Problem 12 - University Physics with Modern Physics

Question

Deviations from the Ideal-Gas Equation. For carbon dioxide gas \(\left(\mathrm{CO}_{2}\right),\) the constants in the van der Waals equation are \(a=0.364 \mathrm{J} \cdot \mathrm{m}^{3} / \mathrm{mol}^{2}\) and \(b=4.27 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} .\) (a) If 1.00 mol of \(\mathrm{CO}_{2}\) gas at 350 \(\mathrm{K}\) is confined to a volume of \(400 \mathrm{cm}^{3},\) find the pressure of the gas using the ideal-gas equation and the van der Waals equation. (b) Which equation gives a lower pressure? Why? What is the percentage difference of the van der Waals equation result from the ideal-gas equation result? (c) The gas is kept at the same temperature as it expands to a volume of 4000 \(\mathrm{cm}^{3} .\) Repeat the calculations of parts (a) and (b). (d) Explain how your calculations show that the van der Waals equation is equivalent to the ideal-gas equation if \(n / V\) is small.

Step-by-Step Solution

Answer

The van der Waals equation gives a higher pressure. For case 1, the difference is 10.08%; for case 2, it's 0.19%. Larger volume shows van der Waals approximating ideal gas behavior.

Step 1: Convert Volume to Cubic Meters

The volume is given in cubic centimeters (cm^3), but the van der Waals equation requires volume in cubic meters (m^3). Convert the volumes for both 400 cm^3 and 4000 cm^3 cases.For 400 cm^3:\[ V = 400 \, \mathrm{cm}^3 = 400 \, \times 10^{-6} \, \mathrm{m}^3 = 0.000400 \, \mathrm{m}^3 \]For 4000 cm^3:\[ V = 4000 \, \mathrm{cm}^3 = 4000 \, \times 10^{-6} \, \mathrm{m}^3 = 0.004 \, \mathrm{m}^3 \]

Step 2: Solve Using the Ideal Gas Equation

Use the ideal gas equation to find the pressure for both volumes.Ideal gas equation: \( PV = nRT \).For both cases, we have:- \( n = 1.00 \) mol,- \( R = 8.314 \, \mathrm{J} / (\mathrm{mol} \cdot \mathrm{K}) \),- \( T = 350 \, \mathrm{K} \).### Case 1: Volume = 0.000400 m^3\[ P = \frac{nRT}{V} = \frac{(1.00)(8.314)(350)}{0.000400} = 7276875 \, \mathrm{Pa} \]### Case 2: Volume = 0.004 m^3\[ P = \frac{nRT}{V} = \frac{(1.00)(8.314)(350)}{0.004} = 727687.5 \, \mathrm{Pa} \]

Step 3: Solve Using the van der Waals Equation

Use the van der Waals equation: \[ \left( P + \frac{an^2}{V^2} \right)(V - nb) = nRT \].Constants:- \( a = 0.364 \, \mathrm{J} \cdot \mathrm{m}^3 / \mathrm{mol}^2 \),- \( b = 4.27 \times 10^{-5} \, \mathrm{m}^3 / \mathrm{mol} \).### Case 1: Volume = 0.000400 m^3\[ P = \frac{nRT}{V - nb} - \frac{an^2}{V^2} \]Substitute the values:\[ P = \frac{(1)(8.314)(350)}{0.000400 - (1)(4.27 \times 10^{-5})} - \frac{(0.364)(1)^2}{(0.000400)^2} \approx 8011356.85 \, \mathrm{Pa} \]### Case 2: Volume = 0.004 m^3\[ P = \frac{(1)(8.314)(350)}{0.004 - (1)(4.27 \times 10^{-5})} - \frac{(0.364)(1)^2}{(0.004)^2} \approx 729083.97 \, \mathrm{Pa} \]

Step 4: Compare Pressures and Calculate Percentage Difference

### Calculate the Result for Case 1- Ideal Gas Pressure: \( 7276875 \, \mathrm{Pa} \)- van der Waals Pressure: \( 8011356.85 \, \mathrm{Pa} \)Percentage Difference:\[ \text{Difference} = \left( \frac{8011356.85 - 7276875}{7276875} \right) \times 100 \% \approx 10.08 \% \]### Calculate the Result for Case 2- Ideal Gas Pressure: \( 727687.5 \, \mathrm{Pa} \)- van der Waals Pressure: \( 729083.97 \, \mathrm{Pa} \)Percentage Difference:\[ \text{Difference} = \left( \frac{729083.97 - 727687.5}{727687.5} \right) \times 100 \% \approx 0.19 \% \]In both cases, the van der Waals equation gives a higher pressure than the ideal gas equation.

Step 5: Analyze n/V for Large Volume

When the volume is larger, the term \( nb \) becomes negligibly small compared to \( V \), simplifying the van der Waals equation to closely resemble the ideal gas equation.In the second case with larger volume (0.004 m^3), the pressure difference is relatively small (0.19%), showing that for lower \( n/V \), both equations give similar results.

Key Concepts

ideal gas law

The ideal gas law is a fundamental equation in chemistry and physics. It relates the pressure, volume, temperature, and amount (in moles) of an ideal gas. The formula is given by:

• \[ PV = nRT \]
• where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (8.314 J/mol·K), and \(T\) is the temperature in Kelvin.

This equation assumes that gas particles have no volume and do not interact with each other, scenarios that often do not hold true in reality for real gases. However, it gives a relatively good estimate under many conditions. For example, CO2 gas at standard conditions can sometimes deviate from these assumptions significantly. Thus, while useful, the ideal gas law might not give entirely accurate results for gases deviating due to intermolecular forces and finite particle volume.

carbon dioxide gas

Carbon dioxide, often referred to by its chemical formula \( ext{CO}_2 \), is a colorless and odorless gas that plays a significant role in our planet's ecological and chemical processes. As it deviates from ideal behavior at certain conditions, more complex equations like the van der Waals equation are used to model its behavior. This is primarily because:

• \( ext{CO}_2 \) has stronger intermolecular forces than an ideal gas due to its polar nature.
• Its molecules occupy a significant volume compared to the gases assumed in the ideal gas law.

Understanding the behavior of carbon dioxide under varying conditions is essential in environments where pressure and temperature differ from standard conditions, such as in certain industrial processes or atmospheric studies.

pressure calculations

Calculating pressure for gases involves understanding both ideal and non-ideal behaviors. Using various gas equations helps determine the actual pressure.

• The ideal gas law can give initial estimates, especially at higher volumes and lower pressures.
• The van der Waals equation takes into account the attraction between molecules (using constant \(a\)) and the volume occupied by gas molecules themselves (using constant \(b\)). This leads to corrected pressure calculations when compared to lower pressures guessed by the ideal gas law.

In the provided exercise, for instance,

• with a smaller volume, the van der Waals calculation showed higher pressure comparing to ideal gas law, indicating significance of molecular interactions and volume.
• With a larger volume, pressure calculated by both methods was more similar because the volume and molecular interaction corrections become less significant.

volume conversions

Understanding measurements in the correct units is crucial for accurate calculations. The van der Waals equation requires volumes in cubic meters (m³) instead of the typical laboratory unit of cubic centimeters (cm³).

• The conversion process is simple: since 1 meter equals 100 centimeters, 1 cubic meter equals 1,000,000 cubic centimeters.
• This means that to convert, you multiply the volume in cm³ by \(10^{-6}\) to get m³, as evident in the original exercise where 400 cm³ was converted to 0.000400 m³.

Understanding these conversions is important for applying the equations correctly and obtaining accurate results. It also helps ensure consistency in technical and laboratory work, where unit consistency is key to ensuring safety and precision.