In Example \(8-11\) of the text, the molar volume of \(\mathrm{N}_{2}(g)\) at \(\mathrm{STP}\) is given as \(22.42 \mathrm{L} / \mathrm{mol} \mathrm{N}_{2} .\) How is this number calculated? How does the molar volume of \(\mathrm{He}(g)\) at \(\mathrm{STP}\) compare to the molar volume of \(\mathrm{N}_{2}(g)\) at \(\mathrm{STP}\) (assuming ideal gas behavior)? Is the molar volume of \(\mathrm{N}_{2}(g)\) at 1.000 atm and \(25.0^{\circ} \mathrm{C}\) equal to, less than, or greater than 22.42 L/mol? Explain. Is the molar volume of \(\mathrm{N}_{2}(g)\) collected over water at a total pressure of 1.000 atm and \(0.0^{\circ} \mathrm{C}\) equal to, less than, or greater than \(22.42 \mathrm{L} / \mathrm{mol} ?\) Explain.

STP conditions are defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm. We are required to calculate the molar volume of N2(g) at STP (22.42 L/mol) using the Ideal Gas Law: PV = nRT where P = pressure, V = volume, n = moles of gas, R = ideal gas constant, and T = temperature. We need to solve for V when n=1 mol (so we'll get the volume for 1 mol of gas) 1 atm * V = 1 mol * 0.0821 L atm/mol K * 273.15 K V = \(\frac{1 mol * 0.0821 L atm/mol K * 273.15 K}{1 atm}\) V = 22.42 L/mol

Since we are assuming ideal gas behavior, all ideal gases will have the same molar volume at STP. Therefore, the molar volume of He(g) at STP is also 22.42 L/mol.

We need to determine if the molar volume of N2(g) at 1 atm and 25°C is equal to, less than, or greater than 22.42 L/mol. We will use the Ideal Gas Law to calculate the molar volume at the given conditions: 1 atm * V = 1 mol * 0.0821 L atm/mol K * 298.15 K V = \(\frac{1 mol * 0.0821 L atm/mol K * 298.15 K}{1 atm}\) V = 24.45 L/mol Since 24.45 L/mol is greater than 22.42 L/mol, the molar volume of N2(g) at 1 atm and 25°C is greater than 22.42 L/mol.

When a gas is collected over water, the total pressure is the sum of the partial pressure of the dry gas and the partial pressure of the water vapor. In this case, the total pressure is 1 atm. Using Dalton's Law (Ptotal = Pdry gas + Pwater vapor), we can find the partial pressure of N2(g). The vapor pressure of water at 0°C is 0.006 atm. Pdry gas = Ptotal - Pwater vapor Pdry gas = 1 atm - 0.006 atm = 0.994 atm Now we can use the Ideal Gas Law to calculate the molar volume of N2(g) at the given conditions: 0.994 atm * V = 1 mol * 0.0821 L atm/mol K * 273.15 K V = \(\frac{1 mol * 0.0821 L atm/mol K * 273.15 K}{0.994 atm}\) V = 22.61 L/mol Since 22.61 L/mol is greater than 22.42 L/mol, the molar volume of N2(g) collected over water at a total pressure of 1 atm and 0°C is greater than 22.42 L/mol.