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Chapter 9: Problem 4

Obtain the Clausius-Clapeyron equation starting from the \(p-V\) diagram.

Short Answer

Expert verified
The Clausius-Clapeyron equation comes from the relationship \(\dfrac{dp}{dT} = \dfrac{L}{T(v2-v1)}\) and, after integration, it becomes \(\ln (\dfrac{p2}{p1}) = \dfrac{L}{R} (\dfrac{1}{T1}-\dfrac{1}{T2})\), which describes the phase boundary on a p-T diagram. This derivation comes from applying the first law of thermodynamics to a phase transition process.

Step by step solution

01

Understand the Problem

The Clausius-Clapeyron equation is a way of characterizing a phase diagram in terms of temperature and pressure. The equation relates the vapor pressure of a substance at a particular temperature to the heat of vaporization of that substance.
02

Apply First Law of Thermodynamics

Start by applying the first law of thermodynamics, which states that in any cyclic process, the net heat supplied to the system equals the net work done by the system. For a phase transition from a lower phase (liquid) to an upper phase (vapor), the differential form of the first law of thermodynamics for a reversible process gives \(dQ = TdS\). Since the transition takes place at constant temperature and pressure, we can write \(dQ = LdN\), where \(L\) is the latent heat involved in the transition and \(dN\) is the number of moles transforming. We then equate these to get \(LdN = TdS\).
03

Replace Entropy Difference

At constant pressure, the entropy difference between two phases equals the heat of transformation divided by the absolute temperature, \(T\). This gives \(S = \dfrac{L}{T}N\), where \(N\) denotes the number of moles. Differentiating this expression gives \(dS = \dfrac{L}{T}dN\). Substituting this into the equation from Step 2 gives \(LdN = LdN\), which is consistent.
04

Derive Clausius-Clapeyron Equation

Recall that \(\dfrac{dp}{dT} = \dfrac{L}{T(v2-v1)}\), where \(v2\) and \(v1\) are the molar volumes of the vapor and liquid phases, respectively. This can be rearranged to be \(dp = \dfrac{L}{T(v2-v1)}dT\). Integrating this from pressure 1 to pressure 2 and temperature T1 to T2 gives \(\ln (\dfrac{p2}{p1}) = \dfrac{L}{R} (\dfrac{1}{T1}-\dfrac{1}{T2})\), which is the integrated form of the Clausius-Clapeyron equation. This equation describes the phase boundary on a p-T diagram.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Law of Thermodynamics
The First Law of Thermodynamics, also known as the principle of the conservation of energy, asserts that energy cannot be created or destroyed in an isolated system. The energy of the system is constant. When energy passes, as work, as heat, or with matter, into or out from a system, the system's internal energy changes in accord with the law of conservation of energy. It can be summarized in the simple equation: \[\begin{equation} \triangle U = Q - W, ewline ewline \end{equation}\] where
  1. \triangle U is the change in the internal energy of the system,
  2. Q is the heat added to the system,
  3. W is the work done by the system.

In the context of phase transitions, such as from liquid to vapor, the First Law implies that the energy required for the phase change, known as latent heat, must be supplied to the system in the form of heat without a change in temperature.
Phase Transition
A phase transition refers to the transformation of a substance from one state of matter to another, such as solid to liquid (melting), liquid to gas (vaporization), or the reverse processes: freezing and condensation. During a phase transition, the physical properties of a substance change dramatically with the input or removal of energy. This process occurs at characteristic temperatures and pressures, and at these specific points, the two phases can exist in equilibrium.
Latent Heat
Latent heat is defined as the amount of heat required to convert a substance from one phase to another at constant temperature and pressure. This heat is 'latent', meaning it does not cause a temperature change in the substance. There are two main types of latent heat:
  1. Latent heat of fusion 鈥� the energy required for changing from solid to liquid or vice versa,
  2. Latent heat of vaporization 鈥� the energy required for changing from liquid to gas or vice versa.

The Clausius-Clapeyron equation utilizes the concept of latent heat to describe the relationship between the vapor pressure and temperature during a phase change. Specifically, it relates the change in pressure with respect to temperature to the latent heat and the volume change during the phase transition.

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Most popular questions from this chapter

Prove that (a) \(\mathrm{d} U=\left(C_{p}-p V \alpha\right) \mathrm{d} T+V\left(\beta_{T} p-\alpha T\right) \mathrm{d} p\) (b) \(\mathrm{d} H=C_{p} \mathrm{~d} T+V(1-\alpha T) \mathrm{d} p\) (c) \(\mathrm{d} F=-(p V \alpha+S) \mathrm{d} T+p V \beta_{T} \mathrm{~d} p\)

The Gibbs free energy of a system is given by \(S=-\left(\frac{\partial G}{\partial T}\right)_{p}\). If \(G=R T \ln \left[\frac{a p}{(R T)^{5 / 2}}\right]\) show that \(C_{p}=\frac{5}{2} R\)

Starting from second latent heat equation, show that $$ \frac{\mathrm{d} L}{\mathrm{~d} T}=C_{p_{2}}-C_{p_{1}}+L\left[\frac{1}{T}-\frac{v_{2} \beta_{2}-v_{1} \beta_{1}}{v_{2}-v_{1}}\right] $$ where \(\beta_{1}\) and \(\beta_{2}\) are isothermal compressibilities of phase 1 and 2 respectively.

The Helmholtz free energy of a system is given by $$ F=A+B T(1-\ln T)-C T \ln V $$ where \(A, B, C\) are constant. Obtain expressions for pressure, entropy, internal energy, enthalpy and Gibbs energy. Also discuss the significance of the constants \(B\) and \(C\).

The weight on the steam exhaust opening of a pressure cooker can maintain the highest pressure of \(220 \mathrm{kPa}\) inside it. Calculate the temperature at which water will boil in the cooker. Given, latent heat of vaporisation of water is \(2257 \mathrm{~kJ} \mathrm{~kg}^{-1}\) at \(100^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}=1.01325 \times 10^{5} \mathrm{~Pa}\)
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