Question

An ideal gas is expanding such that \(P T^{2}=\) constant. The coefficient of volume expansion of the gas is (A) \(\frac{1}{T}\) (B) \(\frac{2}{T}\) (C) \(\frac{3}{T}\) (D) \(\frac{4}{T}\)

Short answer

\(\frac{2}{T}\)

Step-by-step solution

Understanding the Coefficient of Volume Expansion

The coefficient of volume expansion, denoted by \(\beta\), is defined for an ideal gas by the relationship \(\beta = \frac{1}{V} \left(\frac{\partial V}{\partial T}\right)_P\), where \(V\) is the volume, \(T\) is the temperature, and the partial derivative is taken at constant pressure.

Relate the Volume to Pressure and Temperature

Given the relation \(P T^{2} =\) constant, we rearrange to express volume as a function of temperature at constant pressure: \(PV = nRT\) (from the ideal gas law). Substituting \(P\) from the relation, we get \(V = \frac{nRT}{P}\). We find \(V\) as a function of \(T\) while keeping \(P\) constant.

Differentiating Volume with respect to Temperature

To find \(\left(\frac{\partial V}{\partial T}\right)_P\), differentiate the volume function with respect to temperature: \(V = \frac{nRT}{P}\) where \(P\) is considered constant and \(P\) is a function of \(T\) itself due to the condition \(PT^2 = \text{constant}\). We get \(\left(\frac{\partial V}{\partial T}\right)_P = \frac{nR}{P} + \frac{(-2nRT)}{PT^2}\).

Simplify the Expression

Solve the derivative to simplify the expression: \(\left(\frac{\partial V}{\partial T}\right)_P = \frac{nR}{PT^{-2}} - \frac{2nR}{P}\) which simplifies to \(\left(\frac{\partial V}{\partial T}\right)_P = nRT^{-3} - 2nRT^{-1}\). Since \(nR\) and \(P\) are constants, the expression further simplifies to a function of temperature only.

Calculating the Coefficient of Volume Expansion

To find the coefficient \(\beta\), we substitute \(\left(\frac{\partial V}{\partial T}\right)_P\) back into the definition, yielding \(\beta = \frac{1}{V} (nRT^{-3} - 2nRT^{-1})\). Since \(V\) is proportional to \(nRT^{-1}\), we can replace \(V\) with \(nRT^{-1}\) to get \(\beta = \frac{1}{nRT^{-1}} (nRT^{-3} - 2nRT^{-1}) = T^{-2} - 2 = -2T^{-2}\).

Interpreting the Result and Matching the Options

The coefficient \(\beta\) is \(-2T^{-2}\). However, coefficients are defined to be positive, and the temperature is always positive for an ideal gas, which means we need to take the absolute value. This gives us \(\beta = 2T^{-2}\) or \(\beta = \frac{2}{T}\), which corresponds to option (B).

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is a cornerstone in the study of thermodynamics and is pivotal to solving a range of problems involving gases. This law is usually stated as PV = nRT, where P stands for pressure, V represents volume, n is the amount of substance in moles, R is the ideal, or universal, gas constant, and T is the absolute temperature in Kelvin.

The beauty of this equation is that it gives a simple relationship between the four key properties of an ideal gas where intermolecular forces and volume taken up by the gas molecules are negligible. This allows us to predict changes to a gas when subjected to alterations in temperature, pressure, or volume. Exercise Improvement Tip: It is essential to emphasize that the Ideal Gas Law applies to ideal scenarios and real gases might exhibit deviations, especially at high pressures or low temperatures, due to intermolecular forces.

Thermodynamic Processes

• Constant Pressure Process (Isobaric)
• Constant Volume Process (Isochoric)
• Constant Temperature Process (Isothermal)
• Adiabatic Process (No heat exchange)

When discussing thermodynamic processes concerning gases, we often consider the gas laws and principles, such as the one mentioned in the exercise: P T^{2} = constant. This points towards an isobaric process, where the gas expands at a constant pressure.

Exercise Improvement Tip: For a comprehensive understanding, it's beneficial to explain that each thermodynamic process follows specific laws and equations, and demonstrating these processes graphically (like on a P-V diagram) can often make the concepts more accessible.

Partial Differentiation

Partial differentiation is a fundamental concept in calculus, particularly useful in thermodynamics, which involves variables dependent on two or more variables. This mathematical tool allows us to see how a multivariable function changes as only one of the variables is allowed to increase, holding others constant.

In the context of our exercise, we apply partial differentiation to find how the volume of an ideal gas changes with temperature while keeping the pressure constant. This is denoted as \begin{align*}\left(\frac{\partial V}{\partial T}\right)_P\end{align*}Here, \(V\) is the variable we want to differentiate with respect to temperature \(T\), and \(P\) is the parameter we hold constant.

Exercise Improvement Tip: Clarifying the concept of holding one variable constant (in this case, pressure) while varying another (temperature) aids in bridging the gap between the mathematical process and its application to physical problems like gas expansion.