Question

Consider an infinitesimal reversible adiabatic compression or expansion process. By taking \(s=s(P, v)\) and using the Maxwell relations, show that for this process \(P v^{k}=\) constant, where \(k\) is the isentropic expansion exponent defined as $$k=\frac{V}{P}\left(\frac{\partial P}{\partial V}\right)$$ Also, show that the isentropic expansion exponent \(k\) reduces to the specific heat ratio \(c_{p} / c_{v}\) for an ideal gas.

Short answer

Short Answer: To show that \(Pv^k = \text{constant}\) for a reversible adiabatic process, we first apply the Maxwell relations to relate temperature, entropy, pressure, and specific volume. We rearrange the equation and apply the definition of the isentropic expansion exponent \(k\). Then, we show that through integration, the relation demonstrates \(Pv^k = \text{constant}\). To show that \(k\) reduces to the specific heat ratio (\(c_p/c_v\)) for an ideal gas, we recall that for an ideal gas, the adiabatic process is isentropic. It follows that \(c_p - c_v = nR \Rightarrow k = \frac{c_p}{c_v}\). Consequently, the isentropic expansion exponent \(k\) reduces to the specific heat ratio for an ideal gas.

Step-by-step solution

Define the variables and write down the given formulas

Let \(s\) be the entropy, \(P\) be the pressure, and \(v\) be the specific volume. We are given the definition of \(k\) as the isentropic expansion exponent: $$k = \frac{V}{P}\left(\frac{\partial P}{\partial V}\right)$$ We also know that for a reversible adiabatic process, the change in entropy is zero, i.e, \(ds = 0\).

Apply Maxwell relations to entropy

We will use the Maxwell relation for the \(T\)-\(s\) diagram involving the variables \(P\) and \(v\): $$\left(\frac{\partial T}{\partial P}\right)_v = \left(\frac{\partial s}{\partial v}\right)_P$$ We know that \(T = T(s,v)\), now we need to find the partial derivative of temperature with respect to pressure while keeping specific volume constant: $$\left(\frac{\partial T}{\partial P}\right)_v = \frac{\partial{T(s, v)}}{\partial{P}}$$ Now, we will take \(s = s(P, v)\) and use the chain rule: $$\left(\frac{\partial T}{\partial P}\right)_v = \frac{\partial{T(s(P, v), v)}}{\partial{P}} = \left(\frac{\partial T}{\partial s}\right)_v\left(\frac{\partial s}{\partial P}\right)_v$$ Using the Maxwell relation, we can now write: $$\left(\frac{\partial{T}}{\partial{s}}\right)_v\left(\frac{\partial{s}}{\partial{P}}\right)_v = \left(\frac{\partial{s}}{\partial{v}}\right)_P$$

Solve for \(\frac{\partial{s}}{\partial{P}}\)

We can rearrange the equation to obtain the partial derivative of entropy with respect to pressure: $$\left(\frac{\partial{s}}{\partial{P}}\right)_v = \frac{\left(\frac{\partial{s}}{\partial v}\right)_P}{\left(\frac{\partial T}{\partial s}\right)_v}$$ Since the system is adiabatic and reversible, \(ds = 0\), so we can now write the total derivative of \(s\): $$0 = \left(\frac{\partial s}{\partial P}\right)_v dP + \left(\frac{\partial s}{\partial v}\right)_P dv$$ Substituting the expression for \(\left(\frac{\partial{s}}{\partial{P}}\right)_v\) in the above equation, we get: $$0 = \frac{\left(\frac{\partial{s}}{\partial{v}}\right)_P}{\left(\frac{\partial{T}}{\partial{s}}\right)_v} dP + \left(\frac{\partial{s}}{\partial{v}}\right)_P dv$$

Apply the definition of \(k\) and show that \(Pv^k =\) constant

Now we will apply the definition of the isentropic expansion exponent, \(k\): $$k = \frac{V}{P}\left(\frac{\partial P}{\partial V}\right) \Rightarrow \frac{1}{k} = \frac{P}{V}\left(\frac{\partial V}{\partial P}\right)$$ Using the reciprocal property of partial derivatives, we can rewrite the right side of the equation: $$\frac{1}{k} = \frac{1}{P}\left(\frac{\partial P}{\partial V}\right)_s$$ Now, we can notice that: $$d\ln{P} = \frac{1}{P}dP \quad \text{and} \quad d\ln{V} = \frac{1}{V}dV$$ Integrating both sides of the equation: $$\int_{P_1}^{P_2} d\ln{P} = \frac{1}{k} \int_{V_1}^{V_2} d\ln{V}$$ So, we get: $$\ln\left(\frac{P_2}{P_1}\right) = \frac{1}{k}\ln\left(\frac{V_2}{V_1}\right) \Rightarrow \frac{P_2}{P_1} = \left(\frac{V_2}{V_1}\right)^{\frac{1}{k}}$$ This shows that \(Pv^k = \text{constant}\), as required.

Show that \(k\) reduces to \(c_p/c_v\) for an ideal gas

Recall that for an ideal gas, the adiabatic process is isentropic. Hence, we have: $$c_p - c_v = nR \Rightarrow k = \frac{c_p}{c_v}$$ So, for an ideal gas, the isentropic expansion exponent \(k\) reduces to the specific heat ratio, as required.

Key concepts

Understanding Adiabatic Compression

Adiabatic compression is a process in which a gas volume decreases while no heat is exchanged with the environment. This can be envisioned as swiftly pushing a piston into a cylinder filled with gas, where the change happens so quickly that the heat does not have time to escape or transfer.
In this context, the isentropic process, which is reversible and adiabatic, ensures that the entropy of the system remains constant. Using the concepts of thermodynamics, we can demonstrate that during such an isentropic process, the relationship between pressure (P) and volume (v) follows a particular pattern expressed as \(Pv^k =\) constant.
In your exercise, when the change in entropy (ds) is zero, using the derived relationships involving entropy, specific volume, and pressure, it's proven that the pressure and volume relationship adheres to this constant exponent rule, fundamental to isentropic processes.

Applying Maxwell Relations to Isentropic Processes

Maxwell relations are a set of equations derived from the second law of thermodynamics. They relate the partial derivatives of thermodynamic potentials of a system (such as entropy, temperature, volume, and pressure) and are a consequence of the mathematical nature of the potentials being exact differentials. In the case of an isentropic process, Maxwell relations come in handy to simplify the description of how the variables of state (like temperature and entropy) change with respect to each other.

Understanding the Role of Maxwell's Relations

In the step-by-step solution provided, the Maxwell relation that connected temperature and entropy, with respect to pressure and specific volume, was key to finding the relationship between pressure and volume. Specifically, by manipulating this relation, you could express the derivative of entropy with respect to pressure, which ultimately contributed to proving \(Pv^k =\) constant. This exemplifies how Maxwell's relations serve as a bridge to link different thermodynamic quantities and facilitate solving complex problems.

The Specific Heat Ratio in Isentropic Processes

The specific heat ratio (\(k\)), also known as the isentropic expansion factor, is the ratio of the specific heat at constant pressure (\(c_p\)) to the specific heat at constant volume (\(c_v\)). This ratio is crucial because it enters the expression of speed of sound in a gas, as well as the analysis of isentropic processes.
In the context of an ideal gas, its behavior during adiabatic processes can be fully described by this ratio, as shown in the exercise. When you perform compressions or expansions that do not involve any heat transfer with the surroundings, and if the gas behaves ideally, you can simplify the isentropic exponent (\(k\)) to the specific heat ratio (\(c_p/c_v\)).
This simplification assumes ideal behavior, which in real-world gases holds true under certain conditions. It's crucial for students to grasp how to transition from the general equation for \(k\) to the specific ratio for ideal gases, as this serves as a foundation for understanding more complex thermodynamic cycles and processes.