Problem 27 - Chemistry An Atoms First Approach

Question

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.

Step-by-Step Solution

Answer

The molar volume of N2(g) at STP (22.42 L/mol) is calculated using the Ideal Gas Law: PV=nRT. At STP, both N2(g) and He(g) have the same molar volume of 22.42 L/mol. The molar volume of N2(g) at 1 atm and 25°C is 24.45 L/mol, which is greater than 22.42 L/mol. Finally, the molar volume of N2(g) collected over water at a total pressure of 1 atm and 0°C is 22.61 L/mol, which is also greater than 22.42 L/mol.

Step 1: 1. Molar Volume of N2(g) at STP

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

Step 2: 2. Molar Volume of He(g) at STP

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.

Step 3: 3. Molar Volume of N2(g) at 1 atm and 25°C

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.

Step 4: 4. Molar Volume of N2(g) collected over water at 1 atm and 0°C

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.

Key Concepts

Molar Volume

Molar volume is a term used in chemistry to describe the volume occupied by one mole of a gas under specific conditions. At Standard Temperature and Pressure (STP), which is 0°C (273.15 K) and 1 atm, the molar volume of any ideal gas is approximately 22.42 L/mol. This value is derived from the Ideal Gas Law equation:
\[ PV = nRT \]
Here, \( P \) is pressure, \( V \) is volume, \( n \) is the amount of gas, expressed in moles, \( R \) is the ideal gas constant (0.0821 L atm/mol K), and \( T \) is temperature.
By substituting \( P = 1\) atm, \( T = 273.15\) K, \( n = 1\) mol into the equation, we solve for \( V \), giving us:

• \( V = \frac{1mol \times 0.0821 L \cdot atm/mol \cdot K \times 273.15 K}{1 atm} \)
• \( V = 22.42 \) L/mol

This means that each mole of an ideal gas at STP occupies a volume of 22.42 liters.

STP Conditions

STP stands for Standard Temperature and Pressure, a reference point used in many scientific calculations and experiments. The conditions denote a temperature of 0°C (273.15 K) and an atmospheric pressure of 1 atm. Under these conditions, gases exhibit predictable behavior that simplifies calculations and predictions about gas properties.
Gases behave in a near-ideal manner, meaning they follow the Ideal Gas Law closely. Most notably, the molar volume of an ideal gas is 22.42 L/mol under STP, which provides a straightforward way to calculate the volume of gases in many experiments.
STP is commonly utilized to compare different gases under consistent conditions, allowing scientists to evaluate and calculate differences and similarities, such as the molar volume of helium (He) compared to that of nitrogen (N₂) at STP. Both gases, given ideal behaviors, present the same molar volume of 22.42 L/mol.

Dalton's Law

Dalton's Law of Partial Pressures plays a crucial role when gases are collected over water. This law states that in a gas mixture, the total pressure is the sum of the partial pressures of each individual gas. The equation can be expressed as:
\[ P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water vapor}} \]
Where \( P_{\text{total}} \) is the total pressure of the mixture, \( P_{\text{dry gas}} \) is the pressure exerted by the gas if alone, and \( P_{\text{water vapor}} \) is the pressure exerted by water vapor.
In practical terms, when collecting a gas like nitrogen over water, you must account for the water vapor's pressure to find the partial pressure of the nitrogen gas itself. For example, at 0°C, the vapor pressure of water is roughly 0.006 atm. If we know the total pressure is 1 atm, then:

• \( P_{\text{dry gas}} = P_{\text{total}} - P_{\text{water vapor}} \)
• \( P_{\text{dry gas}} = 1 \text{ atm} - 0.006 \text{ atm} \)
• \( P_{\text{dry gas}} = 0.994 \text{ atm} \)

Using the dry gas pressure and the Ideal Gas Law, we can then correctly calculate the molar volume of the gas at these conditions, considering the presence of water vapor and ensuring accurate results.