12.4: The Volume-Temperature Law
- Understand the relationship between volume of a gas and temperature.
Consider a gas under constant pressure in a cylinder with a piston in Fig. 7.3.1. Increasing temperature increases the average kinetic energy ( KE ) of the gas molecules. This results in more forceful collisions inside the chamber. The gas volume starts to increase since the pressure in the container does not change. In other words, increasing temperature increases the volume of the gas if the pressure and amount of gas are not changed.
If two related parameters increase or decrease together, they are directly proportional to each other.
Charles’s law states that the volume of a given amount of gas is directly proportional to the temperature in the Kelvin scale at constant pressure.
Fig. 12.4.2 demonstrates that the volume of a gas decreases when the gas is cooled down.
The mathematical forms of Charles’s law are the following.
\[V\propto{T}\nonumber\], or \[V=\mathrm{k}T\nonumber\], or \[\frac{V}{T}=\mathrm{k}\nonumber\]
, where k is a constant, V is volume, and T is the temperature (in kelvin scale) of the gas. Since \(\frac{V}{T}\) is a constant, it implies that
\[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}=\mathrm{k}\nonumber\]
where V 1 is the initial volume, T 1 is the initial temperature in Kelvin, V 2 is the final volume, and T 2 is the final temperature in Kelvin, provided the amount of gas and pressure do not change. Note that the kelvin scale is used in Charles’s law because the kelvin scale does not have negative numbers. If the given temperature is not in the kelvin scale, first convert the temperature to the Kelvin scale and then use the gas laws for the calculations.
A sample of CO 2 occupies 3.23 L volume at 25.0 o C. Calculate the volume of the gas at 50.0 o C if pressure and amount of gas do not change?
Solution
Given: T 1 = 25.0 o C + 273.15 = 298.2 K, T 2 = 50.0 o C + 273.15 = 323.2 K, V 1 = 3.23 L, V 2 = ?
Formula: \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\), rearrange the formula to isolate the desired variable: \(V_{2}=\frac{V_{1} T_{2}}{T_{1}}\)
Plug in the values in the rearranged formula and calculate: \(V_{2}=\frac{3.23 \mathrm{~L} \times 323.2 \mathrm{~K}}{298.2 \mathrm{~K}}=3.50 \mathrm{~L}\)
Applications
Car Tires
As winter approaches, the temperature drops rapidly. The volume of the tire decreases as temperature decreases. For every 10 o F decrease in temperature, the tire pressure decreases by 1 psi. It is important to keep your car tires inflated to the correct pressure to maximize gas mileage.
Hot Air Balloons
Charles’s law explains the drifting of warm air upward in the atmosphere. As the gas is warmed, its volume increases and its density decreases which makes the gas drift upward. A hot air balloon, shown in Fig. 12.4.3 operates using hot air.
Summary
The volume of a gas is directly proportional to the temperature in the Kelvin scale provided the pressure and amount of gas is not changed.