Question

A 3.50 mole sample of \(\mathrm{N}_{2}\) in a state defined by \(T_{i}=\) 250\. \(\mathrm{K}\) and \(V_{i}=3.25 \mathrm{L}\) undergoes an isothermal reversible expansion until \(V_{f}=35.5 \mathrm{L}\) Calculate \(w,\) assuming (a) that the gas is described by the ideal gas law, and (b) that the gas is described by the van der Waals equation of state. What is the percent error in using the ideal gas law instead of the van der Waals equation? The van der Waals parameters for \(\mathrm{N}_{2}\) are listed in Table 7.4

Short answer

The work done during the isothermal reversible expansion for an ideal gas is approximately -204.27 atm L, and for a van der Waals gas, it's approximately -206.82 atm L. The percent error in using the ideal gas law instead of the van der Waals equation in this case is approximately 1.23%.

Step-by-step solution

Calculate work done for ideal gas

Recall that the work done due to a reversible isothermal process on an ideal gas can be calculated using the formula: \[ w = -nRT \ln \frac{V_f}{V_i} \] Where: \(n\) is the number of moles of gas, \(R\) is the gas constant, \(T\) is the constant temperature of the process, and \(V_i\) and \(V_f\) are the initial and final volumes of the gas, respectively. We already have all values needed. So, we just need to substitute and find the work done.

Calculate work done for van der Waals gas

For a reversible, isothermal process on a van der Waals gas, the work done is given by: \[ w = -nRT \ln \frac{(V_f - nb)}{(V_i - nb)} \] Where: \(b\) is the van der Waals constant for the gas (specific to each gas) From Table 7.4, we have the values of van der Waals constants \(a \) and \(b\) for nitrogen gas.

Compare the two results and find the percent error

Once we have calculated the work done in both cases, we can find the percent error in using the ideal gas law instead of the van der Waals equation using the formula: \[ \text{Percent error} = \frac{\left| w_{\text{ideal}} - w_{\text{van der Waals}}\right|}{w_{\text{van der Waals}}} \times 100\% \] Now, let's put all the given values and solve.

Substitute the given values

Given: \(n = 3.50 \text{ moles}\), \(T_i = 250 \text{K}\), \(V_i = 3.25 \text{L}\), \(V_f = 35.5 \text{L}\), \(R = 0.0821 \frac{\text{L atm}}{\text{K mol}}\) (gas constant), \(a = 1.390 \frac{\text{L}^2 \text{ atm}}{\text{mol}^2}\) (van der Waals constant for nitrogen gas), and \(b = 0.0391 \text{L/mol}\) (van der Waals constant for nitrogen gas). Let's substitute these values to find the work done in both cases: For ideal gas: \[ w_{\text{ideal}} = -nRT \ln \frac{V_f}{V_i} = -(3.50 \times 0.0821 \times 250) \ln \frac{35.5}{3.25} = -204.27 \text{ atm L} \] Note that work has a negative sign, indicating that it is being done on the system. For van der Waals gas: \[ w_{\text{van der Waals}} = -nRT \ln \frac{(V_f - nb)}{(V_i - nb)} = -(3.50 \times 0.0821 \times 250) \ln \frac{(35.5 - 3.50 \times 0.0391)}{(3.25 - 3.50 \times 0.0391)} = -206.82 \text{ atm L} \] Again, the work has a negative sign, indicating that it is being done on the system. The percent error can now be calculated as: \[ \text{Percent error} = \frac{\left| w_{\text{ideal}} - w_{\text{van der Waals}}\right|}{w_{\text{van der Waals}}} \times 100\% = \frac{\left|-204.27 - (-206.82)\right|}{-206.82} \times 100\% \approx 1.23\% \] So, the percent error in using the ideal gas law instead of the van der Waals equation in this case is approximately 1.23%.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and amount of an ideal gas. The equation is expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant (0.0821 L atm K^-1 mol^-1), and
• \( T \) is the temperature in Kelvin.

The Ideal Gas Law assumes that the gas particles do not interact, except for elastic collisions, and occupy no volume themselves. While this is an approximation, it works well for gases at high temperatures and low pressures. In the context of the exercise, the ideal gas assumption helps simplify the calculation of work during isothermal expansion, providing a straightforward method to compute gas behavior.

van der Waals Equation

The van der Waals equation is a modified version of the Ideal Gas Law that provides a more accurate description of the behavior of real gases. It accounts for the volume occupied by gas particles and the attractive forces between them. The equation is represented as:\[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\]
Here, \( a \) and \( b \) are van der Waals constants unique to each gas, reflecting the strength of intermolecular forces and the volume occupied by gas molecules, respectively. For nitrogen, the values used are \( a = 1.390 \text{ L}^2 \text{ atm/mol}^2 \) and \( b = 0.0391 \text{ L/mol} \).
In the exercise, the van der Waals equation is applied to calculate the work done during the isothermal expansion, providing a more realistic result compared to the ideal gas calculation. This consideration is crucial for gases under conditions where particle interactions cannot be neglected.

Work Calculation

Work in thermodynamics, particularly during an isothermal process, can be calculated differently depending on whether the gas is considered ideal or non-ideal. The formula for work done on an ideal gas during an isothermal, reversible expansion is:\[w = -nRT \ln \frac{V_f}{V_i}\]
For a gas following the van der Waals equation, an adjustment is made to account for molecular volume and interactions:\[w = -nRT \ln \frac{(V_f - nb)}{(V_i - nb)} \]
Where \( V_i \) and \( V_f \) are the initial and final volumes, and \( n \), \( R \), and \( T \) retain their usual meanings. The negative sign indicates that work is done on the gas as it expands.
In our case, we find the work done for nitrogen gas using both methods, with adjustments as needed for van der Waals gases. With these equations, students can clearly see how the additional factors in the van der Waals equation affect the overall calculation.

Percent Error

Percent error provides a measure of how close an approximate measurement is to an exact one, expressed as a percentage. It is particularly useful for comparing theoretical models to real-world results. The formula for calculating percent error is:\[\text{Percent error} = \frac{\left| w_{\text{ideal}} - w_{\text{van der Waals}}\right|}{w_{\text{van der Waals}}} \times 100\%\]
This equation computes the difference between the work calculated using the ideal gas law (\( w_{\text{ideal}} \)) and the van der Waals equation (\( w_{\text{van der Waals}} \)), divided by the value from the van der Waals calculation.
In the given exercise, the percent error determined is about 1.23%. This indicates that the ideal gas law provides a result that is quite close to the more accurate van der Waals equation. However, understanding and calculating this percent error is vital in acknowledging the limitations of simpler models when applied to real gases.