The Ideal Gas Law Info Sheet
(C) - Copyright, 2000 F.W. Boyle, Jr., Ph.D.

Empirical studies have shown the following relationships.

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

```    V ŕ n      P ŕ n
V ŕ T      P ŕ T
V ŕ 1/P    P ŕ 1/V
```
Combining these relationship we find:

PV ŕ n T

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

PV = nRT

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.

Uses of 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:

P1V1 = n1RT1

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

Then set up Equation 2:

P2V2 = n2RT2

(P2)(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 mole-1 K-1)(293K)
------------------ = -----------------------------------
(P2)(22.4 L)      (2)(0.0821 L atm mole-1 K-1)(348K)

```
Cancel equals in the numerator and denominator just as you would in an algebraic expression.

```    (2.00 atm)      (293K)
----------  =  --------
(P2)         (348K)
```
Lastly, solve for P2:

P2 = 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:

```     P1        T1
------ = ------      Constant n & V
P2        T2

V1        T1
------ = ------      Constant n & P
V2        T2

V1        P2
------ = ------      Constant n & T
V2        P1

V1        n1
------ = ------      Constant P & T
V2        n2

P1        n1
------ = ------      Constant V & T
P2        n2
```
By substituting the appropriate numbers, the above equations can be used to solve for any single remaining unknown.

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

PV = nRT
How many moles of Cl2 are in a container (usually called a vessel) if the pressure is 2.5 atm, the temperature is 27 C, and the volume is 50 L? R is, of course, 0.0821 L atm mole-1 K- 1.

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:

(2.5 atm)(50 L) = n(0.0821 L atm mole-1 K-1)(300.15K)

Solve for n:

```                     (2.5 atm)(50 L)
n=  -----------------------------  =  5.1 moles Cl2
(0.0821 L atm mole-1 K-1)(300.15K)
```
FYI, the particular gas is not relevant in this problem because all gases occupy the same volume per mole at the same pressure and temperature.

Concentrations

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

PV = nRT

Divide both sides by V so we get:

```             nRT
P = ---
V
```
Divide both sides by RT to get:

```        P      n
---- = ---
RT     V
```
Look 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 = ---
M
```
If the mass, m, pressure, P, temperature, T, and volume, V, of the gas are known, the molar mass, M, can be calculated.

Density Calculations

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      RT
```
Replace m/V with d, for density.

```                MP
d = -------
RT
```
Knowing molar mass, pressure, and temperature, the density can be calculated. Note the gas with the greatest molar mass will have the highest density at any particular pressure and temperature.

Partial Pressures

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 H2 gas, 5 moles of O2 gas, and 2 moles of N2 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:

Ptot = PA + PB + PC + ...

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

```            3
m.f. H2 = -----  = 0.3
10

5
m.f. O2 = -----  = 0.5
10

2
m.f. N2 = -----  = 0.2
10
```
We can now use the mole fractions to determine the partial pressure of each gas. Lets assume the total pressure is 50 atm. Then:

```
50 atm * 0.3 = 15 atm partial pressure due to H2

50 atm * 0.5 = 25 atm partial pressure due to O2

50 atm * 0.2 = 10 atm partial pressure due to N2
```
The equation for the above is:

PA = XAPtot

Where PA is the partial pressure of gas A (Of course gas A just means any gas.), XA is the mole fraction of gas A, and Ptot is the total pressure.

Some Reaction Stoichiometry with masses

Using the following reaction:

2H2(g) + O2(g) --> 2H2(l)
The reaction tells us that 2 moles of H2 react with 1 mole of O2 to procude 2 moles of H2O.

Converting moles to mass using 2 g/mole for H2, 32 g/mole for O2, and 18 g/mole for H2O, we get:

4 g of H2 reacts with 32 g of O2 to produce 36 g of H2O. Now one could ask, How many g of H2O can be produced with 40 g of H2 ? Also, How many g of O2 would be needed?

First convert 40 g of H2 back to moles using the molar mass.

40 g / 2 g/mole = 20 moles H2

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 H2 from the mass of H2O:

360 g - 40 g = 320 g O2

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

Combining volumes

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 1023. All moles of all substance contain 6.022 x 1023 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 1023 parts of the gas. Look at the reaction:

```
2H2(g) + O2(g) <--> 2H2O(l)
```
The 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.

Looking at 1 L of gas we have:

```                   1 L
----- x 6.022 x 1023 = 2.69 x 1022 parts
22.4L

```
Thus the number of parts for any gas at STP in 1 L would be 2.69 x 1022. Parts react with parts so to make the above H2/O2 reaction proceed stoichiometrically, we can get the proper number of parts by using 2 L of H2(g) and 1 L of O2(g) because 2 L of H contain twice the number of molecules as 1 L of O.

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.

Kinetic Theory of Gases

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.

Graham's Law of Effusion

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 CO2.

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)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

```
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.

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.