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.

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 [link].

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: $$V\;α\;T\quad or\quad V=constant·T\quad or\quad V=k·T\quad or\quad \frac{V_1}{T_1}=\frac{V_2}{T_2}$$

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

For a confined, constant pressure gas sample, $\frac{V}{T}$ is constant (i.e., the ratio = *k*), and as seen with the *P*–*T* relationship, this leads to another form of Charles’s law: $\frac{V_1}{T_1}=\frac{V_2}{T_2}$.

**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: $$\frac{V_1}{T_1}=\frac{V_2}{T_2}\quad which\;means\; that\quad \frac{0.300\;L}{283\;K}=\frac{V_2}{303\;K}$$

Rearranging and solving gives: $$V_2=\frac{0.300\;L\times 303\;\require{enclose}\enclose{horizontalstrike}{K}}{283\; \enclose{horizontalstrike}{K}}=0.321\;L$$

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).

**Check Your Learning**

A sample of oxygen, O_{2}, occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure?

**Answer:**

21.6 mL

**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:

$$\frac{V_1}{T_1}=\frac{V_2}{T_2}\quad which\;means\;that\quad \frac{150.0\;cm^3}{273.15\;K}=131.7\;cm^3$$

Rearrangement gives $$T2=\frac{131.7\;\enclose{horizontalstrike}{cm^3}\times 273.15\;K}{150.0\;\enclose{horizontalstrike}{cm^3}}=239.8\;K$$

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

**Check Your Learning**

What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?

**Answer:**

635 mL