In Sample Exercise 10.16 , we found that one mole of \(\mathrm{Cl}_{2}\) confined to \(22.41 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\) deviated slightly from ideal behavior. Calculate the pressure exerted by \(1.00 \mathrm{~mol} \mathrm{Cl}_{2}\) confined to a smaller volume, \(5.00 \mathrm{~L}\), at \(25^{\circ} \mathrm{C} .\) (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table \(10.3 .)\) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to \(5.00 \mathrm{~L}\) compared to \(22.4 \mathrm{~L} ?\)

The ideal-gas equation is given as follows: \[PV = nRT\] where: - P is the pressure - V is the volume (force and \(V = 5.00 \mathrm{~L}\) - T is the temperature (force and \(T = 25^{\circ} \mathrm{C}\) which is equal to \(T = 298\mathrm{~K}\) when converted to Kelvin) - n is the number of moles (\(n = 1.00 \mathrm{~mol}\)) - R is the gas constant (\(R = 0.08206 \frac{\mathrm{L~atm}}{\mathrm{K~mol}}\)) We need to find the pressure, P. Rearrange the equation to solve for P: \[P = \frac{nRT}{V}\] Now, plug in the values of n, R, T, and V: \[P = \frac{(1.00 \mathrm{~mol})(0.08206 \frac{\mathrm{L~atm}}{\mathrm{K~mol}})(298\mathrm{~K})}{5.00 \mathrm{~L}}\] Calculate the result: \[P = 49.2 \mathrm{~atm}\] So, the pressure using the ideal-gas equation is 49.2 atm.

The van der Waals equation accounts for the deviations from ideal behaviour due to molecular interactions. The equation is given as follows: \[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\] For this exercise, we're given the van der Waals constants for Cl2: - a = 6.49 L²atm/mol² - b = 0.0562 L/mol Plug in the values of n, R, T, V, a, and b: \[\left(P + \frac{(6.49 \frac{\mathrm L^2\mathrm{~atm}}{\mathrm{mol^2}})(1.00 \mathrm{~mol})^2}{(5.00 \mathrm{~L})^2}\right)\{(5.00 \mathrm{~L}) - (0.0562 \frac{\mathrm{L}}{\mathrm{mol}})(1.00 \mathrm{~mol})\} = (1.00 \mathrm{~mol})(0.08206\frac{\mathrm{L~atm}}{\mathrm{K~mol}})(298\mathrm{~K})\] Solve the equation for P: \[P = \frac{nRT + a(\frac{n}{V})^2(V - nb) - nbR}{V - nb}\] Plug in the same values: \[P = \frac{(1.00 \mathrm{~mol})(0.08206 \frac{\mathrm{L~atm}}{\mathrm{K~mol}})(298\mathrm{~K}) + (6.49 \frac{\mathrm{L^2\mathrm{~atm}}{\mathrm{mol^2}})(\frac{1.00 \mathrm{~mol}}{5.00 \mathrm{~L}})^2(5.00 \mathrm{~L} - (0.0562 \frac{\mathrm{L}}{\mathrm{mol}})(1.00 \mathrm{~mol})) - (0.0562 \frac{\mathrm{L}}{\mathrm{mol}})(1.00 \mathrm{~mol})(0.08206\frac{\mathrm{L~atm}}{\mathrm{K~mol}})}{(5.00 \mathrm{~L} - (0.0562\frac{\mathrm{L}}{\mathrm{mol}})(1.00 \mathrm{~mol}))}\] Calculate the result: \[P = 51.9 \mathrm{~atm}\] So, the pressure using the van der Waals equation is 51.9 atm.

The ideal-gas equation gave a result of 49.2 atm, while the van der Waals equation gave a result of 51.9 atm. The difference between the two is greater when the gas is confined to a smaller volume compared to a larger volume (22.4 L) due to the fact that the van der Waals equation accounts for the intermolecular forces. At a smaller volume, the interactions between gas molecules becomes significant as they are now located closer to each other. These forces make the gas deviate from ideal behaviour, as predicted by the ideal-gas law. The van der Waals equation accounts for the molecular size and attractions, therefore providing a better estimation of the pressure exerted by the gas when confined to smaller volumes.