(C) - Copyright, 2000 F.W. Boyle, Jr., Ph.D.

Empirical studies have shown the following relationships.

Please note à is the proportionality symbol but I couldn't get it in HTML.

V à n P à n V à T P à T V à 1/P P à 1/VCombining these relationship we find:

This proportion requires a factor known as the gas constant to complete the equation.

P - pressure in atm

V - volume in L

n - moles

R - gas constant, 0.0821 L atm mole^{-1} K^{-1}

T - temperature, K

This equation is known as the Ideal Gas Law.

The best use of the Ideal Gas Law is in calculating expected results when one of the parameters is changed. These are sometimes referred to as Initial state/final state calculations. For example:

2 Moles of a gas at a pressure of 2.00 atm occupies a volume of 22.4 L. The temperature is 293 K. What will the pressure be if the volume is held constant and the temperature is raised to 348K?

First set up Equation 1:

P_{1}V_{1} = n_{1}RT_{1}

(2.00 atm)(22.4 L) = (2)(0.0821 L atm mole^{-1} K^{-1})(293K)

Then set up Equation 2:

P_{2}V_{2} = n_{2}RT_{2}

(P_{2})(22.4 L) = (2)(0.0821 L atm mole^{-1} K^{-1})(348K)

Divide Equation 1 by Equation 2 (Or Equation 2 by Equation 1, makes no difference):

(2.00 atm)(22.4 L) (2)(0.0821 L atm moleCancel equals in the numerator and denominator just as you would in an algebraic expression.^{-1}K^{-1})(293K) ------------------ = ----------------------------------- (P_{2})(22.4 L) (2)(0.0821 L atm mole^{-1}K^{-1})(348K)

(2.00 atm) (293K) ---------- = -------- (PLastly, solve for P_{2}) (348K)

P_{2} = 2.375 atm

The Ideal Gas Law can be used similarly to calculate n, V, or T when the appropriate other variable is held constant and/or known.

Some forms of the InitialState/Final State Gas Law:

PBy substituting the appropriate numbers, the above equations can be used to solve for any single remaining unknown._{1}T_{1}------ = ------ Constant n & V P_{2}T_{2}V_{1}T_{1}------ = ------ Constant n & P V_{2}T_{2}V_{1}P_{2}------ = ------ Constant n & T V_{2}P_{1}V_{1}n_{1}------ = ------ Constant P & T V_{2}n_{2}P_{1}n_{1}------ = ------ Constant V & T P_{2}n_{2}

Here's an example for calculating the number of moles of a gas knowing P, V, and T:

The temperature MUST BE converted to K from C by adding 273.15 so the new T is 300.15 K.

Now plug in the numbers:

Solve for n:

(2.5 atm)(50 L) n= ----------------------------- = 5.1 moles ClFYI, the particular gas is not relevant in this problem because all gases occupy the same volume per mole at the same pressure and temperature._{2}(0.0821 L atm mole^{-1}K^{-1})(300.15K)

If one looks closely at the Ideal Gas Law one can see the following manipulation.

Divide both sides by V so we get:

nRT P = --- VDivide both sides by RT to get:

P n ---- = --- RT VLook closesly. n/V is moles per volume and volume is measured in L. Thus n/V = moles/L which is molarity. The Ideal Gas Law can be used to determine gas concentrations.

The Ideal Gas Law can be used to find Molar Mass first by understanding that n - the number of moles in the Ideal Gas Law is equal to the mass divided by the molar mass.

So n = m/M

We can substitute m/M for n in the equation.

mRT PV = --- MIf the mass, m, pressure, P, temperature, T, and volume, V, of the gas are known, the molar mass, M, can be calculated.

Density of a gas can be calculated from the Ideal Gas Law. Recall that density is mass/volume. So rearrange the above version of the gas law:

m MP --- = ------- V RTReplace m/V with d, for density.

MP d = ------- RTKnowing molar mass, pressure, and temperature, the density can be calculated.

When more than one gas is contained in a vessel, the total
pressure of the gases is equal to the sum of the individual or
partial pressures. We call them partial pressures because each
gas exerts only a part of the total or a partial pressure.
While there are many ways to determine the partial pressure of a
gas, one of the easiest is through the use of mole fractions. As
an example, in a mixture of 3 moles of H_{2} gas, 5 moles of
O_{2} gas, and 2 moles of N_{2} gas, the mole
fractions are colculated in the following manner.

First total the number of moles: 3 + 5 + 2 = 10 total moles of gases.

In general equation form:

Then divide the moles of each gas by the total to get the mole fraction (m.f.) for each gas.

3 m.f. HWe can now use the mole fractions to determine the partial pressure of each gas. Lets assume the total pressure is 50 atm. Then:_{2}= ----- = 0.3 10 5 m.f. O_{2}= ----- = 0.5 10 2 m.f. N_{2}= ----- = 0.2 10

50 atm * 0.3 = 15 atm partial pressure due to HThe equation for the above is:_{2}50 atm * 0.5 = 25 atm partial pressure due to O_{2}50 atm * 0.2 = 10 atm partial pressure due to N_{2}

Where P_{A} is the partial pressure of gas A (Of course
gas A just means any gas.), X_{A} is the mole fraction of
gas A, and P_{tot} is the total pressure.

Using the following reaction:

Converting moles to mass using 2 g/mole for H_{2}, 32
g/mole for O_{2}, and 18 g/mole for H_{2}O, we
get:

4 g of H_{2} reacts with 32 g of O_{2} to
produce 36 g of H_{2}O.
Now one could ask, How many g of H_{2}O can be produced
with 40 g of H_{2} ? Also, How many g of O_{2}
would be needed?

First convert 40 g of H_{2} back to moles using the molar
mass.

Since hydrogen produces water on a 2 mole:2 mole basis, 20 moles
would produce 20 moles of water which is 360 g (18g/mole * 20
moles). The quickest way to figure the mass of oxygen is to
remember the Law of Conservation of Mass and subtract the mass of
H_{2} from the mass of H_{2}O:

The same procedure is used for every other reaction whether it is for a gas of for solids or dissolved ions.

As the student may recall, one mole of all gases at STP occupies
22.4 L of volume. To grasp the following discussion, recall that
a dozen is a constant of 12. Avogadro's number is a constant
of 6.022 x 10^{23}. All moles of all substance contain
6.022 x 10^{23} parts be the parts atoms or molecules.
Since one mole of all gases occupies 22.4 L then in 22.4 L of
gas at STP there must be 6.022 x 10^{23} parts of the
gas.
Look at the reaction:

2HThe coefficients in front of each substance can apply to molecules or to moles. Since a molecule is merely part of a mole, albeit a very small part, then one can conclude that any part so long as the parts are equal pieces of a mole will react according to this reaction. So I can take some volume, say a L and as long as I maintain the ratio of 2:1, I will maintain the stoichiometry of the reaction._{2}(g) + O_{2}(g) <--> 2H_{2}O(l)

Looking at 1 L of gas we have:

1 L ----- x 6.022 x 10Thus the number of parts for any gas at STP in 1 L would be 2.69 x 10^{23}= 2.69 x 10^{22}parts 22.4L

One can use other volumes just as well because for any volume, so long as the temperatures and pressures are constant, the number of parts is a direct function of the volume.

1) Gases are atoms or molecules in constant random motion.

2) Gas atoms or molecules move in straight lines and have elastic collisions. This means there is no energy change due to the collision.

3) Gases occupy small volumes and are small particles. The distances between particles are large in comparison to the size of the particles themselves.

4) At any given temperature, all gases have the same average translation energy.

5) The average translational energy of a gas particle is directly proporational to the absolute temperature.

6) The attractive force between atoms or molecules are negligible.

Effusion is the movement of a gas through the tiny pores of a
solid. Effusion is the process that allows a "sealed" balloon to
lose gas. Helium balloons lose their pressure faster than a
balloon filled with a large gas such as CO_{2}.

The ratio of the rates of effusion are directly proportional to the square root of the ratio of the densities or the sqaure root of the ratios of the molar masses of the two gases.

The rate of effusion is a measure of how fast gas particles can move through the tiny pores of another substance.

rate of effusion of B --------------------- = (density of A/density of B)To use the above equations, three variables must be known. Once you know the three variables, substitute their values and solve for the last variable.^{1/2}rate of effusion of A rate of effusion of B --------------------- = (molar mass of A/molar mass of B)^{1/2}rate of effusion of A

Please note that should you find any errors in this info please contact Dr. Boyle to get them corrected. My apologies for any errors due to my hurrying to ready this webpage.