Question

A \(2.50-\mathrm{L}\) flask was used to collect a \(5.65-\mathrm{g}\) sample of propane gas, \(\mathrm{C}_{3} \mathrm{H}_{8} .\) After the sample was collected, the gas pressure was found to be \(741 \mathrm{mmHg}\). What was the temperature of the propane in the flask?

Short answer

The temperature of the propane is approximately 232 K.

Step-by-step solution

Identify the Known Values

We know the volume of the flask is 2.50 L, the mass of propane is 5.65 g, and the pressure is 741 mmHg. We are required to find the temperature.

Convert Units where Necessary

We need to convert the pressure from mmHg to atm because the ideal gas constant (R) is usually in atm. The conversion is 1 atm = 760 mmHg.\[ P = \frac{741 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.975 \, \text{atm} \]

Determine the Moles of Propane

Use the molar mass of propane, C₃H₈, which is calculated as:\[ (3 \times 12.01) + (8 \times 1.008) = 44.094 \, \text{g/mol} \] Calculate the moles of propane:\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.65 \, \text{g}}{44.094 \, \text{g/mol}} = 0.128 \, \text{mol} \]

Use the Ideal Gas Law

Apply the ideal gas law \(PV = nRT\) to find the temperature. Solve for T:\[ T = \frac{PV}{nR} \]where \(R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1}\). Substitute the values:\[ T = \frac{(0.975 \, \text{atm})(2.50 \, \text{L})}{(0.128 \, \text{mol})(0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1})} \]

Calculate the Temperature

Calculate the expression obtained:\[ T = \frac{2.4375 \, \text{L atm}}{0.0105088 \, \text{L atm mol}^{-1} \text{K}^{-1}} \approx 232 \, \text{K} \]

Key concepts

Gas Pressure Conversion

When dealing with gas laws, we often need to convert pressure into different units for accurate calculations. In the given exercise, the goal was to determine the temperature of a propane gas sample. Propane gas had its pressure initially measured in mmHg, a commonly used unit. However, the Ideal Gas Law equation typically requires pressure to be in atmospheres (atm).
To convert the pressure from mmHg to atm, recall that 1 atm equals 760 mmHg.
Therefore, divide the pressure in mmHg by 760 to convert it to atm:

• Given pressure: 741 mmHg
• Conversion factor: 1 atm = 760 mmHg
• Calculated pressure: \( \frac{741}{760} = 0.975 \) atm

Knowing how to convert between these units is essential in chemistry, as it allows us to apply the Ideal Gas Law consistently.

Moles Calculation

Understanding moles is a vital part of solving problems involving gases. A mole offers a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. To find moles, use the substance's mass and its molar mass.
In this exercise, we're dealing with propane, which has a specific molar mass. Propane (\(\text{C}_3\text{H}_8\)) consists of carbon and hydrogen. The molar mass of one molecule is calculated using:

• Carbon: 3 atoms \(\times\) 12.01 g/mol = 36.03 g/mol
• Hydrogen: 8 atoms \(\times\) 1.008 g/mol = 8.064 g/mol

Adding these gives propane’s molar mass of 44.094 g/mol. Now, using the mass of propane from the exercise (5.65 g), calculate the moles:

• Number of moles = \( \frac{\text{mass}}{\text{molar mass}} = \frac{5.65}{44.094} = 0.128 \) mol.

This step is necessary as moles are a requirement in the Ideal Gas Law for determining temperature.

Temperature Calculation

Temperature calculations in gas law problems are often the finale in solving for unknown variables. With the Ideal Gas Law, which is a mathematical relationship between pressure, volume, temperature, and moles of a gas, calculations become straightforward. The law is expressed as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.
We solved for temperature in this exercise by rearranging the equation to get:

• \[T = \frac{PV}{nR}\]

Using the known variables:

• \(P = 0.975\) atm
• \(V = 2.50\) L
• \(n = 0.128\) mol
• \(R = 0.0821\, \text{L atm mol}^{-1} \text{K}^{-1}\)

Substitute into the equation:

• \[T = \frac{(0.975 \times 2.50)}{(0.128 \times 0.0821)} \approx 232\, \text{K}\]

Converting to Celsius can often help, remembering that Kelvin is always positive and absolute.

Propane Gas Properties

Propane gas is a hydrocarbon, made of carbon and hydrogen, used in applications like heating and cooking. Knowing its properties helps us understand its behavior under different conditions.

• Propane's chemical formula is \(\text{C}_3\text{H}_8\).
• It is a colorless gas at room temperature and can easily be liquefied under pressure.
• Typically stored under pressure in liquid form for use as a fuel.

When dealing with propane and similar gases, the Ideal Gas Law is an effective tool to predict the gas behavior under various conditions of pressure, volume, and temperature. Additionally, propane’s molar mass (44.094 g/mol), previously calculated, is essential for conversions involving grams to moles, as demonstrated in exercises and real-life applications like fuel estimations.