Question

In an experiment, you fill a heavy-walled 6.00 -L flask with methane gas, \(\mathrm{CH}_{4}\). If the flask contains \(7.13 \mathrm{~g}\) of methane at \(19^{\circ} \mathrm{C}\), what is the gas pressure?

Short answer

The gas pressure is approximately 1.77 atm.

Step-by-step solution

Understanding Given Information

We have a 6.00-L flask filled with methane (CH extsubscript{4}). The mass of CH extsubscript{4} in the flask is 7.13 g, and the temperature of the gas is 19°C.

Convert Temperature to Kelvin

Convert the temperature from Celsius to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \] So the temperature in Kelvin is:\[ 19 + 273.15 = 292.15 \text{ K} \]

Calculate Moles of Methane

First, find the molar mass of methane (CH extsubscript{4}). The molar mass of carbon (C) is approximately 12.01 g/mol, and hydrogen (H) is about 1.01 g/mol. Therefore, the molar mass of CH extsubscript{4} is:\[ 12.01 + (4 \times 1.01) = 16.05 \text{ g/mol} \]Next, calculate the number of moles (n) using the formula: \[ n = \frac{{\text{mass}}}{{\text{molar mass}}} \] So, \[ n = \frac{{7.13 \text{ g}}}{{16.05 \text{ g/mol}}} \approx 0.444 \text{ moles} \]

Use the Ideal Gas Law to Calculate Pressure

Use the Ideal Gas Law: \[ PV = nRT \]Where: - \( P \) is the pressure (in atm), - \( V \) is the volume (in liters), - \( n \) is the number of moles, - \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), - \( T \) is the temperature in Kelvin.Rearrange the equation to solve for pressure (P): \[ P = \frac{{nRT}}{V} \]Substitute the values: \[ P = \frac{{0.444 \times 0.0821 \times 292.15}}{6.00} \approx 1.77 \text{ atm} \]

Conclusion

The pressure of the methane gas in the flask is approximately 1.77 atm.

Key concepts

Moles of Gas

Understanding the concept of moles is essential when working with gases, especially when using the Ideal Gas Law. The mole is a fundamental unit in chemistry that measures the amount of a substance. To determine the number of moles of a gas, we use the formula:\[ n = \frac{\text{mass}}{\text{molar mass}} \]where:

• \( n \) is the number of moles,
• mass is the amount of substance we have (in grams),
• molar mass is the mass of one mole of the substance (in grams per mole). This value can be found on the periodic table or through calculation by adding the atomic masses of all elements in a compound.

In our example with methane \( \text{CH}_4 \), the molar mass is calculated by adding the molar masses of carbon \( (12.01 \text{ g/mol}) \) and four hydrogens \( (4 \times 1.01 \text{ g/mol}) \), giving us \( 16.05 \text{ g/mol} \). By dividing the mass of our methane sample \((7.13 \text{ g})\) by its molar mass \((16.05 \text{ g/mol})\), we find that there are approximately \( 0.444 \text{ moles} \) of methane in the flask.

Temperature Conversion

Temperature plays a crucial role in the behavior of gases. When dealing with gas laws, it's important to express temperature in the Kelvin scale. This is because Kelvin is an absolute temperature scale where zero corresponds to absolute zero, the point at which particles have minimal kinetic energy. Converting from Celsius to Kelvin is straightforward with the following formula:\[ T(K) = T(°C) + 273.15 \]In our exercise, converting the temperature from \( 19^{\circ} \text{C} \) to Kelvin results in:\[ 19 + 273.15 = 292.15 \text{ K} \]This step ensures that all variables in the Ideal Gas Law calculation are in compatible units. Remember, never use Celsius directly when applying the Ideal Gas Law as it can lead to incorrect results.

Gas Pressure Calculation

The Ideal Gas Law is crucial when calculating the pressure of a gas in a container, which can be expressed as:\[ PV = nRT \]where:

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

To find the pressure \( P \), rearrange the equation:\[ P = \frac{nRT}{V} \] Using the values from our experiment:- \( n = 0.444 \text{ moles} \),- \( R = 0.0821 \text{ L} \cdot \text{atm/mol} \cdot \text{K} \),- \( T = 292.15 \text{ K} \), - \( V = 6.00 \text{ L} \).Substituting these into the formula, we calculate:\[ P = \frac{0.444 \times 0.0821 \times 292.15}{6.00} \approx 1.77 \text{ atm} \]This calculation shows that the methane gas exerts a pressure of approximately \(1.77 \text{ atm}\) inside the flask. Understanding this process is vital for predicting and manipulating gas behavior in practical scenarios.