States of Matter: Question 6

Worked solution

Part (a): Calculating the ideal volume from pV = nRT

The pressure and temperature are already given in the SI units required by the ideal gas equation: $p = 6.00\times10^{6}\ \text{Pa}$ and $T = 350\ \text{K}$.

Rearranging $pV = nRT$ for $V$: $V = \frac{nRT}{p}$

Substituting $n = 2.50\ \text{mol}$, $R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}$, $T = 350\ \text{K}$ and $p = 6.00\times10^{6}\ \text{Pa}$: $V = \frac{2.50\times8.31\times350}{6.00\times10^{6}}$

Working out the numerator first: $2.50\times8.31 = 20.775$ $20.775\times350 = 7271.25$

So: $V = \frac{7271.25}{6.00\times10^{6}} = 1.21188\times10^{-3}\ \text{m}^3$

(Check: $6.00\times10^{6}\times1.21188\times10^{-3} \approx 7271.3$, matching the numerator above.)

Converting from m³ to dm³, since $1\ \text{m}^3 = 10^{3}\ \text{dm}^3$: $V_{\text{ideal}} = 1.21188\times10^{-3}\times10^{3}\ \text{dm}^3 = 1.21188\ \text{dm}^3$

To 3 significant figures: $V_{\text{ideal}} = 1.21\ \text{dm}^3$

Part (b): Percentage by which the actual volume exceeds the ideal volume

The percentage difference is calculated relative to the ideal (predicted) volume from part (a), using the unrounded value $V_{\text{ideal}} = 1.21188\ \text{dm}^3$: $\text{percentage difference} = \frac{V_{\text{actual}} - V_{\text{ideal}}}{V_{\text{ideal}}}\times100\%$

$= \frac{1.34 - 1.21188}{1.21188}\times100\%$

$= \frac{0.12812}{1.21188}\times100\%$

$= 10.57\%$

(Check: $1.21188\times1.1057 \approx 1.340$, confirming the actual volume is about $10.6\%$ greater than the ideal volume.)

To 3 significant figures, the actual volume exceeds the ideal volume by $10.6\%$.

Part (c): Explaining the deviation using the kinetic theory assumptions

The ideal gas model assumes that the volume occupied by the gas particles themselves is negligible compared with the total volume of the container, so that the gas can in principle be compressed into an arbitrarily small volume as pressure increases. At the very high pressure in this cylinder, the CO₂ molecules are forced much closer together, so the actual volume taken up by the molecules themselves becomes a significant fraction of the total gas volume. The assumption of negligible particle volume no longer holds. Because the molecules themselves take up real space that cannot be compressed away, the real gas resists compression more than the ideal gas equation predicts, so the actual measured volume is larger than the ideal volume calculated from $pV = nRT$.

Final answers

• (a) $V_{\text{ideal}} = 1.21\ \text{dm}^3$ (3 s.f.)
• (b) The actual volume exceeds the ideal volume by $10.6\%$ (3 s.f.)
• (c) At this high pressure the CO₂ molecules’ own volume is no longer negligible compared with the container volume, so the real gas cannot be compressed as much as the ideal gas equation predicts, making the actual volume larger than the ideal prediction