## Volume and Temperature: Charles’s Law

If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.

### Optional Video:

This video shows how cooling and heating a gas causes its volume to decrease or increase, respectively.

These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in the figure below.

The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. **Charles’s law** states that *the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant*.

Mathematically, this can be written as:

\(V\phantom{\rule{0.2em}{0ex}}\text{α}\phantom{\rule{0.2em}{0ex}}T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}V=\text{constant}\text{·}T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}V=k\text{·}T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}{V}_{1}\text{/}{T}_{1}={V}_{2}\text{/}{T}_{2}\)

with *k* being a proportionality constant that depends on the amount and pressure of the gas.

For a confined, constant pressure gas sample, \(\cfrac{V}{T}\) is constant (i.e., the ratio = *k*), and as seen with the *P*–*T* relationship, this leads to another form of Charles’s law: \(\cfrac{{V}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{V}_{2}}{{T}_{2}}.\)

## Example

### Predicting Change in Volume with Temperature

A sample of carbon dioxide, CO_{2}, occupies 0.300 L at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr?

### Solution

Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking *V*_{1} and *T*_{1} as the initial values, *T*_{2} as the temperature at which the volume is unknown and *V*_{2} as the unknown volume, and converting °C into K we have:

\(\cfrac{{V}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{V}_{2}}{{T}_{2}}\phantom{\rule{0.4em}{0ex}}\text{which means that}\phantom{\rule{0.4em}{0ex}}\cfrac{0.300\phantom{\rule{0.2em}{0ex}}\text{L}}{283\phantom{\rule{0.2em}{0ex}}\text{K}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{V}_{2}}{303\phantom{\rule{0.2em}{0ex}}\text{K}}\)

Rearranging and solving gives: \({V}_{2}=\phantom{\rule{0.2em}{0ex}}\cfrac{0.300\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{303}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{K}}}{283\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{K}}}\phantom{\rule{0.2em}{0ex}}=0.321\phantom{\rule{0.2em}{0ex}}\text{L}\)

This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).

## Example

### Measuring Temperature with a Volume Change

Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm^{3} when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm^{3}. Find the temperature of boiling ammonia on the kelvin and Celsius scales.

### Solution

A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking *V*_{1} and *T*_{1} as the initial values, *T*_{2} as the temperature at which the volume is unknown and *V*_{2} as the unknown volume, and converting °C into K we have:

\(\cfrac{{V}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{V}_{2}}{{T}_{2}}\phantom{\rule{0.4em}{0ex}}\text{which means that}\phantom{\rule{0.4em}{0ex}}\cfrac{150.0\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}}{273.15\phantom{\rule{0.2em}{0ex}}\text{K}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{131.7\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}}{{T}_{2}}\)

Rearrangement gives \({T}_{2}=\phantom{\rule{0.2em}{0ex}}\cfrac{131.7\phantom{\rule{0.2em}{0ex}}{\require{cancel}\cancel{\text{cm}}}^{3}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}273.15\phantom{\rule{0.2em}{0ex}}\text{K}}{150.0\phantom{\rule{0.2em}{0ex}}{\require{cancel}\cancel{\text{cm}}}^{3}}\phantom{\rule{0.2em}{0ex}}=239.8\phantom{\rule{0.2em}{0ex}}\text{K}\)

Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.