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Chemistry · Ch 6 — Gaseous State

Charles Law (Volume-temperature relationship)

6.2.2

Charles Law (Volume-temperature relationship)

J. A. C. Charles studied how the volume of a fixed mass of gas responds to a change in temperature at constant pressure, and found that volume is directly proportional to the Kelvin temperature (at constant P and n):

V=kT(6.5)orVT=constantV=kT\qquad(6.5)\qquad\text{or}\qquad\frac{V}{T}=\text{constant}

For the same system at constant pressure, comparing two states,

V1T1=V2T2=constant(6.6)\frac{V_1}{T_1}=\frac{V_2}{T_2}=\text{constant}\qquad(6.6)

A simple demonstration: a balloon moved from an ice-water bath into boiling water visibly swells (Figure 6.4), because the gas molecules inside move faster at the higher temperature and the balloon must expand to keep the internal pressure matched to the (roughly constant) atmospheric pressure outside.

The V-T graph and absolute zero. Plotting the volume of a fixed mass of gas against its Celsius temperature at constant pressure gives a straight line (called an isobar, Figure 6.5): V=mT+CV=mT+C, where T is in degrees Celsius. At T=0 ∘CT=0\,^\circ\text{C} the volume is some value V0V_0, so C=V0C=V_0, and the slope m=ΔV/ΔTm=\Delta V/\Delta T, giving

V=V0(1+1V0ΔVΔTT)(6.7, 6.8)V=V_0\left(1+\frac{1}{V_0}\frac{\Delta V}{\Delta T}T\right)\qquad(6.7,\,6.8)

Charles and Gay-Lussac found that this relative increase in volume per degree, α=1V0(ΔVΔT)\alpha=\dfrac{1}{V_0}\left(\dfrac{\Delta V}{\Delta T}\right), is essentially the same for every gas, and numerically close to 1/2731/273 -- so at constant pressure, every gas expands by about 1/2731/273 of its 0 ∘^\circC volume for each one-degree rise in temperature:

V=V0(αT+1)(6.9)V=V_0(\alpha T+1)\qquad(6.9)

If the straight line of Figure 6.5 is extrapolated backwards, beyond any real experimental measurement, it crosses the temperature axis (V = 0) at −273 ∘C-273\,^\circ\text{C} (more precisely −273.15 ∘C-273.15\,^\circ\text{C}). A gas cannot physically have negative volume, so this extrapolated point marks a genuine lower limit on temperature -- absolute zero -- and it is exactly this point that Kelvin used as the zero of his own temperature scale, the Kelvin scale. The Kelvin and Celsius scales are identical in size of degree; they differ only in where zero is placed (Table, Kelvin vs Celsius). …

Figure 6.4Air-filled balloon in ice-cold and hot water

What this figure shows. Two identical balloons, side by side: the left one sits in a beaker of ice water and is small and only loosely inflated; the right one sits in a beaker of boiling water and is visibly larger and more taut, illustrating that raising the gas temperature at constant (atmospheric) pressure expands the trapped …

Figure 6.5Plot of volume vs temperature for an ideal gas

What this figure shows. A straight rising line of volume V (y-axis) against temperature T, with the x-axis marked in both Celsius (-400 C to 400 C) and the corresponding Kelvin values (0 K, 73 K, 173 K, 373 K, 473 K, 573 K, 673 K) directly underneath. Extended backwards (dashed), the line crosses V = 0 at -273 C (0 K on the Kelvin scale), showing that the volume-temperature relationship is linear all the way back to absolute zero, where the (physically impossible) volume w …

Table 6.2-kelvin-celsiusKelvin scale vs Celsius scale
Reference pointKelvin ScaleCelsius scale
Absolute Zero0 K-273.15 ∘^\circC
Freezing point of water273.15 K0 ∘^\circC
Misc 6.2.2-worked-charlesWorked example: finding the missing volume/temperature in a three-panel Charles's law figure

Worked out. A fixed mass of gas at a constant pressure of 1 atm is shown in three states (the textbook's figure 6.6): state 1 has V1=0.3 dm3V_1=0.3\ \text{dm}^3, T1=200T_1=200 K; state 2 has V2=?V_2=?, T2=300T_2=300 K; state 3 has V3=0.15 dm3V_3=0.15\ \text{dm}^3, T3=?T_3=?. Applying V1T1=V2T2=V3T3\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}=\dfrac{V_3}{T_3} gives V2=V1T2T1=0.3×300200=0.45 dm3V_2=\dfrac{V_1T_2}{T_1}=\dfrac{0.3\times300}{200}=0.45\ \text{dm}^3 and T3=V3T1V1=0.15×2000.3=100T_3=\dfrac{V_3T_1}{V_1}=\dfrac{0.15\times200}{0.3}=100 K. …