The Ideal Gas Law And Some Applications | Introductory Chemistry

Learning Objectives

1. Learn the ideal gas law.
2. Apply the ideal gas law to any set of conditions of a gas.
3. Apply the ideal gas law to molar volumes, density, and stoichiometry problems.

So far, the gas laws we have considered have all required that the gas change its conditions; then we predict a resulting change in one of its properties. Are there any gas laws that relate the physical properties of a gas at any given time?

Consider a further extension of the combined gas law to include n. By analogy to Avogadro’s law, n is positioned in the denominator of the fraction, opposite the volume. So

Because pressure, volume, temperature, and amount are the only four independent physical properties of a gas, the constant in the above equation is truly a constant; indeed, because we do not need to specify the identity of a gas to apply the gas laws, this constant is the same for all gases. We define this constant with the symbol R, so the previous equation is written as

which is usually rearranged as

PV = nRT

This equation is called the ideal gas law. It relates the four independent properties of a gas at any time. The constant R is called the ideal gas law constant. Its value depends on the units used to express pressure and volume. Table 6.1 “Values of the Ideal Gas Law Constant R“ lists the numerical values of R.

Table 6.1 Values of the Ideal Gas Law Constant R

Numerical Value	Units
0.08205	L·atm/mol·K
62.36	L·torr/mol·K = L·mmHg/mol·K
8.314	J/mol·K

The ideal gas law is used like any other gas law, with attention paid to the units and making sure that temperature is expressed in kelvins. However, the ideal gas law does not require a change in the conditions of a gas sample. The ideal gas law implies that if you know any three of the physical properties of a gas, you can calculate the fourth property.

Example 9

A 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of 34°C. What is its volume?

Solution

The first step is to convert temperature to kelvins:

34 + 273 = 307 K

Now we can substitute the conditions into the ideal gas law:

The atm unit is in the numerator of both sides, so it cancels. On the right side of the equation, the mol and K units appear in the numerator and the denominator, so they cancel as well. The only unit remaining is L, which is the unit of volume we are looking for. We isolate the volume variable by dividing both sides of the equation by 1.21:

Then solving for volume, we get

V = 87.9 L

Test Yourself

A 0.0997 mol sample of O2 has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?

Answer

3.94 L