# 11.5: Universal Gas Law

**At Grade**Created by: CK-12

Compressed gases provide vital fuels for industry and for homes and farms in rural areas.

### Universal Gas Law

The combined gas law, \begin{align*}PV \propto T\end{align*}

Because different gases have different weights per molecule, including a term for mass of gas does not produce a consistent equation. If, however, we include a term expressing the number of moles of gas rather than its mass, we can produce a constant proportionality. A mole is a unit representing the number of atoms present. The letter \begin{align*}n\end{align*}**universal gas law**, \begin{align*}PV = nRT\end{align*}

The unit term for \begin{align*}n\end{align*}

Pressure Units | Volume Units |
Units for \begin{align*}\underline{n}\end{align*} |
Units for \begin{align*}\underline{T}\end{align*} |
Value of \begin{align*}\underline{R}\end{align*} |

atm | liters | moles | Kelvin | 0.0821 L•atm/mol•K |

atm | milliliters | moles | Kelvin | 82.1 mL•atm/mol•K |

Since the product of (liters)(atm) can be converted to joules, we also have a value for \begin{align*}R\end{align*}**universal gas law \begin{align*}R\end{align*} Rconstant** are \begin{align*}0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K}\end{align*}

Most universal gas law problems are calculated at STP. STP stands for *standard temperature and pressure,* which is the most commonly calculated temperature and pressure value. STP is defined as 1.00 atm and 0°C, or 273 K.

**Example Problem:** Determine the volume of 1.00 mol of any gas at STP.

**Solution:** First isolate V from PV=nRT. Then plug in known values and solve.

\begin{align*}V=\frac{nRT}{P}=\frac{(1.00 \ \text{mol})(0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K})(273 \ \text{K})}{(1.00 \ \text{atm})}=22.4 \ \text{liters}\end{align*}

For any gas at STP, one mole has a volume of 22.4 liters. This can be an extremely convenient conversion factor.

**Example Problem:** A sample of oxygen gas occupies 10.0 liters at STP. How many moles of oxygen are in the container?

**Solution:**

\begin{align*}n=\frac{PV}{RT}=\frac{(1.00 \ \text{atm})(10.0 \ \text{L})}{(0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K})(273 \ \text{K})}=0.446 \ \text{moles}\end{align*}

#### Summary

- The universal gas law is \begin{align*}PV = nRT\end{align*}
PV=nRT , where \begin{align*}P\end{align*}P is pressure, \begin{align*}V\end{align*}V is volume, \begin{align*}n\end{align*}n is number of moles, \begin{align*}R\end{align*}R is the universal gas law constant, and \begin{align*}T\end{align*}T is the absolute temperature. - The value of \begin{align*}R\end{align*}
R varies depending on the units used for \begin{align*}P\end{align*}P and \begin{align*}V\end{align*}V . Two common values are \begin{align*}0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K}\end{align*}0.0821 L⋅atm/mol⋅K and \begin{align*}R = 8.314 \ \text{J/mol} \cdot \text{K}\end{align*}R=8.314 J/mol⋅K . - STP is standard temperature and pressure; 273 K and 1.00 atm.
- One mole of a gas at STP has a volume of 22.4 liters.

#### Practice

The following video discusses the constant \begin{align*}R\end{align*}

- Why is it important to have values for R in kPa, atm, and mmHg?
- Why do the units of R include pressure, temperature, volume, and moles?

Instruction and practice problems involving the universal gas law:

http://www.sparknotes.com/testprep/books/sat2/chemistry/chapter5section10.rhtml

#### Review

- The initial pressure in a helium gas cylinder is 30 atm. After many balloons have been blown up, the pressure in the cylinder has decreased to 6 atm while the volume and temperature remain the same. What fraction of the original amount of gas remains in the cylinder?
- Calculate the volume of 8.88 mol of helium gas at 20.0°C and 1.19 atm pressure.

### Image Attributions

- State the universal gas law.
- State the universal gas law constant, R.
- Given three of the four unknowns in the universal gas law, solve for the fourth.