Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(80^{\circ} \mathrm{C}\) if 1.00 mol occupies \(33.3 \mathrm{L},\) assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table \(10.3 .\) ) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4}\) ? Explain.

For this problem, we will need the following information: Constants: - R (the ideal gas constant) = 0.08206 L * atm / mol * K - The van der Waals constants for CCl4 (from Table 10.3): - a = 20.41 L^2 * atm / mol^2 - b = 0.1281 L/mol Variables: - T (temperature) = 80°C = 353.15 K (converted to Kelvin) - n (amount) = 1.00 mol - V (volume) = 33.3 L

The ideal gas equation is given by: \(PV = nRT\) We need to solve for P (pressure), so we can rewrite the equation as: \(P = \frac{nRT}{V}\) Now, we can plug in the values for T, n, and V: \(P = \frac{(1.0\ \mathrm{mol})(0.08206\ \mathrm{L\ atm\cdot K^{-1} mol^{-1}})(353.15\ \mathrm{K})}{33.3\ \mathrm{L}}\) Calculate the pressure P: \(P \approx 0.828\ \mathrm{atm}\)

The van der Waals equation is given by: \[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\] To solve for P, we rewrite the equation as: \[P = \frac{nRT}{V - nb} - \frac{an^2}{V^2}\] Plug in the given values for n, V, a, b, R, and T: \[P = \frac{(1.0\ \mathrm{mol})(0.08206\ \mathrm{L\ atm\cdot K^{-1} mol^{-1}})(353.15\ \mathrm{K})}{(33.3\ \mathrm{L}) - (1.0\ \mathrm{mol})(0.1281\ \mathrm{L/mol})} - \frac{(20.41\ \mathrm{L^2 \cdot atm\cdot mol^{-2}})(1.0\ \mathrm{mol})^2}{(33.3\ \mathrm{L})^2}\] Calculate the pressure P: \(P \approx 0.983\ \mathrm{atm}\)

The deviation from ideal behavior can be attributed to the fact that real gases have volume and exhibit intermolecular forces. In the van der Waals equation, the constant a corresponds to attractive forces, and b corresponds to molecular volume. Larger values for a and b denote greater deviation from ideal gas behavior. Given the values of a and b for Cl2 and CCl4: - Cl2: a = 6.49 L^2 * atm / mol^2, b = 0.05468 L/mol - CCl4: a = 20.41 L^2 * atm / mol^2, b = 0.1281 L/mol Considering that CCl4 has higher values for both a and b than Cl2, we would expect CCl4 to show greater deviation from ideal behavior under these conditions compared to Cl2. #Conclusion# The pressure of CCl4 at 80°C if 1.00 mol occupies 33.3 L is approximately 0.828 atm, assuming ideal gas behavior, and approximately 0.983 atm, assuming van der Waals behavior. We also expect CCl4 to deviate more from ideal gas behavior compared to Cl2 under these conditions.