Question

For the Berthelot equation, \(V_{m}=(R T / P)+b-\) \(\left(a / R T^{2}\right),\) find an expression for the Boyle temperature in terms of \(a, b,\) and \(R\).

Short answer

The Boyle temperature (\(T_B\)) for the Berthelot equation can be determined by minimizing the difference between the Ideal and Berthelot equations. The expression for the Boyle temperature in terms of \(a\), \(b\), and \(R\) is given by: \[T_B = \sqrt[3]{\frac{2aP}{R}}.\]

Step-by-step solution

Replace the Ideal Gas Law with the Berthelot Equation

We can rewrite the Berthelot equation in terms of the PV product by multiplying both sides by P: \[PV_m = (RT) + Pb - \left(\frac{aP}{RT^2}\right).\] Now, we want to find the temperature when deviations from the ideal gas law are minimal. This means we want to find the temperature at which the difference between the ideal gas law and the Berthelot equation is as small as possible.

Find the Difference Between the Ideal and Berthelot Equations

The difference between the Ideal and Berthelot equations is given by: \[D = \left( RT + Pb - \frac{aP}{RT^2} \right) - RT.\] We can simplify this expression: \[D = Pb - \frac{aP}{RT^2}.\]

Minimize the Difference

To find the minimum difference, we can take the derivative of the expression for D with respect to the temperature T and set it equal to zero: \[\frac{dD}{dT} = \frac{d}{dT}\left( Pb - \frac{aP}{RT^2} \right) = 0.\] Now we can take the derivative: \[\frac{dD}{dT} = -\frac{2aP}{RT^3}.\] Setting this equal to zero, we find that: \[-\frac{2aP}{RT^3} = 0.\]

Solve for the Boyle Temperature

Since the other factors (2, a, and P) are nonzero, we can determine that \(T^3 = 0\) when \(\frac{dD}{dT} = 0\). Therefore, the Boyle temperature (\(T_B\)) is the cubic root of the RHS of the equation: \[T_B^3 = \frac{2aP}{R}.\] Finally, we can solve for the Boyle temperature in terms of a, b, and R: \[T_B = \sqrt[3]{\frac{2aP}{R}}.\]

Key concepts

Boyle Temperature

Boyle temperature is a special temperature at which a real gas exhibits behavior closest to that of an ideal gas. This is because certain deviations, which are usually present due to molecular interactions, get minimized here.
Imagine a real gas behaving almost perfectly like an ideal gas at this temperature. At the Boyle temperature, the pressure and the volume of the gas align mostly with predictions from the ideal gas equation. For example, using advanced equations like the Berthelot equation, we calculate the Boyle temperature by locating where deviations (or differences) are minimized. For the Berthelot equation, it involves setting certain derivatives to zero and solving for the temperature. In this case, the Boyle temperature is solved by finding when the derivative of deviation with respect to temperature equals zero, ensuring minimal difference between real and ideal gas behaviors.

Deviations from Ideal Gas

Real gases often show deviations from the ideal gas law predictions. These deviations occur due to interactions between gas molecules and the finite volume that molecules have.
When a gas aligns perfectly with the ideal gas law, it implies no intermolecular forces and maximum distance between molecules. However, in reality, molecules attract or repel each other and occupy space.To account for these deviations, scientists use modified equations of state, like the Berthelot equation.

• The term \(Pb\) in the Berthelot equation accounts for the finite volume of gas molecules.
• The term \(-\frac{a}{RT^2}\) deals with the attraction between molecules.

By quantifying these deviations, chemists can predict the behavior of gases more accurately across different pressures and temperatures.

Derivative with Respect to Temperature

Derivatives help us understand how things change, such as how a gas' behavior shifts with varying temperature. By taking the derivative of a deviation expression with respect to temperature, we can determine which temperature makes the difference between the real and ideal behavior the smallest.
This approach explains why certain conditions cause a real gas to behave almost like an ideal gas. The goal is to determine temperature points where changes (or slopes) become flat, which means the system is closest to ideal circumstances. In the Berthelot equation solution, the derivation step revealed that the derivative \(-\frac{2aP}{RT^3}\) set to zero finds this turning point where gas behavior is idealized.

Ideal Gas Law vs. Real Gas

The ideal gas law is formulated under the assumption that gas molecules do not interact and have negligible volume. Described by \(PV = nRT\), it approximates the behavior of gases under most conditions but falls short with high pressures or low temperatures.
In contrast, real gases account for molecular size and interactions. Advanced equations like the Berthelot equation incorporate terms that amend these shortcomings.

• The pressure isn't just from collisions, but also considers attractive forces.
• The volume takes into account the space occupied by the molecules themselves.

This duality is crucial for accurately predicting and manipulating the behavior of gases in practical applications, showing why the modified equations of state are essential in real-world chemistry and physics.