Problem 17 - Chemistry and Chemical Reactivity

Question

A \(1.25-\mathrm{g}\) sample of \(\mathrm{CO}_{2}\) is contained in a \(750 .-\mathrm{mL}\) flask at \(22.5^{\circ} \mathrm{C} .\) What is the pressure of the gas?

Step-by-Step Solution

Answer

The pressure of the gas is 0.915 atm.

Step 1: Understanding What is Given

We have a 1.25-g sample of \( \text{CO}_2 \) gas, and it occupies a 750 mL flask. The temperature is given as \( 22.5^{\circ} \text{C} \). The task is to find out the pressure of this gas.

Step 2: Convert Mass to Moles

To find pressure using the ideal gas law, we first need the number of moles. The molecular weight of \( \text{CO}_2 \) is \( 44.01 \text{ g/mol} \). Calculate the moles using:\[\text{Moles of } \text{CO}_2 = \frac{1.25 \text{ g}}{44.01 \text{ g/mol}} \approx 0.0284 \text{ moles}\]

Step 3: Convert Temperature to Kelvin

The temperature should be in Kelvin for the ideal gas law calculation. Convert from Celsius to Kelvin:\[T = 22.5 + 273.15 = 295.65 \text{ K}\]

Step 4: Convert Volume to Liters

The volume should be in liters. Since 1000 mL = 1 L, convert the volume:\[V = 750 \text{ mL} = 0.750 \text{ L}\]

Step 5: Use the Ideal Gas Law to Find Pressure

The ideal gas law is \( PV = nRT \), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the ideal gas constant (0.0821 L·atm/mol·K), and \(T\) is temperature in Kelvin. Rearrange to find \(P\):\[P = \frac{nRT}{V} = \frac{0.0284 \times 0.0821 \times 295.65}{0.750}\]Calculate:\[P \approx \frac{0.686}{0.750} = 0.915 \text{ atm}\]

Step 6: Conclusion

The calculated pressure of the \( \text{CO}_2 \) gas is approximately \( 0.915 \text{ atm} \).

Key Concepts

Mole Calculations

To determine the amount of substance in a sample, we often convert mass into moles, which is a fundamental concept in chemistry. This involves using the molecular weight (or molar mass) of the compound involved. In the example, we have a sample of carbon dioxide (\(\text{CO}_2\)) weighing 1.25 grams. To find the number of moles, use the formula:

• Number of Moles = \(\frac{\text{mass of sample}}{\text{molar mass}}\)

For \(\text{CO}_2\), the molar mass is 44.01 g/mol. Thus, the moles of \(\text{CO}_2\) are calculated as:\[ \text{Moles of } \text{CO}_2 = \frac{1.25 \text{ g}}{44.01 \text{ g/mol}} \approx 0.0284 \text{ moles} \]This conversion is crucial, as the molar quantity gives us a way to further compute other properties of the gas, such as pressure when using gas laws. Understanding mole concepts helps in linking mass with the number of particles in a sample, enabling detailed chemical analysis.

Temperature Conversion

Temperature conversion is often necessary when using scientific formulas, especially when dealing with gases. For many gas calculations, including the Ideal Gas Law, temperature must be in Kelvin. The Kelvin scale is an absolute temperature scale where 0 Kelvin is the lowest theoretically possible temperature, known as absolute zero.
To convert Celsius (°C) to Kelvin (K), use the formula:

• Temperature in Kelvin \( T = \text{Temperature in Celsius} + 273.15 \)

Using this conversion, the given temperature of 22.5°C becomes:\[ T = 22.5 + 273.15 = 295.65 \text{ K} \]Using Kelvin ensures that our gas law calculations are consistent, as the Ideal Gas Law depends directly on temperature, involving proportionality that starts from absolute zero.

Pressure Calculation

Pressure calculation in gases is typically done using the Ideal Gas Law, a central formula in chemistry describing the relationships among the pressure, volume, temperature, and moles of a gas. The Ideal Gas Law is expressed as:

• \( PV = nRT \)

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)To find pressure \( P \), rearrange the formula:\[ P = \frac{nRT}{V} \]In the example, by plugging in the values:- \( n = 0.0284 \text{ moles} \)- \( R = 0.0821 \text{ L·atm/mol·K} \)- \( T = 295.65 \text{ K} \)- \( V = 0.750 \text{ L} \)Calculate:\[ P = \frac{0.0284 \times 0.0821 \times 295.65}{0.750} \approx 0.915 \text{ atm} \]Calculating pressure helps us understand how dense or packed the gas molecules are within the container.

Volume Conversion

Volume conversion is primarily about ensuring units are consistent when performing calculations. For example, in the case of the Ideal Gas Law, volume should be in liters rather than milliliters. This is because the ideal gas constant \( R \) is defined to work with liters.
To convert from milliliters to liters:

• 1 liter = 1000 milliliters

To convert the volume of 750 milliliters to liters, divide by 1000:\[ V = 750 \text{ mL} = 0.750 \text{ L} \]The correct unit conversion is essential to apply the Ideal Gas Law accurately. Without this step, calculations could yield erroneous or contradictory results due to unit mismatches, underscoring the importance of keen attention to units in all scientific computations.