**What chemical reactions require ammonia?**

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:

\begin{align*}\frac{P_1 \times V_1}{T_1 \times n_1}=\frac{P_2 \times V_2}{T_2 \times n_2}\end{align*}

As with the other gas laws, we can also say that \begin{align*}\frac{\left(P \times V \right)}{\left(T \times n \right)}\end{align*}

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 \begin{align*}R\end{align*}

\begin{align*}\frac{P \times V}{T \times n}=R\end{align*}

The ideal gas law is conventionally rearranged to look this way, with the multiplication signs omitted:

\begin{align*}PV=nRT\end{align*}

The variable \begin{align*}R\end{align*}**ideal gas constant**.

#### Evaluating the Ideal Gas Constant

The value of \begin{align*}R\end{align*}

We will demonstrate how \begin{align*}R\end{align*}

\begin{align*}R=\frac{PV}{nT}=\frac{101.325 \ \text{kPa} \times 22.414 \text{ L}}{1.000 \ \text{mol} \times 273.15 \text{ K}}=8.314 \ \text{kPa} \cdot \text{L/K} \cdot \text{mol}\end{align*}

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

Unit of \begin{align*}P\end{align*} |
Unit of \begin{align*}V\end{align*} |
Unit of \begin{align*}n\end{align*} |
Unit of \begin{align*}T\end{align*} |
Value and unit of \begin{align*}R\end{align*} |

kPa | L | mol | K | \begin{align*}8.314 \ \text{J/K} \cdot \text{mol}\end{align*} |

atm | L | mol | K | \begin{align*}0.08206 \ \text{L} \cdot \text{atm/K} \cdot \text{mol}\end{align*} |

mmHg | L | mol | K | \begin{align*}62.36 \text{ L} \cdot \text{mmHg/K} \cdot \text{mol}\end{align*} |

Notice that the unit for \begin{align*}R\end{align*} when the pressure is in kPa has been changed to J/K • mol. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule (J).

#### Sample Problem: Ideal Gas Law

What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19°C? Assume the oxygen is ideal.

*Step 1: List the known quantities and plan the problem.*

Known

*\begin{align*}P = 88.4 \ \text{kPa}\end{align*}*- \begin{align*}T = 19^\circ \text{C} = 292 \text{ K}\end{align*}
- \begin{align*}\text{mass} \ O_2 = 3.760 \text{ g}\end{align*}
- \begin{align*}O_2 = 32.00 \text{ g/mol}\end{align*}
- \begin{align*}R = 8.314 \text{ J/K} \cdot \text{mol}\end{align*}

Unknown

*\begin{align*}V = ? \text{ L}\end{align*}*

In order to use the ideal gas law, the number of moles of O_{2} \begin{align*}(n)\end{align*} must be found from the given mass and the molar mass. Then, use \begin{align*}PV = nRT\end{align*} to solve for the volume of oxygen.

*Step 2: Solve*.

\begin{align*}3.760 \text{ g} \times \frac{1 \ \text{mol} \ O_2}{32.00 \text{ g} \ O_2}=0.1175 \ \text{mol} \ O_2\end{align*}

Rearrange the ideal gas law and solve for \begin{align*}V\end{align*}.

\begin{align*}V=\frac{nRT}{P}=\frac{0.1175 \ \text{mol} \times 8.314 \text{ J/K} \cdot \text{mol} \times 292 \text{ K}}{88.4 \ \text{kPa}}=3.23 \text{ L } O_2\end{align*}

*Step 3: 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 (22.4 L/mol) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for \begin{align*}T\end{align*} and \begin{align*}P\end{align*}. Since a joule (J) = kPa • L, the units cancel correctly, leaving a volume in liters.

### Summary

- The ideal gas constant is calculated.
- An example of calculations using the ideal gas law is shown.

### Review

- Which value of \begin{align*}R\end{align*} will you use if the pressure is given in atm?
- You are doing a calculation where the pressure is given in mm Hg. You select 8.314 J/K • mol as your value for \begin{align*}R\end{align*}. Will you get a correct answer?
- How would you check that you have chosen the correct value of \begin{align*}R\end{align*} for your problem?