## Pressure and Temperature: Amontons’s Law

Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (see the figure below) and the pressure increases.

This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in the figure below. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then *P* and *T* are directly proportional (again, when *volume and moles of gas are held constant*); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.

Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the *P*–*T* relationship for gases is known as either **Amontons’s law** or **Gay-Lussac’s law**. Under either name, it states that *the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant*. Mathematically, this can be written:

\(P\propto T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}P=\text{constant}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}P=k\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}T\)

where ∝ means “is proportional to,” and *k* is a proportionality constant that depends on the identity, amount, and volume of the gas.

For a confined, constant volume of gas, the ratio \(\cfrac{P}{T}\) is therefore constant (i.e., \(\cfrac{P}{T}\phantom{\rule{0.2em}{0ex}}=k\)). If the gas is initially in “Condition 1” (with *P* = *P*_{1} and *T* = *T*_{1}), and then changes to “Condition 2” (with *P* = *P*_{2} and *T* = *T*_{2}), we have that \(\cfrac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=k\) and \(\cfrac{{P}_{2}}{{T}_{2}}\phantom{\rule{0.2em}{0ex}}=k,\) which reduces to \(\cfrac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{2}}{{T}_{2}}.\)

This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called **absolute zero**). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)

## Example

### Predicting Change in Pressure with Temperature

A can of hair spray is used until it is empty except for the propellant, isobutane gas.

(a) On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?

(b) The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?

### Solution

(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)

(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking *P*_{1} and *T*_{1} as the initial values, *T*_{2} as the temperature where the pressure is unknown and *P*_{2} as the unknown pressure, and converting °C to K, we have:

\(\require{cancel}\cfrac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{2}}{{T}_{2}}\phantom{\rule{0.4em}{0ex}}\text{which means that}\phantom{\rule{0.4em}{0ex}}\cfrac{360\phantom{\rule{0.2em}{0ex}}\text{kPa}}{297\phantom{\rule{0.2em}{0ex}}\text{K}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{2}}{323\phantom{\rule{0.2em}{0ex}}\text{K}}\)

Rearranging and solving gives: \({P}_{2}=\phantom{\rule{0.2em}{0ex}}\cfrac{360\phantom{\rule{0.2em}{0ex}}\text{kPa}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}323\phantom{\rule{0.2em}{0ex}}\cancel{\text{K}}}{297\phantom{\rule{0.2em}{0ex}}\cancel{\text{K}}}\phantom{\rule{0.2em}{0ex}}=390\phantom{\rule{0.2em}{0ex}}\text{kPa}\)