Chemistry » Gases » Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law

Volume and Temperature: Charles’s Law

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.

This figure includes a table and a graph. The table has 3 columns and 6 rows. The first row is a header, which labels the columns “Temperature, degrees C,” “Temperature, K,” and “Pressure, k P a.” The first column contains the values from top to bottom negative 100, negative 50, 0, 100, and 200. The second column contains the values from top to bottom 173, 223, 273, 373, and 473. The third column contains the values 14.10, 18.26, 22.40, 30.65, and 38.88. A graph appears to the right of the table. The horizontal axis is labeled “Temperature ( K ).” with markings and labels provided for multiples of 100 beginning at 0 and ending at 300. The vertical axis is labeled “Volume ( L )” with marking and labels provided for multiples of 10, beginning at 0 and ending at 30. Five data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. The graph shows a positive linear trend.

The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph’s origin, representing a temperature of absolute zero.

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 PT 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, CO2, 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 V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 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 cm3 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 cm3. 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 V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 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.

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