100.
We need two pieces of information to calculate the percent by mass of a solute in a solution:
The mass of the solute in the solution.
The mass of the solution.

Molarity
Molarity tells us the number of moles of solute in exactly one liter of a solution. (Note that molarity is spelled with an "r" and is represented by a capital M.)
We need two pieces of information to calculate the molarity of a solute in a solution:
The moles of solute present in the solution.
The volume of solution (in liters) containing the solute.

Molality

Molality, m, tells us the number of moles of solute dissolved in exactly one kilogram of solvent. (Note that molality is spelled with two "l"'s and represented by a lower case m.)
We need two pieces of information to calculate the molality of a solute in a solution:
The moles of solute present in the solution.
The mass of solvent (in kilograms) in the solution.

Mole Fraction

The mole fraction, X, of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution.
The number of moles of each component present in the solution.

MASS PERCENT CONCENTRATION PROBLEMS

1.) A 4 g sugar cube (Sucrose: C12H22O11) is dissolved in a 350 ml teacup of 80 °C water. What is the percent composition by mass of the sugar solution?
Given: Density of water at 80 °C = 0.975 g/ml.

2.) Concentrated sulfuric acid is 96.0% H2SO4 by mass. (The remaining 4% is water.) The density of this solution is 1.84 g/cm3. Calculate the number of moles of H2SO4 in a liter of concentrated sulfuric acid

3) What is the weight percent of glucose in a solution made by dissolving 4.6 g of glucose in 145.2 g of water?

4) How would you prepare 400. g of a 2.50% solution of sodium chloride?

5) What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol in 85.0 g of water?

6) How would you make 250. g of a 7.5% solution of glucose in water?

7) A sample of a solution weighing 850.0 g is known to contain .223 moles of potassium chloride. What is the weight percentof potassium chloride in the solution.

MOLARITY PROBLEMS

1.) What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough water to make 125 ml of solution?

2.) How would you prepare 400.0 ml of 1.20 M solution of sodium chloride?

3.) How would you prepare 100.0 mL of 0.25 M KNO3 solution?

4.) A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of solution. What is the molarity of the solution?

5.) How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution?

6.) Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in 400.0 mL of this solution?

7.) Calcuate the molarity when 75.0 grams of MgCl2 is dissolved in 500.0 mL of solution.

8.) How many grams of KMnO4 are needed to make 600.0 mL of a 0.200 M solution?

MOLALITY /MOLARITY/MOLE FRACTION PROBLEMS

1.) Calcuate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent.

2.) What is the molality of a solution made by dissolving 5.0g of toluene (C7H8) in 225g of benzene (C6H6)?

3.) The concentrated sulfuric acid we use in the laboratory is 98.0% H2SO4 by mass. Calculate the molality and molarity of the acid solution. The density of the solution is 0.982g/mL.

4.) If you want a solution that is 0.100m in ions, how many grams of Na2SO4 must you dissolve in 125g of water? (assume total dissociation of the ionic solid)

5.) You want to prepare a solution that is 0.200m in Na2CO3. How many grams of the salt must you add to 125g of water? . What is the mole fraction of Na2CO3 in the resulting solution?

6.) A solution is prepared by dissolving 34.2 g of MgCl2 in 0.430 L of H2O. Calculate the molarity, molality and mole fraction of MgCl2 if the density of water is 1.00 g/cm3 and the density of the solution is 1.089 g/cm3.


STUDENTS:

Select 3 problems in each type of concentration problems.
send me ur answers to malen_0106@yahoo.com