2019 H2 Chem P1 Q6 — Gases Solution

The answer is A.

\(pV = nRT = \dfrac{m}{M_r}RT\)
The mass (0.433 g), volume and temperature are the same for every gas, so \(p \propto n \propto \dfrac{1}{M_r}\): the higher the \(M_r\), the lower the pressure.
• \(\text{CH}_4\): \(M_r = 16.0\), n = 0.433 / 16.0 = 0.0271 mol (highest P)
• \(\text{HCHO}\): \(M_r = 30.0\), n = 0.433 / 30.0 = 0.0144 mol
• \(\text{HCO}_2\text{H}\): \(M_r = 46.0\), n = 0.433 / 46.0 = 0.00941 mol
• \(\text{CH}_3\text{Cl}\): \(M_r = 50.5\), n = 0.433 / 50.5 = 0.00857 mol (lowest P)
So P falls from \(\text{CH}_4\) to \(\text{HCHO}\) to \(\text{CH}_3\text{Cl}\), and \(\text{HCO}_2\text{H}\) is slightly higher than \(\text{CH}_3\text{Cl}\). Graph A shows exactly this.

• B: shows \(\text{HCO}_2\text{H}\) lower than \(\text{CH}_3\text{Cl}\), but \(\text{HCO}_2\text{H}\) has the smaller \(M_r\) (46.0 against 50.5), so it has the higher pressure.
• C: equal pressures would need equal amounts in mol; equal masses of gases with different \(M_r\) give different amounts.
• D: shows P increasing with \(M_r\) (\(\text{CH}_4\) lowest, \(\text{CH}_3\text{Cl}\) highest), the reverse of \(p \propto \dfrac{1}{M_r}\).