Charles's Law: MDCAT Chemistry notes

Statement

Charles's law: the volume of a fixed amount of gas is directly proportional to its absolute temperature at constant pressure.

$$V\propto T\quad\Rightarrow\quad \frac{V}{T}=k\quad\Rightarrow\quad \frac{V_1}{T_1}=\frac{V_2}{T_2}$$

An older form of the law: at constant pressure, the volume of a given mass of gas increases or decreases by $\frac{1}{273}$ of its volume at 0 °C for every 1 °C rise or fall in temperature:

$$V_t=V_0\left(1+\frac{t}{273}\right)$$

Why it works (kinetic theory)

Raising the temperature increases the average kinetic energy and speed of the molecules. They strike the walls harder and more often. To keep the pressure constant, the gas must expand so that the collisions are spread over a larger area.

Graphs

• V against T (kelvin): a straight line through the origin.
• V against t (°C): a straight line which, when extrapolated, cuts the temperature axis at −273.15 °C. This temperature, where the volume would theoretically become zero, is absolute zero (0 K).
• V/T against T: a horizontal straight line (constant).
• Lines for higher pressure have smaller slopes, but all meet at −273.15 °C.

Real gases liquefy before reaching absolute zero, so zero volume is never actually reached.

Kelvin conversion

$$T(\text{K})=t(^\circ\text{C})+273$$

For example, 70 °C = 343 K, 10 °C = 283 K, 0 °C = 273 K. Always convert to kelvin before using Charles's law; using °C gives wrong answers.

Worked examples

Example 1. Doubling the volume. A gas at 0 °C (273 K) is heated at constant pressure until its volume doubles. $T_2=2\times273=546$ K $=273$ °C. (Not 0 × 2 = 0 °C, and not 200 °C.)

Example 2. Halving the volume. 10 dm³ of gas at 10 °C (283 K) is cooled to 5 dm³. $T_2=\frac{283}{2}=141.5\approx142$ K.

Example 3. 250 cm³ of gas at 27 °C (300 K) is heated to 127 °C (400 K) at constant pressure. $V_2=250\times\frac{400}{300}=333$ cm³.

Gas law	Constant	Relation
Boyle's law	T, n	$PV=k$
Charles's law	P, n	$V/T=k$
Avogadro's law	P, T	$V/n=k$

Common MDCAT traps

• The correct form is $V/T=k$, not $VT=k$ (that would be an inverse relation).
• Proportionality is with absolute temperature; doubling °C does not double volume.
• The V-t line cuts the axis at −273 °C, not 0 °C.
• Pressure must be constant; if it changes, use the general gas equation.

Quick revision

• $V\propto T$ at constant P and n.
• Volume changes by 1/273 of its 0 °C value per degree.
• Zero volume is predicted at −273.15 °C.
• K = °C + 273.
• Doubling volume from 0 °C needs 273 °C.