Problem 51 - An Introduction to Thermal Physics

Question

When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables, $$t \equiv T / T_{c}, \quad p \equiv P / P_{c}, \quad v \equiv V / V_{c}$$ Rewrite the van der Waals equation in terms of these variables, and notice that the constants \(a\) and \(b\) disappear.

Step-by-Step Solution

Answer

The van der Waals equation in reduced variables is \( (3p + \frac{1}{v^2})(3v - 1) = 8t \).

Step 1: Van der Waals Equation

The van der Waals equation for a real gas is given by the formula: \[\left( P + \frac{a}{V^2} \right)(V - b) = RT\] where \(P\) is the pressure, \(V\) is the volume, \(T\) is the temperature, and \(R\) is the ideal gas constant. The parameters \(a\) and \(b\) account for the interactions between gas molecules.

Step 2: Substitute Reduced Variables

Substitute the reduced variables into the van der Waals equation: \[ p = \frac{P}{P_c}, \quad t = \frac{T}{T_c}, \quad v = \frac{V}{V_c} \]. This leads to \[ P = p \cdot P_c, \quad T = t \cdot T_c, \quad V = v \cdot V_c \].

Step 3: Replace in the Equation

Substitute \(P\), \(V\), and \(T\) in the van der Waals equation:\[\left( pP_c + \frac{a}{(vV_c)^2} \right)(vV_c - b) = R(tT_c)\] Simplifying the terms, we obtain:\[\left( pP_c + \frac{a}{v^2 V_c^2} \right)(vV_c - b) = RtT_c\].

Step 4: Factor and Simplify

Realize that the critical point conditions satisfy:\[P_c = \frac{a}{27b^2}, \ V_c = 3b, \ T_c = \frac{8a}{27Rb} \]. Substitute these critical conditions to simplify terms so the constants \(a\) and \(b\) cancel out.

Step 5: Final Reduced Form

After simplification, the van der Waals equation in terms of reduced variables is:\[\left( 3p + \frac{1}{v^2} \right)(3v - 1) = 8t\]. This form no longer includes the constants \(a\) and \(b\).

Key Concepts

Van der Waals Equation

In thermodynamics, the Van der Waals equation modifies the Ideal Gas Law to better describe the behavior of real gases. It is expressed as:

• \( \left( P + \frac{a}{V^2} \right)(V - b) = RT \)

In this equation, \( P \) stands for pressure, \( V \) for volume, and \( T \) for temperature. The constant \( R \) is the ideal gas constant. What sets the Van der Waals equation apart are the parameters \( a \) and \( b \), which are unique to each gas and account for:

• Molecular interactions (\( a \))
• Molecular volume (\( b \))

The modifications introduced by these parameters help to explain deviations from ideal behavior observed in real gases at high pressures and low temperatures.
In the context of the Van der Waals equation, it is crucial to understand how these constants influence the state of a gas, making it possible to predict its behavior more accurately than with the Ideal Gas Law alone. By re-expressing it in terms of reduced variables, the equation simplifies, and these constants disappear, which we'll explore further.

Reduced Variables

When working with gases, using reduced variables is a method that simplifies calculations. Reduced variables are dimensionless quantities expressed as:

• \( t \equiv \frac{T}{T_{c}} \) for temperature
• \( p \equiv \frac{P}{P_{c}} \) for pressure
• \( v \equiv \frac{V}{V_{c}} \) for volume

In these formulas, \( T_c \), \( P_c \), and \( V_c \) are critical temperature, pressure, and volume, respectively. These represent the specific point at which distinct gas and liquid phases cease to exist. By reducing the Van der Waals equation using these values, the equation becomes dimensionless and independent of the gas-specific parameters \( a \) and \( b \), simplifying comparisons between different gases.
The advantage of working with reduced variables is that they create a universal framework applicable to all gases. It reduces complexity in calculations regarding different gases, especially near critical conditions, by scaling them in relation to their critical states.

Critical Point

The critical point of a substance is a unique state where the gas and liquid phase of a substance become indistinguishable from each other. At this point, critical temperature (\( T_c \)), critical pressure (\( P_c \)), and critical volume (\( V_c \)) are defined. For instance, in the Van der Waals context:

• The critical pressure \( P_c = \frac{a}{27b^2} \)
• The critical volume \( V_c = 3b \)
• The critical temperature \( T_c = \frac{8a}{27Rb} \)

These critical properties are significant because they serve as reference points for reducing variables and normalizing the behavior of different gases.
Understanding the critical point allows scientists to predict at what conditions a substance will exhibit specific phase behavior. This is crucial for various applications, such as designing equipment for industries where precise conditions are needed for processes involving gases.