DAT General Chemistry: Charles's Law and Gay-Lussac's Law

DAT High-Yield Concept

Charles's Law (Isobaric): At constant pressure, gas volume is directly proportional to absolute temperature ( $V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ ).
Gay-Lussac's Law (Isochoric): At constant volume, gas pressure is directly proportional to absolute temperature ( $P \propto T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ ).
Absolute Kelvin Scale: Every gas calculation MUST convert Celsius to Kelvin ( $\pu{T_{Kelvin}} = \pu{T_{Celsius}} + 273.15$ ). Doubling Celsius does NOT double volume or pressure.
Absolute Zero: Extrapolating linear volume-versus-temperature or pressure-versus-temperature graphs yields $\pu{0 K}$ ( $\pu{-273.15 degC}$ ), where molecular kinetic energy is minimal.

Gas Laws in Context

Gas behavior depends on four fundamental variables: pressure ( $P$ ), volume ( $V$ ), temperature ( $T$ ), and mole amount ( $n$ ). Simple empirical gas laws fix two variables to observe the direct relationship between the remaining two. (Note: Boyle's law covers the inverse relationship between pressure and volume at constant temperature and is discussed in depth in its own lesson).

Charles's Law: Constant Pressure (Isobaric)

Charles's law states that at constant pressure and mole amount, the volume of a gas is directly proportional to its absolute temperature:

\frac{V_1}{T_1} = \frac{V_2}{T_2}

When a gas is heated, its particles gain kinetic energy and strike the container walls with greater speed and frequency. To maintain constant internal pressure against the surroundings, the container expands, increasing volume.

The relationship between volume and temperature for two different gas samples, according to Charles's law as determined by Joseph Louis Gay-Lussac (1802). The volumes are measured at 0 ºC and 100 ºC; the proportional increase in volume is the same for all gases, and all lines cut the x -axis at the same point. (Source: Physchim62 · CC BY-SA 3.0 · Wikimedia Commons)

Worked Calculation Example — Charles's Law:
A gas balloon occupies

\pu{2.40 L}

\pu{25.0 degC}

. What is its volume if heated to

\pu{75.0 degC}

at constant pressure?
Step 1: Convert temperatures to Kelvin.

T_1 = 25.0 + 273.15 = \pu{298.15 K}

T_2 = 75.0 + 273.15 = \pu{348.15 K}

Step 2: Apply dimensional analysis to solve for $V_2$ .

V_2 = V_1 \times \left(\frac{T_2}{T_1}\right) = \pu{2.40 L} \times \left(\frac{\pu{348.15 K}}{\pu{298.15 K}}\right) = \pu{2.8026 L} \approx \pu{2.80 L}

Common Exam Trap: Students often double the volume when temperature increases from

\pu{25 degC}

\pu{50 degC}

. However,

\pu{298 K}

\pu{323 K}

is only an

8.4\%

increase in absolute temperature, so volume increases by only

8.4\%

Gay-Lussac's Law: Constant Volume (Isochoric)

Gay-Lussac's law states that at constant volume (rigid container) and mole amount, the pressure exerted by a gas is directly proportional to its absolute temperature:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

As thermal energy increases, gas particles move faster. Because the container walls are rigid and cannot expand, higher collision momentum translates directly into elevated internal pressure.

Molecular collisions within a closed container (the propane tank) are shown (right). The arrows represent the random motions and collisions of these molecules. The pressure and temperature of the gas are directly proportional: as the temperature is increased, the pressure of the propane increases by the same factor. A simple consequence of this proportionality is that on a hot summer day, the propane tank pressure will be elevated, and thus propane tanks must be rated to withstand such increases in pressure. (Source: BlyumJ · CC BY-SA 4.0 · Wikimedia Commons)

Worked Calculation Example — Gay-Lussac's Law:
A rigid metal tank contains gas at

\pu{1.50 atm}

and

\pu{20.0 degC}

. If heated to

\pu{120.0 degC}

, what is the final pressure?
Step 1: Convert temperatures to Kelvin.

T_1 = 20.0 + 273.15 = \pu{293.15 K}

T_2 = 120.0 + 273.15 = \pu{393.15 K}

Step 2: Solve for $P_2$ .

P_2 = P_1 \times \left(\frac{T_2}{T_1}\right) = \pu{1.50 atm} \times \left(\frac{\pu{393.15 K}}{\pu{293.15 K}}\right) = \pu{2.0117 atm} \approx \pu{2.01 atm}

Common Exam Trap: Rigid containers imply an isochoric path (constant volume). Recognizing container geometry avoids choosing an incorrect gas formula.

The Kelvin Requirement and Absolute Zero

According to kinetic molecular theory, average kinetic energy is directly proportional to absolute temperature ( $E_k = \frac{3}{2}RT$ ). The Celsius scale sets an arbitrary zero at the freezing point of water, yielding negative values that produce mathematically unphysical negative volumes or pressures. Only the Kelvin scale sets its origin at absolute zero ( $\pu{0 K}$ or $\pu{-273.15 degC}$ ), where molecular kinetic energy reaches its theoretical minimum.

Volume against temperature: why the scale must be kelvin — For a fixed amount of gas held at constant pressure, the volume measured at several temperatures falls on a straight line, and lines taken at different pressures extrapolate back to the same intercept near minus 273.15 degrees Celsius. Shifting the scale so that point becomes zero gives the kelvin scale, on which volume is directly proportional to temperature. That is why gas-law calculations require kelvin: on the Celsius scale the line misses the origin, so ratios of Celsius temperatures are meaningless.

DAT Memory Aid:

References

Sources returned by Google Search grounding during lesson verification.

14.4: Charles's Law - Chemistry LibreTexts — chem.libretexts.org
Chad's Prep - Gas Laws Including the Ideal Gas Law — chadsprep.com

Charles's Law and Gay-Lussac's Law

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