# 11.9: The Ideal Gas Law- Pressure, Volume, Temperature, and Moles

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##### Learning Objectives
• Explain the Ideal Gas Law.

There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.

## Ideal Gas Law

The Combined Gas Law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro's Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation:

$\dfrac{P_1 \times V_1}{T_1 \times n_1} = \dfrac{P_2 \times V_2}{T_2 \times n_2} \nonumber$

As with the other gas laws, we can also say that $$\dfrac{\left( P \times V \right)}{\left( T \times n \right)}$$ is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.

The Ideal Gas Law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable $$R$$ for the constant, the equation becomes:

$\dfrac{P \times V}{T \times n} = R \nonumber$

The Ideal Gas Law is conveniently rearranged to look this way, with the multiplication signs omitted:

$PV = nRT \nonumber$

The variable $$R$$ in the equation is called the ideal gas constant.

## Evaluating the Ideal Gas Constant

The value of $$R$$, the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: $$\text{kPa}$$, $$\text{atm}$$, or $$\text{mm} \: \ce{Hg}$$. Therefore, $$R$$ can have three different values.

We will demonstrate how $$R$$ is calculated when the pressure is measured in $$\text{kPa}$$. The volume of $$1.00 \: \text{mol}$$ of any gas at STP (Standard temperature, 273.15 K and pressure, 1 atm) is measured to be $$22.414 \: \text{L}$$. We can substitute $$101.325 \: \text{kPa}$$ for pressure, $$22.414 \: \text{L}$$ for volume, and $$273.15 \: \text{K}$$ for temperature into the ideal gas equation and solve for $$R$$.

\begin{align*} R &= \dfrac{PV}{nT} \\[4pt] &= \dfrac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} \\[4pt] &= 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol} \end{align*} \nonumber

This is the value of $$R$$ that is to be used in the ideal gas equation when the pressure is given in $$\text{kPa}$$. The table below shows a summary of this and the other possible values of $$R$$. It is important to choose the correct value of $$R$$ to use for a given problem.

Unit of $$P$$ Unit of $$V$$ Unit of $$n$$ Unit of $$T$$ Value and Unit of $$R$$
Table $$\PageIndex{1}$$: Values of the Ideal Gas Constant
$$\text{kPa}$$ $$\text{L}$$ $$\text{mol}$$ $$\text{K}$$ $$8.314 \: \text{J/K} \cdot \text{mol}$$
$$\text{atm}$$ $$\text{L}$$ $$\text{mol}$$ $$\text{K}$$ $$0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}$$
$$\text{mm} \: \ce{Hg}$$ $$\text{L}$$ $$\text{mol}$$ $$\text{K}$$ $$62.36 \: \text{L} \cdot \text{mm} \: \ce{Hg}/\text{K} \cdot \text{mol}$$

Notice that the unit for $$R$$ when the pressure is in $$\text{kPa}$$ has been changed to $$\text{J/K} \cdot \text{mol}$$. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule $$\left( \text{J} \right)$$.

##### Example $$\PageIndex{1}$$ Oxygen Gas

What volume is occupied by $$3.76 \: \text{g}$$ of oxygen gas at a pressure of $$88.4 \: \text{kPa}$$ and a temperature of $$19^\text{o} \text{C}$$? Assume the oxygen is ideal.

###### Solution
Solutions to Example 11.5.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."

Given:

• $$P = 88.4 \: \text{kPa}$$
• $$T = 19^\text{o} \text{C} = 292 \: \text{K}$$

Mass $$\ce{O_2} = 3.76 \: \text{g}$$

Find: V = ? L

List other known quantities.

$$\ce{O_2} = 32.00 \: \text{g/mol}$$

$$R = 8.314 \: \text{J/K} \cdot \text{mol}$$

Plan the problem.
1. First, determine the number of moles of O2 from the given mass and the molar mass.
2. Then, rearrange the equation algebraically to solve for V

$V = \dfrac{nRT}{P} \nonumber$

Calculate.

1.

$3.76 \: \cancel{\text{g}} \times \dfrac{1 \: \text{mol} \: \ce{O_2}}{32.00 \: \cancel{\text{g}} \: \ce{O_2}} = 0.1175 \: \text{mol} \: \ce{O_2} \nonumber$

2. Now substitute the known quantities into the equation and solve.

$V = \dfrac{nRT}{P} = \dfrac{0.1175 \: \cancel{\text{mol}} \times 8.314 \: \cancel{\text{J/K}} \cdot \cancel{\text{mol}} \times 292 \: \cancel{\text{K}}}{88.4 \: \cancel{\text{kPa}}} = 3.23 \: \text{L} \: \ce{O_2} \nonumber$

Think about your result. The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume $$\left( 22.4 \: \text{L/mol} \right)$$ since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for $$T$$ and $$P$$. Since a joule $$\left( \text{J} \right) = \text{kPa} \cdot \text{L}$$, the units cancel out correctly, leaving a volume in liters.
##### Example $$\PageIndex{2}$$: Argon Gas

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
Solutions to Example 11.5.2
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."

Given:

n = 4.22 mol

P = 1.21 atm

T = 34°C

Find: V = ? L

List other known quantities. none
Plan the problem.

1. The first step is to convert temperature to Kelvin.

2. Then, rearrange the equation algebraically to solve for V

$V = \dfrac{nRT}{P} \nonumber$

Calculate.

1. 34 + 273 = 307 K

2. Now substitute the known quantities into the equation and solve.

\begin{align*} V&=\dfrac{(4.22\, \cancel{mol})(0.08205\dfrac{L.\cancel{atm}}{\cancel{mol.K}})(307\, \cancel{K)}}{1.21\cancel{atm}} \\[4pt] &= 87.9 \,L \end{align*}

Think about your result. The number of moles of Ar is large so the expected volume should also be large.
##### Exercise $$\PageIndex{1}$$

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

3.94 L

##### Exercise $$\PageIndex{2}$$

For a 0.00554 mol sample of H2, P = 23.44 torr and T = 557 K. What is its volume?