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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 \(\mathrm{mol}\) of \(\mathrm{CO}_{2} \mathrm{gas}\) 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 ideat-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 ideat-gas equation if \(n / V\) is small.

Step by step solution

01

Convert Units

Convert volumes from cm鲁 to m鲁 since the constants for the van der Waals equation are given in m鲁. For part (a), 400 cm鲁 is converted to m鲁 by dividing by 1,000,000: V = 400 cm鲁 = 400 脳 10鈦烩伓 m鲁 = 4 脳 10鈦烩伌 m鲁. For part (c), 4000 cm鲁 is converted similarly: V = 4000 cm鲁 = 4000 脳 10鈦烩伓 m鲁 = 4 脳 10鈦宦� m鲁.

02

Calculate Pressure Using Ideal Gas Equation

Use the ideal gas equation, \[ PV = nRT \] where R = 8.314 J/mol路K, n = 1 mol, and T = 350 K. For part (a) with V = 4 脳 10鈦烩伌 m鲁: \[ P = \frac{nRT}{V} = \frac{1 \, \text{mol} \times 8.314 \, \text{J/mol路K} \times 350 \, \text{K}}{4 \times 10^{-4} \, \text{m}^3} = 7.272 \times 10^{6} \, \text{Pa} \]For part (c) with V = 4 脳 10鈦宦� m鲁: \[ P = \frac{1 \, \text{mol} \times 8.314 \, \text{J/mol路K} \times 350 \text{K}}{4 \times 10^{-3} \, \text{m}^3} = 7.272 \times 10^{5} \, \text{Pa} \]

03

Calculate Pressure Using van der Waals Equation

The van der Waals equation is: \[ \left( P + \frac{a n^2}{V^2} \right) (V - nb) = nRT \]For part (a), substitute known values: \[ P + \frac{0.364 \times (1)^2}{(4 \times 10^{-4})^2} = \frac{1 \times 8.314 \times 350}{4 \times 10^{-4} - (1 \times 4.27 \times 10^{-5})} \]Solve for P: \[ P \approx 9.275 \times 10^{6}\, \text{Pa} \]For part (c), use V = 4 脳 10鈦宦� m鲁: \[ P + \frac{0.364 \times (1)^2}{(4 \times 10^{-3})^2} = \frac{1 \times 8.314 \times 350}{4 \times 10^{-3} - (1 \times 4.27 \times 10^{-5})} \]Solve for P: \[ P \approx 7.327 \times 10^{5} \, \text{Pa} \]

04

Compare Pressures and Calculate Percentage Difference

For part (a), compare ideal gas pressure (\(7.272 \times 10^{6}\) Pa) with van der Waals pressure (\(9.275 \times 10^{6}\) Pa). Since the van der Waals pressure is higher, calculate percentage difference: \[ \text{Percentage difference} = \frac{|7.272 - 9.275|}{7.272} \times 100\% \approx 27.57\% \]For part (c), compare ideal gas pressure (\(7.272 \times 10^{5}\) Pa) with van der Waals pressure (\(7.327 \times 10^{5}\) Pa): \[ \text{Percentage difference} = \frac{|7.272 - 7.327|}{7.272} \times 100\% \approx 0.76\% \]

05

Analyze Results for Equivalent Behavior

In part (d), at higher volumes (lower \(\frac{n}{V}\)), the deviation between the van der Waals and ideal gas pressures is smaller. This is evident in part (c), where the percentage difference decreases to about 0.76%. This shows that as \(\frac{n}{V}\) becomes smaller, the influence of the correction terms in the van der Waals equation diminishes, and it approaches ideal gas behavior.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law

The ideal gas law is a simple equation that describes the relationship between pressure, volume, and temperature for an ideal gas. It is expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas in pascals (Pa),
• \( V \) is the volume of the gas in cubic meters (m鲁),
• \( n \) is the amount of gas in moles (mol),
• \( R \) is the universal gas constant, 8.314 J/mol路K,
• \( T \) is the temperature in kelvins (K).

This equation assumes that gas molecules do not interact with each other and occupy no space. It works well under conditions of high temperature and low pressure, where intermolecular forces are negligible. For carbon dioxide gas at 350 K in this exercise, the ideal gas law helps us calculate the pressure efficiently in its ideal state before considering real gas behavior.

Real Gases

Real gases deviate from ideal behavior due to interactions between gas molecules and their finite volumes. At lower temperatures and higher pressures, these effects become significant. The van der Waals equation helps adjust for these deviations. This equation includes two correction factors to account for:

• The volume occupied by gas molecules, and
• The attractive forces between them.

These corrections are represented in the van der Waals equation: \[\left( P + \frac{a n^2}{V^2} \right) (V - nb) = nRT\]where \( a \) is the measure of attraction between particles, and \( b \) is the volume occupied by a mole of gas molecules. For carbon dioxide, these corrections significantly impact calculated pressures, especially when the gas is confined to smaller volumes.

Molar Volume

Molar volume is the volume occupied by one mole of a substance and is determined under specific conditions of temperature and pressure. For ideal gases, it can be computed using the ideal gas law. Simply put, molar volume shows how much space one mole of gas occupies.
In real gases, due to molecular interactions and finite space they occupy, the effective molar volume differs from that predicted by the ideal gas law. The van der Waals equation accounts for these differences. As the equation includes volume adjustments with \( b \), it helps us understand how molar volume variations affect gas properties. With molar volume data, we can derive the deviating behavior of gases under various conditions and analyze the impact of confinement on gas expansion.

Pressure Calculation

Pressure calculation is vital in both the ideal and real gas scenarios. For gases like carbon dioxide, pressure is significantly influenced by volume and temperature. Using the ideal gas law, the calculation simplifies to \[P = \frac{nRT}{V},\]assuming ideal conditions. However, real gases require more complex calculations, as indicated by the van der Waals equation. Here, pressure is determined by substituting:

• Compressibility factor additions from molecular attractions, and
• Volume adjustments from molecular size.

In exercises, calculating the pressure for different volumes reveals the equation's adjustments. A lower real pressure compared to an ideal one indicates strong molecular attractions, while higher pressures suggest significant volume exclusion. These calculations highlight the precision needed to measure differences in pressures, showing why the van der Waals equation provides a deeper insight into real gas behavior.