Problem 43 - 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

29.50 g of \(\mathrm{Cl}_2\); 9.46 L at STP; 237.82°C for 15.00 L; 228.30 kPa for 5.00 L at 58°C.

Step 1: Calculate Moles of Chlorine Gas

First, we use the ideal gas law to find the number of moles of chlorine gas (\(\mathrm{Cl}_2\)). The ideal gas law equation is \(PV = nRT\), where \(P\) is pressure (in kPa), \(V\) is volume (in L), \(n\) is moles, \(R\) is the ideal gas constant (8.314 J/mol·K), and \(T\) is temperature in Kelvin. Convert the temperature from Celsius to Kelvin: \(T = 24 + 273.15 = 297.15\,\mathrm{K}\). Rearrange the ideal gas law to solve for \(n\): \[n = \frac{PV}{RT} = \frac{119.3 \times 8.70}{8.314 \times 297.15} = 0.416\,\mathrm{mol}\].

Step 2: Convert Moles to Grams

To find how many grams of \(\mathrm{Cl}_2\) we have, we multiply the moles (calculated above) by the molar mass of \(\mathrm{Cl}_2\) (70.90 g/mol):\[\text{mass in grams} = 0.416 \times 70.90 = 29.50\,\mathrm{g}\].

Step 3: Calculate Volume at STP

Standard Temperature and Pressure (STP) is defined as 0°C (273.15 K) and 100 kPa. Using the moles from Step 1 and the ideal gas law, solve for the new volume \(V_{STP}\): \[V_{STP} = \frac{nRT_{STP}}{P_{STP}} = \frac{0.416 \times 8.314 \times 273.15}{100} = 9.46\,\mathrm{L}\].

Step 4: Find Temperature for Given Volume and Pressure

Using the ideal gas law, rearrange to find temperature \(T\) when volume is 15.00 L and pressure is 116.8 kPa: \[T = \frac{PV}{nR} = \frac{116.8 \times 15.00}{0.416 \times 8.314} = 510.97\,\mathrm{K}\]. Convert back to Celsius: \(T = 510.97 - 273.15 = 237.82\,\degree\mathrm{C}\).

Step 5: Calculate Pressure for Given Volume and Temperature

To find pressure when the volume is 5.00 L and the temperature is 58°C (331.15 K), use the ideal gas law: \[P = \frac{nRT}{V} = \frac{0.416 \times 8.314 \times 331.15}{5.00} = 228.30\,\mathrm{kPa}\].

Key Concepts

Chlorine gas

Chlorine gas, also known as  \( \text{Cl}_2\), is a diatomic molecule, meaning that each molecule consists of two chlorine atoms. This gas has numerous applications, particularly in disinfection. It is commonly used in water treatment facilities to purify water, making it safe for consumption, and in swimming pools to keep them sanitized. Chlorine gas is known for its pale green color and pungent smell.
Special precautions need to be taken when handling chlorine gas because it can be harmful or even dangerous at high concentrations. Its gas state allows it to be compressed and stored in pressurized containers, making it convenient for industrial and municipal use.

• Chlorine gas is denoted chemically as \( \text{Cl}_2\).
• It is essential for water purification and sanitation.
• Safety measures are crucial when handling chlorine gas due to its toxic nature in high amounts.

Molar mass

The molar mass of a substance is a simple yet essential concept in chemistry. It refers to the mass of one mole of a chemical compound, expressed in grams per mole (g/mol). In this exercise, the molar mass of chlorine gas, \( \text{Cl}_2\), is calculated by summing up the molar masses of two chlorine atoms. Each chlorine atom has a molar mass of about 35.45 g/mol.
Therefore, for chlorine gas, the total molar mass comes out to be approximately 70.90 g/mol. Understanding molar mass is crucial because it allows us to convert between moles and grams, facilitating calculations in chemical reactions and processes.

• Molar mass is used to convert moles of a substance to grams.
• The molar mass of \( \text{Cl}_2\) is approximately 70.90 g/mol.

Pressure calculations

Pressure calculations are integral to working with gases, especially when using the ideal gas law. This law enables the determination of unknown variables like pressure, volume, or temperature when the other conditions are known. The key idea in pressure calculations is how pressure, along with volume and temperature, impacts the behavior of a gas.
In this exercise, we determine the pressure of chlorine gas when the volume and temperature change. The formula \(P = \frac{nRT}{V}\) allows us to find the new pressure by substituting known values of moles, temperature in Kelvin, and volume. Through this process, we calculate the adjusted pressure conditions for a given volume and temperature.

• Pressure is measured in kilopascals (kPa) in this context.
• The ideal gas law is used: \(PV=nRT\).
• Tools like the ideal gas law are essential for predicting how gases behave under different conditions.

Temperature conversion

Temperature conversion is a necessary step in calculations involving gases because the ideal gas law needs temperature values in Kelvin, not Celsius.
To convert a temperature in Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. This conversion ensures that we're using an absolute temperature scale, which is necessary for accurate gas calculations.

• For example, \(24\, \degree\text{C}\) is converted to \(297.15\, \text{K}\).
• Always add 273.15 to Celsius temperatures to get Kelvin.

Kelvin is preferred in gas law calculations because zero Kelvin represents absolute zero, the point where molecular motion stops. This makes it a more appropriate scale for a range of scientific calculations.