Problem 44 - Chemistry The Central Science In Si Units

Question

Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of \(\mathrm{Cl}_{2}\) gas is \(8.70 \mathrm{~L}\) at \(119.3 \mathrm{kPa}\) and \(24^{\circ} \mathrm{C}\). (a) How many grams of \(\mathrm{Cl}_{2}\) are in the sample? (b) What volume will the \(\mathrm{Cl}_{2}\) occupy at STP? (c) At what temperature will the volume be \(15.00 \mathrm{~L}\) if the pressure is \(116.8 \mathrm{kPa}\) (d) At what pressure will the volume equal \(5.00 \mathrm{~L}\) if the temperature is \(58^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) 25.3 g of Cl2. (b) 8.33 L at STP. (c) 281.58°C for 15.00 L at 116.8 kPa. (d) 196.57 kPa for 5.00 L at 58°C.

Step 1: Calculate the moles of Cl2 in the sample

To find the moles of chlorine gas in the sample, we can use the ideal gas law: \[ PV = nRT \]Where - \( P \) is the pressure \( 119.3 \, \text{kPa} \) (converted to \( 1.193 \, \text{atm} \)), - \( V \) is the volume \( 8.70 \, \text{L} \), - \( R \) is the ideal gas constant \( 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \), and - \( T \) is the temperature in Kelvin \( 24 + 273.15 = 297.15 \, \text{K} \).First, convert the pressure from kPa to atm:\[ 119.3 \, \text{kPa} \times \frac{1 \, \text{atm}}{101.325 \, \text{kPa}} = 1.178 \, \text{atm} \]Plugging in the values to solve for \( n \):\[ n = \frac{PV}{RT} = \frac{1.178 \, \text{atm} \times 8.70 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \times 297.15 \, \text{K}} \approx 0.357 \, \text{mol} \]

Step 2: Determine the grams of Cl2

Use the moles of chlorine gas from Step 1 to calculate the mass in grams:\[ \text{Molar mass of Cl}_2 = 2 \times 35.453 \, \text{g/mol} = 70.906 \, \text{g/mol} \]Mass calculation:\[ \text{Mass} = n \times \text{Molar mass} = 0.357 \, \text{mol} \times 70.906 \, \text{g/mol} \approx 25.3 \, \text{g} \]

Step 3: Calculate the volume at STP

Standard temperature and pressure (STP) is defined as 0°C (273.15 K) and 1 atm.Using the number of moles calculated and the ideal gas law: \[ V = \frac{nRT}{P} \]Plug in \( n = 0.357 \, \text{mol} \), \( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \), \( T = 273.15 \, \text{K} \), and \( P = 1 \, \text{atm} \):\[ V = \frac{0.357 \, \text{mol} \times 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \times 273.15 \, \text{K}}{1 \, \text{atm}} \approx 8.33 \, \text{L} \]

Step 4: Find the temperature for volume 15.00 L and pressure 116.8 kPa

First, convert the pressure from kPa to atm:\[ 116.8 \, \text{kPa} \times \frac{1 \, \text{atm}}{101.325 \, \text{kPa}} = 1.153 \, \text{atm} \]Use the ideal gas law to solve for the new temperature:\[ T = \frac{PV}{nR} \]Substitute \( P = 1.153 \, \text{atm} \), \( V = 15.00 \, \text{L} \), \( n = 0.357 \, \text{mol} \), and \( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \):\[ T = \frac{1.153 \, \text{atm} \times 15.00 \, \text{L}}{0.357 \, \text{mol} \times 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol}} \approx 554.73 \, \text{K} \]Convert Kelvin to Celsius:\[ 554.73 \, \text{K} - 273.15 = 281.58 \, ^{\circ} \text{C} \]

Step 5: Calculate the pressure for 5.00 L at 58°C

First, convert the temperature from Celsius to Kelvin:\[ 58 + 273.15 = 331.15 \, \text{K} \]Use the ideal gas law to find the pressure:\[ P = \frac{nRT}{V} \]Substitute \( n = 0.357 \, \text{mol} \), \( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \), \( T = 331.15 \, \text{K} \), and \( V = 5.00 \, \text{L} \):\[ P = \frac{0.357 \, \text{mol} \times 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \times 331.15 \, \text{K}}{5.00 \, \text{L}} \approx 1.94 \, \text{atm} \]Convert atm to kPa:\[ 1.94 \, \text{atm} \times 101.325 \, \text{kPa / atm} = 196.57 \, \text{kPa} \]

Key Concepts

Chlorine gas calculations

Chlorine gas, or \(\text{Cl}_2\), is frequently utilized in disinfection processes due to its effectiveness in purifying water. Calculating the quantity of chlorine gas often involves using the Ideal Gas Law formula: \[ PV = nRT \] where:

• \(P\) denotes pressure
• \(V\) is the volume
• \(n\) is the number of moles of gas
• \(R\) represents the ideal gas constant \(0.0821\, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol}\)
• \(T\) is the temperature in Kelvin

To calculate the grams of chlorine in a sample, it's necessary to first determine the number of moles using the above formula. For example, given a sample with a known volume, temperature, and pressure, we convert all units to match the formula's requirements. After finding the moles, we multiply by the molar mass of \(\text{Cl}_2\) (approximately \(70.906 \, \text{g/mol}\)) to yield the mass.
Tip: Make sure all your units are consistent. Convert pressure to atmospheres and temperature to Kelvin before solving.

Standard Temperature and Pressure (STP)

Understanding Standard Temperature and Pressure (STP) is crucial for comparing gas volumes under different conditions. STP is defined as a temperature of \(0^\circ \text{C}\) or \(273.15 \, \text{K}\) and a pressure of \(1\, \text{atm}\). These conditions are used as a standard reference to measure gas properties.
Let's say you need to find out how much volume a gas will occupy at STP. Using the moles calculated from a previous step and applying the Ideal Gas Law, you would substitute the STP values into the equation to solve for the volume. This is crucial for standardizing gas measurements.
Tip: Always remember that \(R\), the ideal gas constant, remains the same for any of these calculations, ensuring reliable and consistent results every time.

Gas volume and temperature relationships

Gas volume is deeply interconnected with temperature, following Charles's Law. This principle states that, at constant pressure, the volume of a gas is directly proportional to its absolute temperature (in Kelvin). Hence, as the temperature of a gas increases, so does its volume, provided that the pressure remains unchanged.
When tasked with calculating what temperature or volume a gas will reach under varying conditions, the Ideal Gas Law again comes into play. Imagine you need to find the temperature at which a gas expands to a new volume; rearranging the equation to solve for the unknown variable allows us to determine the necessary conditions.
Tip: Ensure temperature is consistently calculated in Kelvin to avoid unexpected discrepancies. This law underscores the dynamic nature of gases, reacting predictably to changes in temperature.