5.2 Applications

Many chemical reactions involve the generation of gases capable of doing mechanical work or the generation of heat. It is important for us to quantify these changes and relate them to the changes in the internal energy. 

Work

First of all, let us concentrate on the nature of the work a system can do. We will consider only mechanical work i.e., pressure-volume work.

For understanding pressure-volume work, let us consider a cylinder that contains one mole of an ideal gas fitted with a frictionless piston. The total volume of the gas is \(V_i\) and the pressure of the gas inside is \(p\). If external pressure is \(p_{\text {ex }}\) which is greater than \(p\), the piston is moved inward till the pressure inside becomes equal to \(p_{\mathrm{ex}}\). Let this change be achieved in a single step and the final volume be \(V_f\). During this compression, suppose the piston moves a distance, \(l\) and is the cross-sectional area of the piston is A [Fig. 6.5(a)].

then, volume change \(=l \times \mathrm{A}=\Delta V=\left(V_f-V_i\right)\)
We also know, pressure \(=\frac{\text { force }}{\text { area }}\)

We also know, pressure \(=\frac{\text { force }}{\text { area }}\)
Therefore, force on the piston \(=p_{e x}\). A
If \(\mathrm{w}\) is the work done on the system by the movement of the piston then
\(
\begin{aligned}
& \mathrm{w}=\text { force } \times \text { distance }=p_{e x} \cdot \text { A } \cdot l \\
& =p_{e x} \cdot(-\Delta V)=-p_{\text {ex }} \Delta V=-p_{\text {ex }}\left(V_f-V_i\right) \dots(6.2)
\end{aligned}
\)
The negative sign of this expression is required to obtain conventional sign for \(w\), which will be positive. It indicates that in the case of compression, work is done on the system. Here \(\left(V_f-V_i\right)\) will be negative and negative multiplied by negative will be positive. Hence the sign obtained for the work will be positive.

If the pressure is not constant at every stage of compression, but changes in number of finite steps, work done on the gas will be summed over all the steps and will be equal to \(-\sum p \Delta V\) [Fig. 6.5 (b)]

If the pressure is not constant but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, \(d V\). In such a case we can calculate the work done on the gas by the relation
\(
\mathrm{w}=-\int_{V_i}^{V_f} p_{e x} d V \dots(6.3)
\)
Here, \(p_{e x}\) at each stage is equal to \(\left(p_{i n}+d p\right)\) in case of compression [Fig. 6.5(c)]. In an expansion process under similar conditions, the external pressure is always less than the pressure of the system i.e., \(p_{e x}=\left(p_{\text {in }}-d p\right)\).
In general case we can write, \(p_{e x}=\left(p_{i n} \pm d p\right)\). Such processes are called reversible processes.

A process or change is said to be reversible, if a change is brought out in such a way that the process could, at any moment, be reversed by an infinitesimal change. A reversible process proceeds infinitely slowly by a series of equilibrium states such that system and the surroundings are always in near equilibrium with each other. Processes other than reversible processes are known as irreversible processes.

In chemistry, we face problems that can be solved if we relate the work term to the internal pressure of the system. We can relate work to the internal pressure of the system under reversible conditions by writing equation 6.3 as follows:
\(
\mathrm{w}_{\text {rev }}=-\int_{V_i}^{V_f} p_{e x} d V=-\int_{V_i}^{V_f}\left(p_{i n} \pm d p\right) d V
\)

Since \(d p \times d V\) is very small we can write
\(
\mathrm{w}_{r e v}=-\int_{V_i}^{V_f} p_{i n} d V \dots(6.4)
\)

Now, the pressure of the gas ( \(p_{i n}\) which we can write as \(p\) now) can be expressed in terms of its volume through gas equation. For \(n \mathrm{~mol}\) of an ideal gas i.e., \(p V=n R T\)
\(
\Rightarrow p=\frac{n R T}{V}
\)

Therefore, at constant temperature (isothermal process),
\(
\begin{aligned}
& \mathrm{w}_{\mathrm{rev}}=-\int_{V_i}^{V_f} n \mathrm{R} T \frac{d V}{V}=-n \mathrm{R} T \ln \frac{V_f}{V_i} \\
& =-2.303 n \mathrm{R} T \log \frac{V_f}{V_i} \dots(6.5)
\end{aligned}
\)

Free Expansion:

Free expansion: Expansion of a gas in vacuum \(\left(p_{e x}=0\right)\) is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible (equations 6.2 and 6.3).

Now, we can write equation 6.1 in a number of ways depending on the type of processes.

Let us substitute \(\mathrm{w}=-p_{e x} \Delta V\) (eq. 6.2) in equation 6.1, and we get
\(
\Delta U=q-p_{e x} \Delta V
\)

If a process is carried out at constant volume \((\Delta V=0)\), then
\(
\Delta U=q_V
\)
the subscript \(V\) in \(q_V\) denotes that heat is supplied at constant volume.

Thermodynamic Processes

Isothermal Process:

It is a thermodynamic process in which temperature remains constant. We know,

According to Gas law,
\(
P V=n R T
\)
\(
P=\frac{n R T}{V}
\)

Using this value of \(\mathrm{P}\) in work done, we get,
\(
\begin{aligned}
& W=n R T \int_{V_i}^{V_f} \frac{d V}{V} \\
& W=n R T \ln \frac{V_f}{V_i}=2.303 n R T \log \frac{V_f}{V_i}
\end{aligned}
\)
Key Points

• \({W}=+\mathrm{ve}(\mathrm{Vf}>\mathrm{Vi})\)
• \(W=-v e(V i>V f)\)
• \(\Delta U=0\)
  
  • Internal energy only depends on temperature
  • If temperature \(=\) Constant
  • Internal Energy = Constant.

Isothermal and free expansion of an ideal gas

For isothermal ( \(T=\) constant) expansion of an ideal gas into vacuum ; \(\mathrm{w}=0\) since \(p_{e x}=0\). Also, Joule determined experimentally that \(q=0\); therefore, \(\Delta U=0\)

Equation 6.1, \(\Delta U=q+\mathrm{w}\) can be expressed for isothermal irreversible and reversible changes as follows:

For an isothermal irreversible change
\(
q=-\mathrm{w}=p_{e x}\left(V_f-V_i\right)
\)

For an isothermal reversible change
\(
\begin{aligned}
& q=-\mathrm{w}=n \mathrm{R} T \ln \frac{V_f}{V_i} \\
& =2.303 \mathrm{nR} T \log \frac{V_f}{V_i}
\end{aligned}
\)

Adiabatic Process:

It is a thermodynamic process in which no heat is exchanged between the system and the surroundings. So, \(q=0\). Mathematically this process is represented as

\(
P V^\gamma=K(\text { constant })
\)
\(
W=\int P d V
\)

Substituting \(P\), we get,
\(
W=K \int_{V_i}^{V_f} \frac{d V}{V^\gamma}
\)
\(
W=K \frac{\left(V_f^{1-\gamma}-V_i^{1-\gamma}\right)}{1-\gamma}
\)

For adiabatic change, \(q=0\),
\(
\Delta U=\mathrm{w}_{\mathrm{ad}}
\)

Isochoric Process:

In the isochoric process, the change in volume of the thermodynamic system is zero. A volume change is zero, so the work done is zero.

• Volume of the system \(=\) Constant
• Change in volume \(=0\)
• If, the change in volume \(=0\), then the work done is zero.

According to the first law of thermodynamic law

• \(\Delta U=q+\mathrm{W}\)
• If \(W=0\)
• \(\mathrm{q}=\mathrm{\Delta U}\)

Isobaric Process:

The pressure remains constant during this process. So,
\(
W=P\left(V_f-V_i\right)
\)
So if volume increases, work done is positive, else negative.

Summary of Some of the Important Results:

(i) For isothermal irreversible expansion,
\(
\mathrm{q}=-\mathrm{W}=\mathrm{P}_{\text {ex }}\left(\mathrm{V}_f-\mathrm{V}_i\right)
\)
(ii) For isothermal reversible expansion or compression from volume \(\mathrm{V}_i\) to \(\mathrm{V}_f\)
\(
\mathrm{q}=-\mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{V}_f}{\mathrm{~V}_i}
\)
(iii) For isothermal expansion of an ideal gas against vacuum i.e. for free expansion,
\(
\Delta \mathrm{U}=0, \mathrm{~W}=0, \mathrm{q}=0
\)
(iv) Since internal energy of an ideal gas is a function of temperature, for all isothermal processes involving ideal gas, \(\Delta \mathrm{U}=0\), whether the process is reversible or irreversible.
(v) For adiabatic change, \(q=0\), therefore \(\Delta U=W_{\text {ad }}\)
(vi) For isochoric process, \(\mathrm{W}=0\)
\(\therefore \Delta \mathrm{U}=\mathrm{q}_{\mathrm{V}}+0\); i.e., heat given to system at constant volume changes internal energy.

Example 6.2: Two litres of an ideal gas at a pressure of \(10 \mathrm{~atm}\) expands isothermally at \(25^{\circ} \mathrm{C}\) into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Answer: We have \(q=-\mathrm{w}=p_{e x}(10-2)=0(8)=0\) No work is done; no heat is absorbed.

Example 6.3: Consider the same expansion, but this time against a constant external pressure of \(1 \mathrm{~atm}\).

Answer: We have \(q=-\mathrm{w}=p_{e x}(8)=8\) litre-atm

Example 6.4: Consider the expansion given in example 6.2, for \(1 \mathrm{~mol}\) of an ideal gas conducted reversibly.

Answer:

\(\begin{aligned} & \text { We have } q=-\mathrm{w}=2.303 \mathrm{nRT} \log \frac{V_f}{V_i} \\ & \quad=2.303 \times 1 \times 0.8206 \times 298 \times \log \frac{10}{2}\end{aligned}\)
\(
\begin{aligned}
& =2.303 \times 0.8206 \times 298 \times \log 5 \\
& =2.303 \times 0.8206 \times 298 \times 0.6990 \\
& =393.66 \mathrm{~L} \mathrm{~atm}
\end{aligned}
\)

Enthalpy, \(H\)

The total heat content of a system at constant pressure is known as its enthalpy.

(a) A Useful New State Function

We know that the heat absorbed at constant volume is equal to the change in the internal energy i.e., \(\Delta U=q_{v}\). But most of chemical reactions are carried out not at constant volume, but in flasks or test tubes under constant atmospheric pressure. We need to define another state function that may be suitable under these conditions.

From the first law of thermodynamics,
\(
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{W} \\
& \text { As } W=-P \Delta V \\
& \therefore \quad \mathrm{q}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V} \\
&
\end{aligned}
\)
We may write the above equation as
\(\Delta U=q_p-p \Delta V\) at constant pressure, where \(q_p\) is the heat absorbed by the system and \(-p \Delta V\) represents expansion work done by the system.

Let us represent the initial state by subscript 1 and the final state by 2
We can rewrite the above equation as
\(
U_2-U_1=q_p-p\left(V_2-V_1\right)
\)

On rearranging, we get
\(
q_p=\left(U_2+p V_2\right)-\left(U_1+p V_1\right) \dots(6.6)
\)
Now we can define another thermodynamic function, the enthalpy \(H\) [Greek word enthalpien, to warm or heat content] as :
\(
H=U+p V \dots(6.7)
\)
so, equation (6.6) becomes
\(
q_p=H_2-H_1=\Delta H
\)
Although \(q\) is a path-dependent function, \(H\) is a state function because it depends on \(U\), \(p\) and \(V\), all of which are state functions. Therefore, \(\Delta H\) is independent of the path. Hence, \(q_p\) is also independent of the path.

For finite changes at constant pressure, we can write equation 6.7 as
\(
\Delta H=\Delta U+\Delta p V
\)

Since \(p\) is constant, we can write
\(
\Delta H=\Delta U+p \Delta V \dots(6.8)
\)

It is important to note that when heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy.
Remember \(\Delta H=q_p\), heat absorbed by the system at constant pressure.

\(\Delta H\) is negative for exothermic reactions which evolve heat during the reaction and \(\Delta H\) is positive for endothermic reactions which absorb heat from the surroundings.

At constant volume \((\Delta V=0), \Delta U=q_{\mathrm{V}}\), therefore equation 6.8 becomes
\(
\Delta H=\Delta U=q_V
\)

The difference between \(\Delta H\) and \(\Delta U\) is not usually significant for systems consisting of only solids and/or liquids. Solids and liquids do not suffer any significant volume changes upon heating. The difference, however, becomes significant when gases are involved. Let us consider a reaction involving gases. If \(V_{\mathrm{A}}\) is the total volume of the gaseous reactants, \(V_{\mathrm{B}}\) is the total volume of the gaseous products, \(n_{\mathrm{A}}\) is the number of moles of gaseous reactants and \(n_B\) is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,

\(
\begin{aligned}
& \qquad p V_{\mathrm{A}}=n_{\mathrm{A}} \mathrm{R} T \\
& \text { and } \quad p V_{\mathrm{B}}=n_{\mathrm{B}} \mathrm{R} T \\
& \text { Thus, } p V_{\mathrm{B}}-p V_{\mathrm{A}}=n_{\mathrm{B}} \mathrm{R} T-n_{\mathrm{A}} \mathrm{R} T=\left(n_{\mathrm{B}}-n_{\mathrm{A}}\right) \mathrm{R} T \\
& \text { or } \quad p\left(V_{\mathrm{B}}-V_{\mathrm{A}}\right)=\left(n_{\mathrm{B}}-n_{\mathrm{A}}\right) \mathrm{R} T \\
& \text { or } \quad p \Delta V=\Delta n_g \mathrm{R} T \dots(6.9)
\end{aligned}
\)
Here, \(\Delta n_g\) refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.
Substituting the value of \(p \Delta V\) from equation 6.9 in equation 6.8 , we get
\(
\Delta H=\Delta U+\Delta n_g R T \dots(6.10)
\)
The equation 6.10 is useful for calculating \(\Delta H\) from \(\Delta U\) and vice versa

Example 6.5: If water vapour is assumed to be a perfect gas, the molar enthalpy change for vapourisation of \(1 \mathrm{~mol}\) of water at \(1 \mathrm{bar}\) and \(100^{\circ} \mathrm{C}\) is \(41 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Calculate the internal energy change, when \(1 \mathrm{~mol}\) of water is vapourised at \(1 \mathrm{bar}\) pressure and \(100^{\circ} \mathrm{C}\).

Answer: (i) The change \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)
\(
\Delta H=\Delta U+\Delta n_g \mathrm{R} T
\)
or \(\Delta U=\Delta H-\Delta n_g \mathrm{R} T\), substituting the values, we get
\(
\begin{aligned}
& \Delta U=41.00 \mathrm{~kJ} \mathrm{~mol}^{-1}-1 \\
& \times 8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 373 \mathrm{~K} \\
& =41.00 \mathrm{~kJ} \mathrm{~mol}^{-1}-3.096 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& =37.904 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&
\end{aligned}
\)

(b) Extensive and Intensive Properties

In thermodynamics, a distinction is made between extensive properties and intensive properties. An extensive property is a property whose value depends on the quantity or size of matter present in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc. are extensive properties.

Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example, temperature, density, pressure, etc. are intensive properties. A molar property, \(\chi_{\mathrm{m}}\), is the value of an extensive property \(\chi\) of the system for \(1 \mathrm{~mol}\) of the substance. If \(n\) is the amount of matter, \(\chi_{\mathrm{m}}=\frac{\chi}{n}\) is independent of the amount of matter. Other examples are molar volume, \(V_{\mathrm{m}}\) and molar heat capacity, \(C_{\mathrm{m}}\). Let us understand the distinction between extensive and intensive properties by considering a gas enclosed in a container of volume \(V\) and at temperature \(T\) [Fig. 6.6(a)]. Let us make a partition such that the volume is halved, each part [Fig. 6.6 (b)] now has one-half of the original volume, \(\frac{V}{2}\), but the temperature will still remain the same i.e., \(T\). It is clear that volume is an extensive property and temperature is an intensive property.

(c) Heat Capacity

It is the amount of heat required to raise the temperature of a system through \(1^{\circ} \mathrm{C}\).

The increase of temperature is proportional to the heat transferred
\(
q=\text { coeff } \times \Delta T
\)
The magnitude of the coefficient depends on the size, composition and nature of the system.
We can also write it as \(q=C \Delta T\)
The coefficient, \(C\) is called the heat capacity.
Thus, we can measure the heat supplied by monitoring the temperature rise, provided we know the heat capacity. When \(C\) is large, a given amount of heat results in only a small temperature rise. Water has a large heat capacity i.e., a lot of energy is needed to raise its temperature. \(C\) is directly proportional to the amount of substance. 

Molar Heat Capacity

It is the amount of heat required to raise the temperature of \(1 \mathrm{~mole}\) of a substance through \(1^{\circ} \mathrm{C}\).
i.e., \(\quad C_m=\frac{C}{n}\)

The molar heat capacity of a substance, \(C_m=\left(\frac{C}{n}\right)\), is the heat capacity for one mole of the substance and is the quantity of heat needed to raise the temperature of one mole by one-degree celsius (or one kelvin).

Specific Heat Capacity

Specific heat, also called specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one-degree Celsius (or one kelvin). To find out the heat, \(q\), required to raise the temperatures of a sample, we multiply the specific heat of the substance, \(\mathrm{c}\), by the mass \(\mathrm{m}\), and temperatures change, \(\Delta T\) as
\(
q=c \times m \times \Delta T=C \Delta T \dots(6.11)
\)

Types of heat capacities or molar heat capacities

Heat capacity at constant volume \(\left(\mathrm{C}_{\mathrm{V}}\right)\)
At constant volume, \(\mathrm{q}_{\mathrm{v}}=\mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}=\Delta \mathrm{U}\)
Heat capacity at constant pressure \(\left(\mathrm{C}_{\mathrm{p}}\right)\)
At constant pressure, \(\mathrm{q}_{\mathrm{p}}=\mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}=\Delta \mathrm{H}\)

(d) The Relationship between \(C_p\) and \(C_V\) for an Ideal Gas

The difference between \(C_p\) and \(C_V\) can be derived for an ideal gas as:

For a mole of an ideal gas,
\(
\begin{aligned}
\Delta H & =\Delta U+\Delta(p V) \\
& =\Delta U+\Delta(\mathrm{R} T) \\
& =\Delta U+\mathrm{R} \Delta T
\end{aligned}
\)
\(
\therefore \Delta H=\Delta U+\mathrm{R} \Delta T \dots(6.12)
\)

On putting the values of \(\Delta H\) and \(\Delta U\), we have
\(
\begin{aligned}
& C_p \Delta T=C_V \Delta T+\mathrm{R} \Delta T \\
& C_p=C_V+\mathrm{R} \\
& C_p-C_V=\mathrm{R} \dots(6.13)
\end{aligned}
\)

Relation between ratio \(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\) and atomicity of a gas

• For monoatomic gases \(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{V}}=1.66\)
• For Diatomic gases \(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}=1.40\)
• For Triatomic gases \(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{V}}=1.33\)