- Explain all the quantities involved in the ideal gas law.
- Evaluate the gas constant R from experimental results.
- Calculate
*T, V, P,*or n of the ideal gas law,*P V = n R T*. - Describe the ideal gas law using graphics.

At a temperature much higher than the critical temperature and at low pressures, however, the ideal gas law is a very good model for gas behavior. When dealing with gases at low temperature and at high pressure, correction has to be made in order to calculate the properties of a gas in industrial and technological applications. One of the common corrections made to the ideal gas law is the van der Waal's equation, but there are also other methods dealing with the deviation of gas from ideality.

P V1 atm 22.4 LR= --- = ------------n T1 mol 273 K = 0.08205 L atm / (mol·K)

When SI units are desirable, *P* = 101325 N/m^{2} (Pa for pascal)
instead of 1 atm. The volume is 0.0224 m^{3}.
The numberical value and units for *R* is

101325 N/m^{2}0.0224 m^{3}R= ---------------------- 1 mol 273 K

= 8.314 J / (mol·K)

For your information, the gas constant can be expressed in the following values and units.

R= 0.08205 L atm / mol·K Notes: = 8.3145 L kPa / mol·K 1 atm = 101.32 kPa = 8.3145 J / mol·K 1 J = 1 L kPa = 1.987 cal / mol·K 1 cal = 4.182 J = 62.364 L torr/ mol·K 1 atm = 760 torr

The volume occupied by one mole, *n* = 1, of substance is called
the **molar volume**, *V*_{molar} = *V / n*.
Using the molar volume notation, the ideal gas law is:

The ideal gas law has four parameters and a constant, *R*,

Furthermore, *n / V* is number of moles per unit volume,
and this quantity has the same units as the concentration (*C*).
Thus, the concentration is a function of pressure and temperature,

The Avogadros law can be further applied to correlate gas density
*d* (weight per unit volume or *n M / V*) and molecular mass
*M* of a gas. The following equation is easily derived from the
ideal gas law:

Thus, we haven M P M = --- R TV

**Example 1
**

*Solution*

From the density *d*, we can evaluate an average molecular weight
(also called molar mass).

M = 22.4 * d

= 22.4 L/mol * 1.3393 g/L

= 30.0 g / mol

- 4

Mass of 0.5 mol oxygen = 0.5 * 32.0 = 16.0 g

Percentage of nitrogen = 100 * 14.0 / 30.0 = 46.7 % Percentage of oxygen = 100 * 16.0 / 30.0 = 100 - 46.7 = 53.3 %

*Discussion*

We can find the density of pure nitrogen and oxygen first and evaluate the
fraction from the density.

1.2500

Solving for

*Exercise*

Now, repeat the calculations for a mixture whose density is 1.400 g/L.

**Example 2
**

*Solution*

The molar mass of acetone = 3*12.0 + 6*1.0 + 16.0 = 58.0. Thus,

= (1.0 * 58.0 atm g/mol) / (0.08205 L atm / (mol K) * 400 K)

= 1.767 g / L

*Exercise*

The density of acetone is 1.767 g/L, calculate its molar mass.