Charles’s law describes the relationship between the volume and temperature of a gas. The law was ﬁrst published by Joseph Louis Gay-Lussac in 1802, but he referenced unpublished work by Jacques Charles from around 1787.

This law states that at constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature (in kelvin) increases or decreases. Another way of saying this is that temperature and volume are directly proportional

**Deﬁnition**: Charles’s Law The volume of an enclosed sample of gas is directly proportional to its absolute temperature provided the pressure is kept constant.

**Fact:****Charles’s Law is also known as Gay-Lussac’s Law. This is because Charles did not publish his discovery, and it was rediscovered independently by another French Chemist Joseph Louis Gay-Lussac some years later**

Mathematically, the relationship between temperature and pressure can be represented as follows:

V ∝ T or V = kT

If the equation is rearranged, then…

V T = k

and, following the same logic that was used for Boyle’s law:

V1/T1 = V2/ T2

The equation relating volume and temperature produces a straight line graph (refer back to the notes on proportionality if this is unclear).

This relationship is shown in the figure below.

The volume of a gas is directly proportional to its temperature, provided the pressure of the gas is constant.

However, if this graph is plotted on a celsius temperature scale, the zero point of temperature doesn’t correspond to the zero point of volume. When the volume is zero, the temperature is actually -273.150C

A new temperature scale, the Kelvin scale must be used instead. Since zero on the Celsius scale corresponds with a Kelvin temperature of -273.150C, it can be said that:

Kelvin temperature (T) = Celsius temperature (t) + 273.15 133

The relationship between volume and temperature, shown on a Celsius temperature scale.

At school level, you can simplify this slightly and convert between the two temperature scales as follows:

T = t + 273 or t = T – 273

Can you explain Charles’s law in terms of the kinetic theory of gases?

When the temperature of a gas increases, so does the average speed of its molecules. The molecules collide with the walls of the container more often and with greater impact. These collisions will push back the walls, so that the gas occupies a greater volume than it did at the start. We saw this in the ﬁrst demonstration. Because the glass bottle couldn’t expand, the gas pushed out the balloon instead.

**Calculations**

**Worked Example 1:**

Question: Ammonium chloride and calcium hydroxide are allowed to react. The ammonia that is released in the reaction is collected in a gas syringe and sealed in. This gas is allowed to come to room temperature which is 32◦C. The volume of the ammonia is found to be 122 ml. It is now placed in a water bath set at 7◦C. What will be the volume reading after the syringe has been left in the bath for a good while (assume the plunger moves completely freely)?

**Answer Step 1** : Write down all the information that you know about the gas. V1 = 122 ml and V2 = ? T1 = 320C and T2 = 70C

**Step 2** : Convert the known values to SI units if necessary. Here, temperature must be converted into Kelvin, therefore: T1 = 32 + 273 = 305 K T2 = 7 + 273 = 280 K

**Step 3**: Choose a relevant gas law equation that will allow you to calculate the unknown variable.

V1/ T1 = V2/ T2

Therefore,

V2 = (V1 ×T2) /T1

**Step 4** : Substitute the known values into the equation. Calculate the unknown variable.

V2 =122×280/ 305

V2 = 112ml

Note: The temperature must be converted to Kelvin (SI) since the change from degrees Celcius involves addition, not multiplication by a ﬁxed conversion ratio (as is the case with pressure and volume.)

**Worked Example 2:**

Question: At a temperature of 298 K, a certain amount of CO2 gas occupies a volume of 6 l. What volume will the gas occupy if its temperature is reduced to 273 K?

Answer

**Step 1**: Write down all the information that you know about the gas. V1 = 6 l and V2 = ? T1 = 298 K and T2 = 273 K

**Step 2** : Convert the known values to SI units if necessary. Temperature data is already in Kelvin, and so no conversions are necessary.

**Step 3**** :** Choose a relevant gas law equation that will allow you to calculate the unknown variable.

V1 /T1 = V2 /T2

Therefore,

V2 = (V1 ×T2) / T1

**Step 4** : Substitute the known values into the equation. Calculate the unknown variable.

V2 = (6×273) /298

V2 = 5.5l

**Exercise**

- 1. A gas of unknown volume has a temperature of 14◦C. When the temperature of the gas is increased to 100◦C, the volume is found to be 5.5 L. What was the initial volume of the gas?
- A gas has an initial volume of 2600 mL and a temperature of 350 K. (a) If the volume is reduced to 1500 mL, what will the temperature of the gas be in Kelvin? (b) Has the temperature increased or decreased? (c) Explain this change, using the kinetic theory of matter.

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