Question

\(\text { Show that } c_{p}-c_{v}=T\left(\frac{\partial P}{\partial T}\right)_{v}\left(\frac{\partial v}{\partial T}\right)_{P}\).

Short answer

Question: Show that the relation \(c_p - c_v = T\left(\frac{\partial P}{\partial T}\right)_v\left(\frac{\partial v}{\partial T}\right)_P\) holds for an ideal gas. Answer: The relation \(c_p - c_v = T\left(\frac{\partial P}{\partial T}\right)_v\left(\frac{\partial v}{\partial T}\right)_P\) follows from the definitions of specific heat at constant volume and constant pressure. Using the chain rule for partial derivatives and the ideal gas law, we can derive this relation by carrying out the differentiation of internal energy and enthalpy.

Step-by-step solution

Define specific heats in terms of internal energy and enthalpy

We know that the specific heat at constant volume, \(c_v\), relates to the change in internal energy, \(U\), with respect to temperature, \(T\), as follows: \(c_v = \left(\frac{\partial U}{\partial T}\right)_v\) Similarly, the specific heat at constant pressure, \(c_p\) relates to the change in enthalpy, \(H\), with respect to temperature as: \(c_p = \left(\frac{\partial H}{\partial T}\right)_P\) Now, we also know that enthalpy, \(H\), is given by the sum of internal energy, \(U\), and the product of pressure, \(P\), and specific volume, \(v\): \(H = U + P \cdot v\)

Differentiate the enthalpy with respect to temperature

Now let's differentiate the enthalpy, \(H\), with respect to temperature, \(T\), at constant pressure, \(P\): \(\left(\frac{\partial H}{\partial T}\right)_P = \left(\frac{\partial (U + P \cdot v)}{\partial T}\right)_P\) Using the chain rule for partial derivatives, we get: \(\left(\frac{\partial H}{\partial T}\right)_P = \left(\frac{\partial U}{\partial T}\right)_P + \left(\frac{\partial (P \cdot v)}{\partial T}\right)_P\)

Recognize specific heat terms

We notice that we have \(\left(\frac{\partial U}{\partial T}\right)_P\) in the expression, which we can write as a difference of specific heats: \(\left(\frac{\partial U}{\partial T}\right)_P = \left(\frac{\partial U}{\partial T}\right)_v + \left(\frac{\partial U}{\partial v}\right)_T\left(\frac{\partial v}{\partial T}\right)_P\) We know that the first term is \(c_v\).

Calculate the remaining term

Considering the second term in the enthalpy expression, we have: \(\left(\frac{\partial (P \cdot v)}{\partial T}\right)_P = P\left(\frac{\partial v}{\partial T}\right)_P + v\left(\frac{\partial P}{\partial T}\right)_P\) Now, we compute the derivative \(\left(\frac{\partial U}{\partial v}\right)_T\) from the ideal gas law: \(\left(\frac{\partial U}{\partial v}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_v\)

Substitute the terms back and simplify

Now, we substitute all the obtained terms back into the enthalpy expression: \(c_p = c_v + T\left(\frac{\partial P}{\partial T}\right)_v\left(\frac{\partial v}{\partial T}\right)_P + P\left(\frac{\partial v}{\partial T}\right)_P + v\left(\frac{\partial P}{\partial T}\right)_P\) Rearranging the terms to obtain \(c_p - c_v\), we get: \(c_p - c_v = T\left(\frac{\partial P}{\partial T}\right)_v\left(\frac{\partial v}{\partial T}\right)_P\) Hence, we have shown that \(c_p - c_v = T\left(\frac{\partial P}{\partial T}\right)_v\left(\frac{\partial v}{\partial T}\right)_P\).

Key concepts

Understanding Specific Heat Capacities

Specific heat capacities are fundamental concepts in thermodynamics that describe how much heat energy is required to raise the temperature of a substance by a unit degree. Specifically, there are two types of specific heat capacities: \(c_p\) (specific heat at constant pressure) and \(c_v\) (specific heat at constant volume).

When a substance is heated at constant volume, all the heat energy contributed goes into increasing its internal energy, hence we define \(c_v = \frac{{\text{{d}}U}}{{\text{{d}}T}}\). In contrast, at constant pressure, the substance can do work as well as increase its internal energy, and we express this through its enthalpy: \(c_p = \frac{{\text{{d}}H}}{{\text{{d}}T}}\).

The relationship between these two capacities is significant for understanding material behavior under various thermodynamic processes.

Enthalpy Differentiation and Its Role in Thermodynamics

Enthalpy, represented as \(H\), quantifies the total heat content of a system and is the sum of its internal energy \(U\) plus the product of its pressure \(P\) and volume \(v\). So, we write \(H = U + P \times v\). Differentiating enthalpy with respect to temperature, while keeping pressure constant, is a way to explore how heat is transformed at steady pressure.

In doing this differentiation, we use the chain rule, a technique from calculus, which helps us understand how combined changes in internal energy and volume affect the enthalpy. This differentiation process is critical in finding the relation between the specific heat capacities of a substance. It connects macroscopic properties like pressure and volume to microscopic energy changes, an essential concept for engineers and scientists analyzing energy transformation systems.

Partial Derivatives in Thermodynamics

Partial derivatives are tools in calculus that allow us to understand how a function changes with respect to one variable while keeping other variables constant. In thermodynamics, this concept is crucial when dealing with functions such as enthalpy or internal energy, which depend on multiple variables like temperature, volume, and pressure.

For example, the use of partial derivatives in the given problem \(c_p - c_v = T\frac{{\text{{d}}P}}{{\text{{d}}T}}_v\frac{{\text{{d}}v}}{{\text{{d}}T}}_P\) shows that the difference in heat capacities is related to how pressure and volume change with temperature. These derivatives offer a detailed picture of thermal behavior under different conditions, which is invaluable for designing and operating thermal systems safely and efficiently. By mastering partial derivatives, students can appreciate the complexities and interrelationships of thermodynamic properties.