Question

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

Short answer

Question: Prove the expressions for specific heat capacities, \(c_v\) and \(c_p\), using thermodynamic relations and Maxwell relations. Answer: The expressions for specific heat capacities \(c_v\) and \(c_p\) can be derived as follows: 1. Define specific heat capacity: $$ c_v = \left(\frac{\partial q}{\partial T}\right)_{v} $$ and $$ c_p = \left(\frac{\partial q}{\partial T}\right)_{p} $$ 2. Recall thermodynamic relation: $$ T\left(\frac{\partial s}{\partial T}\right)_{v} = \left(\frac{\partial q}{\partial T}\right)_{v} $$ 3. Use Maxwell relation: $$ \left(\frac{\partial P}{\partial T}\right)_{v} = \left(\frac{\partial s}{\partial v}\right)_{T} $$ 4. Combine the relations to obtain \(c_v\): $$ c_v = -T\left(\frac{\partial v}{\partial T}\right)_{s}\left(\frac{\partial P}{\partial T}\right)_{v} $$ 5. Use the thermodynamic relation for \(c_p\): $$ T\left(\frac{\partial s}{\partial T}\right)_{p} = \left(\frac{\partial q}{\partial T}\right)_{p} = c_p $$ 6. Use the Maxwell relation for \(c_p\): $$ \left(\frac{\partial v}{\partial T}\right)_{p} = \left(\frac{\partial s}{\partial P}\right)_{T} $$ 7. Combine the relations to obtain \(c_p\): $$ c_p = T\left(\frac{\partial P}{\partial T}\right)_{s}\left(\frac{\partial v}{\partial T}\right)_{p} $$

Step-by-step solution

Define Specific Heat Capacities

Specific heat capacities are defined as the heat needed to raise the temperature of a substance by 1 degree, without changing the volume (\(c_v\)) or the pressure (\(c_p\)). Mathematically, they are given by: $$ c_v = \left(\frac{\partial q}{\partial T}\right)_{v} $$ and $$ c_p = \left(\frac{\partial q}{\partial T}\right)_{p} $$ where \(q\) is the heat added, \(T\) is the temperature, \(v\) is the volume, and \(p\) is the pressure.

Recall the thermodynamic relation

We use the following thermodynamic relation based on entropy (\(s\)) and volume (\(v\)): $$ T\left(\frac{\partial s}{\partial T}\right)_{v} = \left(\frac{\partial q}{\partial T}\right)_{v} $$

Use the Maxwell relation

Now, we can make use of Maxwell relation, which relates the partial derivatives of different thermodynamic properties: $$ \left(\frac{\partial P}{\partial T}\right)_{v} = \left(\frac{\partial s}{\partial v}\right)_{T} $$

Combine the relations to obtain \(c_v\)

Substitute the Maxwell relation (Step 3) into the thermodynamic relation (Step 2) to get the expression for \(c_v\): $$ c_v = -T\left(\frac{\partial v}{\partial T}\right)_{s}\left(\frac{\partial P}{\partial T}\right)_{v} $$

Use the thermodynamic relation for \(c_p\)

We can similarly use the thermodynamic relation for specific heat capacity at constant pressure: $$ T\left(\frac{\partial s}{\partial T}\right)_{p} = \left(\frac{\partial q}{\partial T}\right)_{p} = c_p $$

Use the Maxwell relation for \(c_p\)

We use the following Maxwell relation for \(c_p\): $$ \left(\frac{\partial v}{\partial T}\right)_{p} = \left(\frac{\partial s}{\partial P}\right)_{T} $$

Combine the relations to obtain \(c_p\)

Substitute the Maxwell relation (Step 6) into the thermodynamic relation (Step 5) to get the expression for \(c_p\): $$ c_p = T\left(\frac{\partial P}{\partial T}\right)_{s}\left(\frac{\partial v}{\partial T}\right)_{p} $$ Now we have derived the expressions for \(c_v\) and \(c_p\) as given in the exercise.

Key concepts

Thermodynamic Properties

Understanding thermodynamic properties is crucial when studying the energy transfer within physical systems. These properties include temperature (T), pressure (P), volume (v), and entropy (s), each playing a pivotal role in describing the state of a system. Temperature gauges the thermal energy of a substance; pressure is the force applied by the substance over an area; volume represents the space a substance occupies, and entropy measures the disorder within a system.

Specific heat capacities, such as the heat required to change the temperature of a unit mass of a substance by one-degree Celsius, are intrinsic thermodynamic properties. They come in two variants: the specific heat at constant volume (\(c_v\)) and at constant pressure (\(c_p\)). These capacities are vital for understanding how substances respond to thermal energy, acting as a benchmark for predicting temperature changes during thermal processes.

Role in Energy Transfer

Thermodynamic properties are central to calculating energy transfer. For example, in a calorimetric experiment, the amount of heat absorbed or released (\(q\)) is connected with corresponding temperature changes, making the understanding of \(c_v\) and \(c_p\) essential for precise measurement of heat transfer.

Maxwell Relations

Maxwell relations are a set of equations derived from the second law of thermodynamics. They bridge the gap between different partial derivatives of thermodynamic properties, making them a powerful tool for simplifying complex thermodynamic analyses. Named after James Clerk Maxwell, these relations are a product of the state function properties of thermodynamic potentials.

When examining the problem of specific heat capacities, the Maxwell relation that connects \(\frac{\partial P}{\partial T}\right)_v\) with \(\frac{\partial s}{\partial v}\right)_T\) becomes extremely useful. This relationship indicates that the changes in entropy with volume at constant temperature are directly related to the changes in pressure with temperature at constant volume. Maxwell relations simplify the calculations by providing a method to substitute one difficult-to-measure derivative with another more accessible derivative.

Applications of Maxwell Relations

Practical applications of Maxwell relations extend to various fields such as chemical engineering, meteorology, and materials science, where analyzing the changes and interrelations of thermodynamic properties is necessary.

Partial Derivatives

In thermodynamics, partial derivatives are used to describe how a particular property changes with respect to one variable while keeping other variables constant. They are essential in explaining the behavior of systems when subject to small changes, such as temperature or volume fluctuations. A partial derivative is represented with the symbol \(\frac{\partial}{\partial x}\right)_y\), indicating the rate of change of a function with respect to variable x while holding variable y constant.

The use of partial derivatives allows us to deduce the specific heat capacities \(c_v\) and \(c_p\) from the problem statement. By understanding how to take these derivatives properly, one can explore how the internal energy of a system varies with temperature at constant volume, or understand how the enthalpy changes with temperature at constant pressure. These insights are fundamental to predicting and controlling thermodynamic processes.

Importance in Thermodynamics

Proficiency in calculating partial derivatives is essential for students of physics and engineering, as it underpins the analysis of heat transfer, work done by systems, and the overall energy balance in thermodynamic cycles.