Enthalpy and First Law

Section 24.3 Enthalpy and First Law

Enthalpy of a thermodynamic system is defined by adding the product of presure \((p)\) and volume \((V)\) to the internal energy \((U)\text{.}\) We denote enthalpy by letter \(H\text{.}\)

\begin{equation} H = U + pV.\label{eq-enthalpy-definition}\tag{24.3.1} \end{equation}

Since \(U\text{,}\) \(p\text{,}\) and \(V\) are well-defined quantities for each state of the system, \(H\) has definite values in each state. That is, \(H \) is a state function and can be used to track changes in the thermodynamic state, e.g., what happens when we heat or cool an object.

What is the use of enthalpy? Turns out that many experiments are conducted at constant pressure, e.g., chemical and biological reactions conducted at atmospheric pressure. We will show below that heat exchanged with the environment in these processes give us change in enthapy of the system. For these state changes, the energy exchange with the environment is captured more effectively by enthalpy change rather than internal energy change.

\begin{equation} \Delta H = Q\ \ \ \text{(constant pressure process)}\tag{24.3.2} \end{equation}

For instance, if a chemical reaction that releases \(10\text{kJ}\) of energy takes place at atmospheric pressure, we say that the enthalpy of the products is less than those of the reactants by \(10\text{kJ}\text{.}\)

The heat of transformation or latent heat is another example of enthalpy since latent heat is usually measured at atmospheric pressure.

For situations, such as experiments with solids, where volume change may be negligible, the change in internal energy is approximately the heat exchanged with the environment`.

\begin{equation} \Delta U = Q\ \ \ \text{(constant volume process)}\tag{24.3.3} \end{equation}

By using Eq. (24.3.1) in the First Law statement, viz., \(\Delta U = Q - p\Delta V\text{,}\) we find

\begin{equation*} \Delta H = \Delta U + p \Delta V + V \Delta p = Q + V \Delta p. \end{equation*}

The change in enthlapy

\begin{equation} \Delta H = Q + V \Delta p\tag{24.3.4} \end{equation}

gives us an alternate form of the First Law.

Subsection 24.3.1 (Calculus) First Law of Thermodynamics Using Enthalpy Change

The first law of thermodynamics can also be written in terms of enthalpy change instead of the internal energy change. In this section we will work out the first law in the infinitesimal form in terms of the enthalpy. We will use \(pdV\) for the work associated with the infinitesimal process.

From the definition of the enthaply, \(H = U + pV\text{,}\) we find that the change in enthalpy for an infinitesimal process will be the difference,

\begin{equation*} (H+dH) - H =\left[(U+dU) +(p+dp)(V+dV \right] - (U+pV) \end{equation*}

Expanding and dropping quadratic terms in infinitesimals, we find the following.

\begin{equation*} dH = dU + p dV + V dp \end{equation*}

Putting \(dQ\) for \(dU+pdV\) from the expression for the first law of thermodynamics we find an alternative statement of the first law.

\begin{equation} dH = dQ + V dp.\tag{24.3.5} \end{equation}

This relation says that, for a constant-pressure quasi-static process, i.e., when \(dp=0\text{,}\) the enthalpy change is equal to the heat exchanged.

\begin{equation} dH = dQ \ \ \ (\text{const}\, p)\tag{24.3.6} \end{equation}

Many engineering processes and chemical reactions are conducted at constant pressures. For these systems enthalpy is a more appropriate quantity for analysis.

Checkpoint 24.3.1. Enthalpy Change in Heating Water at Constant Pressure.

One liter of water in a beaker is heated from \(25^{\circ}\text{C}\) to \(30^{\circ}\text{C}\) at constant pressure of 1 atm. Find the change in enthalpy of water.

Hint

Use \(\Delta H = Q\) for a constant pressure process.

Answer

\(20,900\ \text{J}\text{.}\)

Solution

The enthalpy can be obtained from heat transfered at constant pressure.

\begin{equation*} \Delta H = Q = m c_P \Delta T. \end{equation*}

Using the numerical values, we get

\begin{equation*} \Delta H = 1.0\ \text{kg}\ 4180\ \frac{\text{J}}{ \text{kg.} ^{\circ}\text{C}}\times 5^{\circ}\text{C} = 20,900\ \text{J}. \end{equation*}

Checkpoint 24.3.2. Changes in Inernal Energy and Enthalpy in a Gas Expansion Process.

Four moles of a monatomic ideal gas in a cylinder at \(27^{\circ}\)C is expanded at constant pressure, which is equal to 1 atm, till its volume doubles. (a) What are the changes in its internal energy and enthalpy? (b) How much work was done by the gas in the process? (c) How much heat was transferred to the gas?

Formulas for internal energy of an ideal gas = \(\dfrac{3}{2}RT\text{,}\) enthalpy of an ideal gas = \(\dfrac{5}{2}RT\text{.}\)

Hint

Use the formulas for the internal energy and enthalpy along with the equation of state of the ideal gas.

Answer

(a) \(\Delta U = 18,700\ J\text{,}\) \(\Delta H = 31,200\ J\text{,}\) (b) \(10,000\ \text{J} \text{,}\) (c) \(24,900\ \text{J}\text{.}\)

Solution 1 (a)

(a) We will use the formulas for the internal energy and enthalpy along with the equation of state of the ideal gas.

\begin{align*} \Delta U \amp = \frac{3}{2} \left( n R T_2 - n R T_1\right) \\ \amp = \frac{3}{2} \left( p_2 V_2 - p_1 V_1\right). \end{align*}

Since the pressure is constant we get

\begin{align*} \Delta U \amp = \frac{3}{2} p \left(V_2 - V_1\right) \\ \amp = \frac{3}{2} p V_1 \left(\frac{V_2}{V_1} - 1\right). \end{align*}

The data does not give \(V_1\) or \(V_2\) but instead \(T_1\) and \(p\) and the ration \(V_2/v_1\text{.}\)

\begin{equation*} \Delta U = \frac{3}{2}n RT_1 \left(\frac{V_2}{V_1} - 1\right). \end{equation*}

This form of \(\Delta U\) is ready for the data given.

\begin{align*} \Delta U \amp = \frac{3}{2}\times 4\ \text{mol} \times 8.31\ \frac{\text{J}}{\text{mol.K}} \times 300.15\ \text{K} \left(2 - 1\right)\\ \amp = 15,000\ \text{J}. \end{align*}

For the enthalpy change we get the following from the definition \(H = U + pV\text{.}\)

\begin{align*} \Delta H \amp = \Delta U + \Delta (pV) = \Delta U + p \Delta V \\ \amp = \Delta U + \frac{2}{3} \Delta U =\frac{5}{3} \Delta U = 24,900\ \text{J}. \end{align*}

Solution 2 (b)

(b) The work by the gas will be

\begin{equation*} W = p(V_2-V_1) = p V_1 \left(\frac{V_2}{V_1} - 1\right). \end{equation*}

Comparing to the expression for \(\Delta U\text{,}\) we find that this is

\begin{equation*} W = \frac{2}{3} \Delta U = 10,000\ \text{J}. \end{equation*}

Solution 3 (c)

\begin{align*} Q \amp = \Delta U + W = \Delta U + \frac{2}{3} \Delta U \\ \amp = \frac{5}{3} \Delta U= \Delta H = 24,900\ \text{J}. \end{align*}