Question

Calculate the pressure exerted by Ar for a molar volume of \(1.31 \mathrm{L} \mathrm{mol}^{-1}\) at \(426 \mathrm{K}\) using the van der Waals equation of state, The van der Waals parameters \(a\) and \(b\) for Ar are 1.355 bar \(\mathrm{dm}^{6} \mathrm{mol}^{-2}\) and \(0.0320 \mathrm{dm}^{3} \mathrm{mol}^{-1},\) respectively. Is the attractive or repulsive portion of the potential dominant under these conditions?

Short answer

The pressure exerted by Argon under the given conditions, calculated using the van der Waals equation of state, is approximately \(26.457\,\text{bar}\). The attractive forces are dominant under these conditions, as the ratio of attractive to repulsive portions is \(2.008\).

Step-by-step solution

Recall the van der Waals equation of state

The van der Waals equation of state for a gas is given by: \[ \left( P + a\left( \frac{n}{V} \right)^2 \right)(V-nb) = nRT \] Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature. The parameters \(a\) and \(b\) are specific to a particular gas and are given in the problem for Argon. Since we are given the molar volume, we can rewrite the van der Waals equation as: \[ \left( P + a\left(\frac{1}{V_m} \right)^2 \right)(V_m - b) = RT \] Where \(V_m\) is the molar volume, which is equal to \( \frac{V}{n} \).

Substitute the given values into the equation

We are given the molar volume \(V_m = 1.31\,\text{L/mol} = 1.31\,\text{dm}^3\text{/mol}\), temperature \(T = 426\,\text{K}\), and the van der Waals parameters \(a = 1.355\,\text{bar}\,\text{dm}^{6}\,\text{mol}^{-2}\) and \(b = 0.0320\,\text{dm}^3\,\text{mol}^{-1}\). We will also use the gas constant \( R = 0.08314 \,\text{bar}\,\text{dm}^3\,\text{mol}^{-1}\text{K}^{-1} \). Now we can substitute these values into the van der Waals equation: \[ \left( P + 1.355\left(\frac{1}{1.31} \right)^2 \right)(1.31 - 0.0320) = (0.08314)(426) \]

Solve for P

Now, we solve the equation for the pressure \(P\): \[ \left( P + 1.355\left(\frac{1}{1.31} \right)^2 \right)(1.31 - 0.0320) = (0.08314)(426) \] Simplifying, we get: \[ \left( P+ 0.7857 \right)(1.278) = 35.296 \] Expanding and solving for \(P\), we find: \[P = \frac{35.296}{1.278} - 0.7857 = 26.457\,\text{bar}\]

Determine if attractive or repulsive forces are dominant

To determine if attractive or repulsive forces are dominant, we first need to compare the terms in the van der Waals equation corresponding to each of these forces: 1. The attractive portion corresponds to \(a\left( \frac{1}{V_m} \right)^2\). 2. The repulsive portion corresponds to \( b \). We can now calculate the ratio of these portions: \[\frac{\text{attractive}}{\text{repulsive}} = \frac{a\left( \frac{1}{V_m} \right)^2}{b} = \frac{1.355\left(\frac{1}{1.31} \right)^2}{0.0320}\] Calculating the ratio, we get: \[\frac{\text{attractive}}{\text{repulsive}} = 2.008 \] Since the ratio is greater than 1, the attractive portion of the potential is dominant under these conditions. In conclusion, the pressure exerted by Argon under the given conditions and using the van der Waals equation of state is approximately \(26.457\,\text{bar}\). The attractive forces are dominant under these conditions, as the ratio of attractive to repulsive portions is \(2.008\).

Key concepts

Gas Constant

The gas constant, often symbolized as \( R \), plays a vital role in gas laws, including the van der Waals equation. It is a universal constant used to relate various properties of gases such as pressure, volume, temperature, and amount (in moles). In gas law calculations, \( R \) allows us to link these properties mathematically in equations.

For the van der Waals equation, the value of \( R \) used depends on the units of pressure, volume, and temperature involved in the problem. It may differ based on the unit systems: for example, \( R = 0.08314 \, \text{bar}\,\text{dm}^3\,\text{mol}^{-1}\text{K}^{-1} \) when working with volumes in \( \text{dm}^3 \) and pressures in bar.

With \( R \), scientists can determine how a gas's behavior deviates from the ideal by accommodating real-world variables that influence gas particles, thus offering a deeper understanding of their characteristics.

Molar Volume

Molar volume \( (V_m) \) is a crucial concept when analyzing gases using the van der Waals equation. It is defined as the volume occupied by one mole of a gas under specified conditions of temperature and pressure. The molar volume gives insight into how much space an individual gas molecule occupies on average.

In the given problem, the molar volume is provided as \( 1.31 \,\text{L/mol} \), which is equivalent to \( 1.31 \,\text{dm}^3/ ext{mol} \). This value is directly substituted into the van der Waals equation to facilitate calculations concerning the behavior of Argon at the specified temperature and pressure.

Understanding molar volume aids in evaluating the density and spacing of particles within a gas, thereby allowing us to predict and manipulate how the gas will behave in various environments.

Pressure Calculation

The calculation of pressure in the context of the van der Waals equation involves adjusting the ideal gas law to account for real gases. This process considers the volume and temperature provided, along with specific parameters unique to each gas ( \( a \) and \( b \)). These parameters adjust for real conditions by incorporating intermolecular forces and the actual volume occupied by gas molecules.

In this exercise, pressure \( P \) is determined using: \[ \left( P + a\left(\frac{1}{V_m} \right)^2 \right)(V_m - b) = RT \]
By substituting the given argon-specific parameters \( a = 1.355 \,\text{bar}\,\text{dm}^6\,\text{mol}^{-2} \) and \( b = 0.0320 \,\text{dm}^3\/\text{mol} \), in addition to the molar volume and temperature values, we solve for \( P \), obtaining a result of approximately \( 26.457 \,\text{bar} \). This calculation underscores the special adjustments necessary when measuring real gas pressures beyond simple ideal gas law assumptions.

Attractive and Repulsive Forces

In the van der Waals equation, real gas behavior is explained by incorporating two correcting factors: attractive and repulsive forces between molecules. These factors adjust the simple ideal gas calculations to reflect the reality of molecular interactions more accurately.

• Attractive forces: These are long-range forces between particles that pull molecules together, influencing gas pressure when molecules are closer together. In the van der Waals equation, these forces are captured in the term \( a\left( \frac{1}{V_m} \right)^2 \), which reduces the calculated pressure.
• Repulsive forces: These come into play at short distances where molecules push against each other. The parameter \( b \), representing the volume occupied by the gas particles themselves, helps model these forces by reducing the effective volume available for motion.

In this problem, by calculating the ratio of attractive to repulsive forces (\( \frac{2.008}{1} \)), it becomes clear that attractive forces are the dominant interaction. This comparison allows better comprehension of how real gases deviate from ideal behavior and guides us in predicting gas properties under various conditions.