Charles’s Law (constant pressure)
When the amount of gas and the pressure remain constant, the volume of a gas is
directly proportional to its absolute temperature (Kelvin). In symbols:
\[ V \propto T \]
\[ V = b\,T \]
Origin from the ideal-gas law
Starting from \(PV=nRT\) and keeping \(n\) and \(P\) constant,
\[
V=\frac{nR}{P}\,T
\]
\[
b=\frac{nR}{P}
\]
Here \(b\) has units of volume per kelvin (e.g., m³·K⁻¹ or L·K⁻¹).
Two-state (ratio) form
For an initial state \( (V_i,T_i) \) and a final state \( (V_f,T_f) \) at the same pressure:
\[ \frac{V_i}{T_i}=\frac{V_f}{T_f} \]
\[ V_f = V_i\,\frac{T_f}{T_i} \]
\[ T_f = T_i\,\frac{V_f}{V_i} \]
All temperatures in these equations must be in Kelvin and strictly \( > 0 \).
Temperature scales
\[ T_{\mathrm{K}} = t_{^{\circ}\mathrm{C}} + 273.15 \]
\[ T_{\mathrm{K}} = \left(t_{^{\circ}\mathrm{F}} - 32\right)\frac{5}{9} + 273.15 \]
- Use Kelvin in formulas; you may enter °C or °F and convert to K first.
- On a \(V\) vs \(T\) (°C) plot, the straight line extrapolates to \(T=-273.15\,^{\circ}\mathrm{C}\) at \(V=0\).
Common volume units
- \(1~\mathrm{m^3} = 1000~\mathrm{L}\)
- \(1~\mathrm{L} = 10^{-3}~\mathrm{m^3}\)
- \(1~\mathrm{mL} = 1~\mathrm{cm^3} = 10^{-6}~\mathrm{m^3}\)
Interpreting the slope \(b\)
The constant \(b=V/T\) is the slope of the line in a \(V\)–\(T\) graph (Kelvin on the x-axis).
Larger \(b\) means larger volume at the same temperature (e.g., lower external pressure or more moles).
Typical pitfalls
- Forgetting Kelvin. Never plug °C or °F directly into the equations.
- Zero or negative K. \(T\) must be \( > 0~\mathrm{K}\); values near 0 K are unphysical for real gases.
- Changing pressure or moles. Charles’s law holds only when \(P\) and \(n\) are constant.
- Unit mismatches. Do the algebra in SI (m³ and K) and then convert final results to the desired units.
Worked example (two-state)
Problem. A sample has \(V_i=2.50~\mathrm{L}\) at \(t_i=24\,^{\circ}\mathrm{C}\). What is \(V_f\) at \(t_f=-25\,^{\circ}\mathrm{C}\) (same \(P\))?
\[ T_i = 24+273.15 = 297.15~\mathrm{K} \]
\[ T_f = -25+273.15 = 248.15~\mathrm{K} \]
\[
V_f = V_i\frac{T_f}{T_i}
\]
\[
V_f = 2.50~\mathrm{L}\times\frac{248.15}{297.15}
\approx 2.09~\mathrm{L}\ (\text{3 s.f.})
\]
Final volume \( \approx 2.09~\mathrm{L} \).
Worked example (finding \(b\))
Given. \(V=2.50~\mathrm{L}\) at \(T=297.15~\mathrm{K}\).
\[
b=\frac{V}{T}
\]
\[
b=\frac{2.50~\mathrm{L}}{297.15~\mathrm{K}}
\approx 8.41\times 10^{-3}~\mathrm{L\cdot K^{-1}}
\]
\[
=\ 8.41\times 10^{-6}~\mathrm{m^3\cdot K^{-1}}
\]
Once \(b\) is known for a sample at fixed \(P\), any state on the same line follows from \(V=bT\).
