Problem 85 - Chemistry Central Science

Question

Based on their respective van der Waals constants (Table 10.3), is Ar or \(\mathrm{CO}_{2}\) expected to behave more nearly like an ideal gas at high pressures? Explain.

Step-by-Step Solution

Answer

Argon (Ar) is expected to behave more nearly like an ideal gas at high pressures compared to Carbon Dioxide \(\mathrm{CO}_2\), as Ar has weaker intermolecular interactions (\(a_{Ar} = 1.34 \ L^2\cdot atm\cdot mol^{-2}\) vs \(a_{CO2} = 3.59 \ L^2\cdot atm\cdot mol^{-2}\)) and smaller molecular size (\(b_{Ar} = 0.0322 \ L\cdot mol^{-1}\) vs \(b_{CO2} = 0.0427 \ L\cdot mol^{-1}\)).

Step 1: Review constants for Ar and CO2

Refer to Table 10.3 to find the van der Waals constants for Argon (Ar) and Carbon Dioxide (\(\mathrm{CO}_2\)): Argon (Ar): - \(a_Ar = 1.34 \ L^2\cdot atm\cdot mol^{-2}\) - \(b_Ar = 0.0322 \ L\cdot mol^{-1}\) Carbon Dioxide (\(\mathrm{CO}_2\)): - \(a_{CO2} = 3.59 \ L^2\cdot atm\cdot mol^{-2}\) - \(b_{CO2} = 0.0427 \ L\cdot mol^{-1}\)

Step 2: Compare intermolecular interaction constants

Compare the van der Waals constant for intermolecular interactions (\(a\)) for both gases: - For Argon: \(a_Ar = 1.34 \ L^2\cdot atm\cdot mol^{-2}\) - For Carbon Dioxide: \(a_{CO2} = 3.59 \ L^2\cdot atm\cdot mol^{-2}\) Since \(a_Ar < a_{CO2}\), this indicates that Argon has weaker intermolecular interactions compared to Carbon Dioxide.

Step 3: Compare molecular size constants

Compare the van der Waals constant for molecular size (\(b\)) for both gases: - For Argon: \(b_Ar = 0.0322 \ L\cdot mol^{-1}\) - For Carbon Dioxide: \(b_{CO2} = 0.0427 \ L\cdot mol^{-1}\) Since \(b_Ar < b_{CO2}\), this indicates that Argon molecules are smaller in size compared to Carbon Dioxide molecules.

Step 4: Determine gas behavior at high pressures

We have learned that Argon has weaker intermolecular interactions and smaller molecular size than Carbon Dioxide. Based on these factors, Argon should behave more like an ideal gas at high pressures compared to Carbon Dioxide.