Problem 1 - Chemistry: Principles and Reactions

Question

A ten-gallon methane tank contains \(1.243\) mol of methane \(\left(\mathrm{CH}_{4}\right)\) at \(74^{\circ} \mathrm{F}\). Express the volume of the tank in liters, the amount of methane in the tank in grams, and the temperature of the tank in Kelvin.

Step-by-Step Solution

Answer

Answer: The volume of the tank is 37.8541 liters, the amount of methane in the tank is 19.955 grams, and the temperature of the tank is 296.4833 Kelvins.

Step 1: Convert the volume of the tank from gallons to liters

To convert the volume from gallons to liters, we can use the conversion factor: 1 gallon = 3.78541 liters. So, we have: $$ \text{Volume in liters} = \text{Volume in gallons} \times \text{Conversion factor} $$ $$ \text{Volume in liters} = 10 \times 3.78541 = 37.8541 \text{ L} $$

Step 2: Convert the amount of methane from moles to grams

To convert the amount of methane from moles to grams, we need to multiply the given moles by the molar mass of methane: $$ \text{Mass in grams} = \text{Moles} \times \text{Molecular weight} $$ Molecular weight of methane (\(\mathrm{CH}_{4}\)) = \(12.01 (\mathrm{C}) + 4 \times 1.008(\mathrm{H}) = 16.042 \text{ g/mol}\) $$ \text{Mass in grams} = 1.243 \text{ mol} \times 16.042 \frac{\text{g}}{\text{mol}} = 19.955 \text{ g} $$

Step 3: Convert the temperature from Fahrenheit to Kelvin

To convert the temperature from Fahrenheit to Kelvin, we need to use the following conversions: 1. Convert Fahrenheit to Celsius: \((°\mathrm{F} - 32) \times \frac{5}{9} = °\mathrm{C}\) 2. Convert Celsius to Kelvin: \(°\mathrm{C} + 273.15 = \mathrm{K}\) First, convert the given temperature from Fahrenheit to Celsius: $$ \text{Temperature in Celsius} = (74 - 32) \times \frac{5}{9} = 23.3333 °\mathrm{C} $$ Next, convert the temperature from Celsius to Kelvin: $$ \text{Temperature in Kelvin} = 23.3333 + 273.15 = 296.4833\mathrm{K} $$ Therefore, the volume of the tank is \(37.8541 \text{ L}\), the amount of methane in the tank is \(19.955 \text{ g}\), and the temperature of the tank is \(296.4833 \mathrm{K}\).

Key Concepts

Mole Conversions

Mole conversions are essential in chemistry. They allow us to connect the number of particles in a substance to its measurable mass. When converting moles to grams, you must know the molar mass of the compound.
For methane (\(\text{CH}_4\)), the molar mass is the sum of the atomic masses of one carbon atom (\(12.01 \text{ g/mol}\)) and four hydrogen atoms (\(1.008 \times 4 = 4.032 \text{ g/mol}\)). Therefore, the molar mass of methane is \(16.042 \text{ g/mol}\).

To convert the amount of methane given in moles to grams:

• Multiply the number of moles by the molar mass.
• For example, \(1.243 \text{ mol} \times 16.042 \text{ g/mol} = 19.955 \text{ g}\).

Once the conversion is complete, you know the mass of the substance in grams. This simple approach is foundational to quantify chemicals for reactions and experiments.

Temperature Conversions

Temperature conversions can be tricky, but they are crucial for understanding physical conditions for gases. The exercise explains how to convert from Fahrenheit to Kelvin, which involves two steps.
The first step involves converting Fahrenheit to Celsius using the formula: \[(°\mathrm{F} - 32) \times \frac{5}{9} = °\mathrm{C}.\]
In our solution, we had \(74 °\mathrm{F}\) converted to Celsius as follows:

• \((74 - 32) \times \frac{5}{9} = 23.3333 °\mathrm{C}.\)

The second step converts Celsius to Kelvin. Simply add 273.15 to the Celsius temperature:

• \(23.3333 + 273.15 = 296.4833\mathrm{K}\).

Now you have the temperature in Kelvin, which is used extensively in gas laws and when dealing with thermodynamic equations since Kelvin is an absolute temperature scale.

Volume Conversions

Volume conversions are common when dealing with gases, as they often require transitioning between different units like gallons and liters. Using the correct conversion factor is crucial.
For converting gallons to liters, you use the factor that 1 gallon equals approximately \(3.78541\) liters.

To find out the volume of the tank in liters:

• Multiply the volume in gallons by 3.78541.
• For a 10-gallon tank, \(10 \times 3.78541 = 37.8541 \, \text{L}.\)

This conversion is particularly important in international contexts where liters are more commonly used. It ensures accurate calculations in gas-related studies, experimental setups, and real-world applications.