Problem 42 - Chemistry The Central Science In Si Units

Question

A \(334-\mathrm{mL}\) cylinder for use in chemistry lectures contains \(5.225 \mathrm{~g}\) of helium at \(23^{\circ} \mathrm{C}\). How many grams of helium must be released to reduce the pressure to 7.60 MPa assuming ideal gas behavior?

Step-by-Step Solution

Answer

1.098 g of helium must be released.

Step 1: Understand the Given Information

We have a cylinder with a volume of \(334\,\text{mL}\) containing \(5.225\,\text{g}\) of helium at \(23^{\circ} \text{C}\). The task is to determine how much helium must be released to achieve a pressure of \(7.60\,\text{MPa}\). Recall that the ideal gas law \( PV = nRT \) relates pressure \(P\), volume \(V\), number of moles \(n\), the ideal gas constant \(R\), and temperature \(T\).

Step 2: Convert Units

First, convert the volume from \(\text{mL}\) to \(\text{L}\):\[ V = 334\, \text{mL} = 0.334\, \text{L} \]. Convert the temperature from degrees Celsius to Kelvin: \[ T = 23 + 273.15 = 296.15\, K \].

Step 3: Calculate Initial Moles of Helium

Use the molar mass of helium \((4.00\, \text{g/mol})\) to find the initial moles of helium: \[ n = \frac{5.225\, \text{g}}{4.00\, \text{g/mol}} = 1.30625\, \text{mol} \].

Step 4: Use Ideal Gas Law to Find Initial Pressure

Using \( R = 8.314\, \text{J/mol} \cdot \text{K} \), the ideal gas law gives the initial pressure:\[ P = \frac{nRT}{V} = \frac{1.30625 \times 8.314 \times 296.15}{0.334} \].

Step 5: Solve for Initial Pressure and Compare

Calculate the initial pressure: \[ P_{initial} = \frac{1.30625 \times 8.314 \times 296.15}{0.334} = 9.628\, MPa \]. We need to reduce this to \(7.60\, MPa\).

Step 6: Determine Final Moles of Helium

Using \(P_{final} = 7.60\, MPa\), find the final moles \(n_{final}\) using \(\frac{nRT}{V} = 7.60\) and solving for \(n\).

Step 7: Calculate Moles to Release

Set up the equation \[ n_{final} = \frac{7.60 \times 0.334}{8.314 \times 296.15} \]. Compare \(n_{final}\) to \(n_{initial}\) to find the moles of He to release: \[ \Delta n = n_{initial} - n_{final} \].

Step 8: Convert Released Moles to Grams

Calculate \(\Delta n \) and then multiply by helium's molar mass to find grams: \[ \Delta mass = \Delta n \times 4.00 \].

Step 9: Solution Verification

Plug values to verify steps: \[ n_{final} = 1.03175\, \text{mol}\], \[ \Delta n = 1.30625 - 1.03175 = 0.2745\, \text{mol} \], \[ \Delta mass = 0.2745 \times 4.00 = 1.098\, \text{g} \]. Thus, \(1.098\,\text{g}\) of helium must be released.

Key Concepts

Helium Properties

Helium is a noble gas, which means it is chemically inert and does not easily react with other elements. This makes it highly valuable in various applications, from cooling superconductor magnets in MRI machines to providing a controlled atmosphere in deep-sea diving to prevent decompression sickness. Helium's atomic number is 2, and it has an atomic weight of approximately 4.00 g/mol, indicating that it is quite light. In the context of the ideal gas law, helium behaves as an ideal gas under a wide range of conditions due to its simple atomic structure and low intermolecular forces, resulting in negligible deviations from ideal behavior. Consequently, helium is often used in classrooms to teach concepts of gas laws, enabling students to explore how changes in pressure, temperature, and volume interact.

Unit Conversion in Chemistry

In chemistry, unit conversion is essential for ensuring that measurements are consistent and calculations are accurate. For instance, in this problem, we converted the volume from milliliters (mL) to liters (L) because the gas constant \( R \) in the ideal gas equation often has a unit of liter-atmospheres per mole-Kelvin (J/mol·K in this case).

• Volume: To go from mL to L, divide by 1000: \( 334 \, \text{mL} = 0.334 \, \text{L} \).
• Temperature: Convert Celsius to Kelvin because the Kelvin scale is absolute and starts at zero, which is essential for calculations involving temperature: \( T = 23 + 273.15 = 296.15 \, \text{K} \).

Using Kelvin ensures that the relationships between properties like volume and pressure are accurate. Additionally, understanding how to properly convert different units enhances accuracy in predictions and experiments, forming a foundational skill for advanced studies in chemistry.

Pressure Calculation

Pressure is a measure of force applied uniformly over a surface and is vital in determining how gases interact in a closed environment. Using the ideal gas law \( PV = nRT \), we can compute the pressure a gas exerts in a given system. Here, we started with calculating the initial pressure of helium in the cylinder. The initial moles of helium were known (\( 1.30625 \, \text{mol} \)), and the temperature was converted to Kelvin. By rearranging the ideal gas equation to solve for pressure \( P = \frac{nRT}{V} \), we found the initial pressure to be \( 9.628 \, \text{MPa} \).To solve the exercise:

• Find the initial pressure using given moles, \( R \), and temperature values.
• Determine how much helium remains at the desired lower pressure \( 7.60 \, \text{MPa} \).
• Calculate the moles and translate into grams to find out how much helium gas should be released.
• The necessary reduction in moles was calculated, leading to the eventual mass of helium to be removed: \( 1.098 \, \text{g} \).

This example illustrates how you can control gas conditions in closed systems by adjusting variables like temperature or amounts of gas, which is crucial in processes like chemical manufacturing and space explorations.