# 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*}, is true for a particular sample of gas. If any gas is added or allowed to leak out, however, the relationship is lost. In order to get a relationship that is true for any sample of gas, it is necessary to incorporate a term for the amount of gas. From observations as simple as blowing up a balloon, it is clear that increasing the amount of gas increases the volume.

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*} is used to represent the moles of substance. Incorporating \begin{align*}n\end{align*} into the equation yields \begin{align*}PV \propto nT\end{align*}. If we insert a letter, \begin{align*}R\end{align*}, to represent the constant of proportionality, we get the normal form of the **universal gas law**, \begin{align*}PV = nRT\end{align*}.

The unit term for \begin{align*}n\end{align*} is always moles and \begin{align*}T\end{align*} is always in Kelvin. The units for pressure and volume, however, may vary. The value of \begin{align*}R\end{align*} depends on the units that are used for pressure and volume.

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*} where liters × atm have been converted to joules, \begin{align*}R = 8.314 \ \text{J/mol} \cdot \text{K}\end{align*}. The two common values of the **universal gas law constant** **\begin{align*}R\end{align*} **are \begin{align*}0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K}\end{align*} and \begin{align*}8.314 \ \text{J/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.

### Examples

**Example 1**

Determine the volume of 1.00 mol of any gas at STP.

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 2**

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

\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*}, where \begin{align*}P\end{align*} is pressure, \begin{align*}V\end{align*} is volume, \begin{align*}n\end{align*} is number of moles, \begin{align*}R\end{align*} is the universal gas law constant, and \begin{align*}T\end{align*} is the absolute temperature.
- The value of \begin{align*}R\end{align*} varies depending on the units used for \begin{align*}P\end{align*} and \begin{align*}V\end{align*}. Two common values are \begin{align*}0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K}\end{align*} and \begin{align*}R = 8.314 \ \text{J/mol} \cdot \text{K}\end{align*}.
- 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.

### 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.

### Explore More

Use this resource to answer the questions that follow.

- 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?

### Notes/Highlights Having trouble? Report an issue.

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### 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.

## Concept Nodes:

**Universal gas law:**@$\begin{align*}PV = nRT \end{align*}@$**Universal gas law constant:**Denoted by @$\begin{align*}R\end{align*}@$, two common values are @$\begin{align*}0.0821 \ \text{L} \cdot \text{atm/mol} \cdot \text{K}\end{align*}@$ and @$\begin{align*}8.314 \ \text{J/mol} \cdot \text{K}\end{align*}@$.

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