Avogadro’s Law (constant T, P)
When the temperature and pressure of a gas are held constant, its
volume is directly proportional to the amount of gas (in moles or particles).
\[ V \propto n \]
\[ V = k\,n \]
From the ideal-gas equation
Starting with \(PV=nRT\) and keeping \(T\) and \(P\) constant:
\[ V=\frac{nRT}{P} \]
\[ k=\frac{RT}{P} \quad \text{(units: volume per mole)} \]
Thus the slope \(k\) of a \(V\)–\(n\) plot equals the molar volume under the given
conditions (gas sample on a straight line through the origin).
Two-state (ratio) form
Comparing initial \((V_i,n_i)\) and final \((V_f,n_f)\) states at the same \(T\) and \(P\):
\[ \frac{V_i}{n_i}=\frac{V_f}{n_f} \]
\[ V_f = V_i\,\frac{n_f}{n_i} \]
\[ n_f = n_i\,\frac{V_f}{V_i} \]
Amounts: moles and particles
\[ N_A = 6.022\,140\,76\times 10^{23}\ \text{mol}^{-1}\ \ (\text{exact}) \]
\[ n\ (\mathrm{mol})=\frac{N\ (\text{particles})}{N_A} \]
\[ 1\ \mathrm{mol} = 1000\ \mathrm{mmol} \]
- You may work in mol, mmol, or in units of \( \times 10^{23} \) particles; convert to mol for formulas.
- Perform the algebra in SI (m³, mol), then convert the final result to the chosen display units.
Molar volume values you’ll see
- STP (IUPAC 1 bar, 273.15 K): \( \approx 22.711\ \mathrm{L\,mol^{-1}} \)
- 0 °C & 1 atm: \( \approx 22.414\ \mathrm{L\,mol^{-1}} \)
Real gases deviate from ideal behavior at high pressure or very low temperature, but for typical lab
conditions the ideal-gas prediction is a good approximation.
Worked example (two-state)
Problem. A gas has \(V_i=2.50\ \mathrm{L}\) at \(n_i=0.100\ \mathrm{mol}\).
What volume will it occupy when \(n_f=0.150\ \mathrm{mol}\) (same \(T,P\))?
\[ V_f = V_i\,\frac{n_f}{n_i} \]
\[
V_f = 2.50\ \mathrm{L}\times\frac{0.150}{0.100}
\]
\[ V_f = 3.75\ \mathrm{L} \]
Final volume \( = 3.75\ \mathrm{L} \).
Worked example (find \(k\) and use it)
Given. \(V=2.50\ \mathrm{L}\) when \(n=0.100\ \mathrm{mol}\).
\[ k = \frac{V}{n} \]
\[ k = \frac{2.50\ \mathrm{L}}{0.100\ \mathrm{mol}} \]
\[ k = 25.0\ \mathrm{L\,mol^{-1}} \]
\[ \text{If } n=0.060\ \mathrm{mol}, \quad V=k\,n \]
\[ V = 25.0\ \mathrm{L\,mol^{-1}}\times 0.060\ \mathrm{mol} \]
\[ V = 1.50\ \mathrm{L} \]
Common pitfalls
- Changing \(T\) or \(P\). Avogadro’s law requires both to be constant.
- Unit mixing. Keep volumes and amounts consistent (convert mmol → mol, particles → mol).
- Forgetting that the line passes through the origin. In a correct \(V\)–\(n\) graph, \(V=0\) when \(n=0\).
Combining gas volumes (reasoning link)
At fixed \(T,P\), equal volumes of gases contain equal numbers of molecules. Thus gaseous reaction
volume ratios mirror mole ratios (e.g., \(2\,\text{vol}\ \mathrm{H_2}+1\,\text{vol}\ \mathrm{O_2}\to 2\,\text{vol}\ \mathrm{H_2O(g)}\)).
