Gases

At STP:
Pressure =1 atm = 760 mmHg
Temperature = 0˚C = 273K (Kelvin used in all calculations)
1 mole of gas = 22.4 L

The Ideal Gas Law expresses the relationship between volume, pressure, amount, and temperature of a gas.

PV=nRT

~R is the gas constant= .0821 L-atm/mol-K

Final and Initial State Problems use Charles’ Law and Boyle’s Law
Charles’ Law:
P1V1=P2V2
Boyle’s Law:
V1/T1=V2/T2

Density can be calculated by manipulating the ideal gas law:

Density= P(MM)/RT

~Volume is proportional to moles of gases at constant temperature and pressure. Therefore, in an equation that only involves gases; the coefficients can represent mole and volume ratios.

Dalton’s Law expresses the total pressure of a gas mixture by adding the partial pressures of the components of the mixture.

Ptot= PA+PB+…

The partial pressure of a gas is directly proportional to the number of moles present in the mixture (the mole fraction)

PA= (XA)(Ptot)

The Kinetic Theory of Gases explains the common features of all gases including behavior and physical properties.
Gases are made up of particles, atoms, or molecules in constant random motion.
No energy is lost in the collisions between gas particles.
The volume occupied by the particles is relatively small.
The attractive forces between particles are negligible.
The average movement or energy of a gas particle is directly proportional to temperature.

Relationships between the properties of gases:
Pressure and temperature are directly proportional when volume is held constant
Volume and pressure are inversely proportional when temperature is held constant.
Temperature and volume are directly proportional when pressure is held constant.
Pressure is directly proportional to the number of particles (because the pressure of a gas is due to the collisions between the gas particles) called Avagadro’s Law.

Kinetic Energy of a gas is directly proportional to absolute temperature. As the temperature increases, so does the average kinetic energy of the gas molecules.

Ke= 3/2 nRT

~all gases at a given temperature will have the same average kinetic energy, regardless of their identity.

Graham’s Law can be used to calculate the average molecular speed of a gas particle at a given temperature. Molecular speeds and rates of effusion are dependent on the gas’s identity. The rate of effusion will be greater for gas molecules that have a smaller molar mass.

Urms= 3RT/MM

Graham’s Law also states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

Rate1/Rate2= MM2/MM1

The Van der Waals Equation takes into consideration that not all gases behave ideally. The ideal gas law is derived from the postulates of the kinetic theory of gases, which states that the volume of a gas and the attractive forces between the particles is negligible. Since gas particles do have volume and small attractive forces between them, deviation occurs. Deviation will occur the most at low temperatures and high pressures.

(P + an2/V2)(V-nb)=RT


Constants a and b are different for each gas
a= a constant that takes into account the attractive forces between molecules,
b=a constant that takes into account the volume of each molecule









Questions:
(taken from the 2008 AP Chemistry Princeton Review)

A mixture of gases at equilibrium over water at 43˚C contains 9.0 moles of nitrogen, 2.0 moles of oxygen, and 1.0 moles of water vapor. If the total pressure exerted by the gases is 780 mmHg, what is the vapor pressure of water at 43˚C?

(A) 65 mmHg
(B) 130 mmHg
(C) 260 mmHg
(D) 580 mmHg
(E) 720 mmHg


2 KClO3(s) à 2 KCl(s) + 3 02(g)
The reaction above took place and 1.45 liters of oxygen gas were collected over water at a temperature of 29˚C and a pressure of 755 mmHg. The vapor pressure of water at 29˚C is 30.0 mmHg.

(a) What is the partial pressure of the oxygen gas collected?

(b) How many moles of oxygen gas were collected?

(c) What would be the dry volume of the oxygen gas at a pressure of 760 mmHg and a temperature of 273K?

(d) What was the mass of the KClO3 consumed in the reaction?





At 25˚C and 1.0 atmospheric pressure, a balloon contains a mixture of four ideal gases: oxygen, nitrogen, carbon dioxide, and helium. The partial pressure due to each gas is .25 atmosphere. Use the ideas of kinetic molecular theory to answer each of the following questions.

(a) Rank the gases in increasing order for each of the following and explain:
(i) Density
(ii) Average Kinetic Energy
(iii) Average molecular velocity
(b) How would the volume of the balloon be affected by each of the following changes to the system?
(i) The temperature of the gases in the balloon were increased at constant pressure
(ii) The gas mixture was replaced with an equal number of molecules of moles of pure oxygen
(c) What changes to the temperature and pressure of the gases would cause deviation from ideal behavior, and which gas would be most affected.








Answers and Explanations:

A
Dalton’s Law states that the partial pressure of one substance in a container is equal to the product of its mole fraction and the total pressure. The sum of all moles of gas is 12 and there is one mole of water vapor. According to Dalton’s equation:
PH20=(XH2O)(Ptot) à PH2O= (1/12)(780 mmHg) = 65 mmHg


(a) Deals with Dalton’s Law:
Ptot= PO2+PH2O
755 mmHg = PO2 + 30.0 mmHg àsolve for PO2
PO2= 725 mmHg

(b) Deals with the ideal gas equation. Remember to convert to correct units.
n= PV/RT= [(725/760 atm)(1.45 L)]/[(.0821)(302 K)]
n= 0.056 moles

(c) At STP, one mole of gas is equal to 22.4 L
Volume = (0.056 moles O2)(22.4 L/mol) = 1.25 L

(d) Since the question includes a balanced equation, and the moles of oxygen have been calculated. The mass of KClO3, can now be determined through dimensional analysis shown below.
Mass=0.056 moles O2 (2 moles KClO3/3 moles O2)(122.6 g KClO3/1 mol KClO3)
Mass KClO3= 4.51 g


(a) (i) Density = P(MM)/RT. In the balloon described in the question, the temperature and pressure are constant. Therefore, the only factor that affects the density of each gas is its molar mass. The gas with the smallest molar mass will have the smallest density and the gas with the largest molar mass will have the greatest density. Increasing order is shown below:
He, N2, O2, CO2
(ii) Ke= 3/2RT. Kinetic energy is dependent only upon temperature, and since all the gases in the balloon are at the same temperature, they all have the same kinetic energy.
(iii) urms= (3RT/MM)1/2. This equation shows that molecular speed has an inverse relationship with molar mass, therefore, the larger the molar mass of the gas the smaller molecular speed it will have. The molecular speed will increase as molar mass decreases as the increasing order shows below:
CO2, O2, N2, He

(b) (i) Volume increases
Volume increases with temperature because temperature is directly proportional to the energy of the particles in the balloon. When the particles have more energy they move more and experience more collisions with the sides of their container, increasing the volume.
(ii) Volume remains constant
Volume is only affected by the number, not the type of molecule, in the container. Therefore, changing the identity of the particles in the balloon will not affect the volume.

(d) Lowering the temperature and raising the pressure will cause deviation from ideal behavior. CO2 will deviate most from ideal behavior.
Lowering the temperature of a gas decreases the movement of particles. The negligible attractive force between gas particles is due to the speed at which they move past each other. With slower movement the molecules will begin to stick together and condense. Raising pressure also pushes the molecules closer together, making it more difficult for the molecules to resist attractive forces between them and also causes condensation. That is why these factors affect ideal behavior.
The ideal gas law assumes that gas particles have negligible volume and attractive forces. Larger molecules will deviate more from this behavior because they have a larger volume. Larger molecules will also have greater attractive forces because their larger electron clouds are easier to manipulate to create momentary dipoles, and therefore have larger gas particles experience greater intermolecular forces.

Links

More Gas Notes
http://antoine.frostburg.edu/chem/senese/101/gases/

Problems
http://www.sparknotes.com/chemistry/gases/ideal/problems_2.html
http://www.sparknotes.com/chemistry/gases/ideal/problems.html
http://www.sparknotes.com/chemistry/gases/ideal/problems_1.html

Spark Notes
http://www.sparknotes.com/chemistry/gases/ideal/