Question

The work of compression. One mole of a van der Waals gas is compressed quasi- statically and isothermally from volume \(V_{1}\) to \(V_{2}\). For a van der Waals gas, the pressure is $$ p=\frac{R T}{V-b}-\frac{a}{V^{2}}, $$ where \(a\) and \(b\) are material constants, \(V\) is the volume, and \(R T\) is the gas constant \(\times\) temperature. (a) Write the expression for the work done. (b) Is more or less work required in the low-density limit than for an ideal gas? What about the highdensity limit? Why?

Short answer

Work done: \( W = -RT \ln \left( \frac{V_2 - b}{V_1 - b} \right) + a \left( \frac{1}{V_1} - \frac{1}{V_2} \right) \). More work is required than ideal gas at low and high densities due to the \(a\) and \(b\) terms.

Step-by-step solution

Understand the Problem

One mole of a van der Waals gas is compressed isothermally from volume \(V_1\) to \(V_2\). We need to find the work done during this compression process and compare it to the work done in the low-density and high-density limits.

Recall the Work Done Formula

For isothermal compression, the work done on the gas is given by the integral: \[W = -\int_{V_1}^{V_2} p dV\]

Substitute Van der Waals Equation

Substitute the given van der Waals equation for pressure into the work formula: \[ W = -\int_{V_1}^{V_2} \left( \frac{RT}{V-b} - \frac{a}{V^2} \right) dV \]

Separate the Integral

Separate the integral into two parts for easier integration: \[ W = -RT \int_{V_1}^{V_2} \frac{1}{V-b} dV + a \int_{V_1}^{V_2} \frac{1}{V^2} dV \]

Evaluate the Integrals

Solve each integral:For the first integral: \[ \int_{V_1}^{V_2} \frac{1}{V-b} dV = \ln \left(\frac{V_2 - b}{V_1 - b}\right) \]For the second integral: \[ \int_{V_1}^{V_2} \frac{1}{V^2} dV = -\frac{1}{V} \bigg|_{V_1}^{V_2} = \left( \frac{1}{V_1} - \frac{1}{V_2} \right) \]

Combine Results

Combine the results of the integrals: \[ W = -RT \ln \left( \frac{V_2 - b}{V_1 - b} \right) + a \left( \frac{1}{V_1} - \frac{1}{V_2} \right) \]

Analyze Low-Density Limit

In the low-density limit, \(V \gg b\) and \(\frac{V-b}{V} \approx 1\), so the work done is similar to that for an ideal gas (ideal gas formula: \(-RT \ln \frac{V_2}{V_1}\)). Hence, more work is required compared to the ideal gas due to the \(a\) term.

Analyze High-Density Limit

In the high-density limit, \(V \approx b\) or \(V < b\), the work done diverges significantly from the ideal gas behavior because the repulsive force dominates. Thus, significantly more work is required in this limit.

Key concepts

Van der Waals Gas

To appreciate the work of compression in a van der Waals gas, it's crucial to understand what a van der Waals gas is. The van der Waals equation is an adjustment of the ideal gas law that accounts for the real behavior of gases. The equation is given by:

\[ p = \frac{RT}{V - b} - \frac{a}{V^2} \]

Here:

• a: represents the magnitude of intermolecular attraction.
• b: accounts for the volume occupied by gas molecules.
• R: is the universal gas constant, and V is the volume.
• The term \(\frac{RT}{V - b}\) accounts for the modified pressure considering the finite size of molecules.
• The term \(\frac{a}{V^2}\) addresses the intermolecular forces.

These modifications bridge the gap between ideal and real gas behaviors, making the van der Waals equation more accurate for real gases, especially under high pressure and low temperature conditions.

Isothermal Compression

Isothermal compression refers to compressing a gas while keeping its temperature constant. For an ideal gas, the work done during isothermal compression from volume \(V_1\) to \(V_2\) is computed using the formula:

\[ W = -RT \ln \frac{V_2}{V_1} \]

However, when dealing with a van der Waals gas, the equation becomes a bit more complex due to the adjustments in pressure from the van der Waals equation. We use the pressure form:

\[ p = \frac{RT}{V - b} - \frac{a}{V^2} \]

The work done during isothermal compression for a van der Waals gas is:

\[ W = -\int_{V_1}^{ V_2} p \, dV \]

Which breaks down into:

\[ W = -RT \int_{V_1}^{ V_2} \frac{1}{V - b} \, dV + a \int_{V_1}^{ V_2} \frac{1}{V^2} \, dV \]

By solving these integrals, we find:

\[ W = -RT \ln \left( \frac{V_2 - b}{V_1 - b} \right) + a \left( \frac{1}{V_1} - \frac{1}{V_2} \right) \]

This formula shows the extra work due to molecular interactions (\(a\) term and molecular size \(b\)) compared to an ideal gas.

Thermodynamic Work

Thermodynamic work involves the energy transfer that occurs when a force moves an object. For gases, it often involves changing volume against external pressure. When we talk about isothermal compression in a van der Waals gas, we're calculating the work done to reduce the gas’s volume while maintaining a constant temperature.

For ideal gases, this is straightforward, but for a van der Waals gas, we must consider additional interactions:

• At low densities, the volume \(V\) is very large compared to \(b\), and the molecules' volume is negligible. Thus, the behavior of the gas approaches that of an ideal gas.
• At high densities, the volume \(V\) is close to \(b\). Here, the finite size of the molecules significantly affects the behavior, requiring more work due to repulsive forces.
• The attractive forces represented by \(a\) decrease the pressure exerted by the gas molecules, hence less work is needed compared to an ideal gas.

In the low-density limit, fewer intermolecular forces are at play, resembling ideal gas compression. However, in the high-density limit, intermolecular forces and molecular sizes play a significant role, diverging from ideal gas behavior and requiring more work.