A quasi-static process refers to an idealized or imagined process where the change in state is made infinitesimally slowly so that at each instant, the system can be assumed to be at a thermodynamic equilibrium with itself and with the environment. For instance, imagine heating \(1 \mathrm{~kg}\) of water from a temperature \(20^{\circ} \mathrm{C}\) to \(21^{\circ} \mathrm{C}\) at a constant pressure of 1 atmosphere. To heat the water very slowly, we may imagine placing the container with water in a large bath that can be slowly heated such that the temperature of the bath can rise infinitesimally slowly from \(20^{\circ} \mathrm{C}\) to \(21^{\circ} \mathrm{C}\). If we put \(1 \mathrm{~kg}\) of water at \(20^{\circ} \mathrm{C}\) directly into a bath at \(21^{\circ} \mathrm{C}\), the temperature of the water will rise rapidly to \(21^{\circ} \mathrm{C}\) in a non-quasistatic way.

Non-equilibrium states of a system are difficult to deal with. It is, therefore, convenient to imagine an idealised process in which at every stage the system is an equilibrium state. Such a process is, in principle, infinitely slow, hence the name quasi-static (meaning nearly static).

In a quasi-static state, the system changes its variables \((P, T, V)\) so slowly that it remains in thermal and mechanical equilibrium with its surroundings throughout. In a quasi-static process, at every stage, the difference in the pressure of the system and the external pressure is infinitesimally small. The same is true of the temperature difference between the system and its surroundings (Fig. 12.7). To take a gas from the state \((P, T)\) to another state \(\left(P^{\prime}, T^{\prime}\right)\) via a quasi-static process, we change the external pressure by a very small amount, allow the system to equalise its pressure with that of the surroundings and continue the process infinitely slowly until the system achieves the pressure \(P^{\prime}\). Similarly, to change the temperature, we introduce an infinitesimal temperature difference between the system and the surrounding reservoirs and by choosing reservoirs of progressively different temperatures \(T\) to \(T^{\prime}\), the system achieves the temperature \(T^{\prime}\).

A quasi-static process is obviously a hypothetical construct. In practice, processes that are sufficiently slow and do not involve accelerated motion of the piston, large temperature gradient, etc., are reasonable approximations of an ideal quasi-static process. Since quasi-static processes cannot be completely realized for any finite change of the system, all processes in nature are non-quasi-static. Examples of quasi-static and non-quasi-static processes are shown in the figure below.

Isothermal process

A process in which the temperature of the system is kept fixed throughout is called an isothermal process. The expansion of a gas in a metallic cylinder placed in a large reservoir of fixed temperature is an example of an isothermal process. (Heat transferred from the reservoir to the system does not materially affect the temperature of the reservoir, because of its very large heat capacity.)

For an isothermal process ( \(T\) fixed), the ideal gas equation gives
\(
P V=\text { constant }
\)
1.e., the pressure of a given mass of gas varies inversely as its volume. This is nothing but Boyle’s Law.
Suppose an ideal gas goes isothermally (at temperature \(T\) ) from its initial state \(\left(P_1, V_1\right)\) to the final state \(\left(P_2, V_2\right)\). At any intermediate stage with pressure \(P\) and volume change from \(V\) to \(V+\Delta V(\Delta V\) small \()\)
\(
\Delta W=P \Delta V
\)
Taking \((\Delta V \rightarrow 0)\) and summing the quantity \(\Delta W\) over the entire process,
\(
\begin{aligned}
W & =\int_{V_1}^{V_2} P \mathrm{~d} V \\
& =\mu R T \int_{V_1}^{V_2} \frac{\mathrm{d} V}{V}=\mu R T \operatorname{In} \frac{V_2}{V_1} \dots(12.12)
\end{aligned}
\)

where in the second step we have made use of the ideal gas equation \(P V=\mu R T\) and taken the constants out of the integral. For an ideal gas, internal energy depends only on temperature. Thus, there is no change in the internal energy of an ideal gas in an isothermal process. The First Law of Thermodynamics then implies that heat supplied to the gas equals the work done by the gas : \(Q=W\). Note from Eq. \((12.12)\) that for \(V_2>V_1, W>0\); and for \(V_2<V_1, W<0\). That is, in an isothermal expansion, the gas absorbs heat and does work while in an isothermal compression, work is done on the gas by the environment and heat is released.

Key Points
\({W}=+{ve}({V_2}>{V_1})\)
\({W}=-{ve}({V_1}>{V_2})\)
If \(\Delta U=0\)

• Internal energy only depends on temperature
• If temperature \(=\) Constant
• Internal Energy = Constant.

From the first law of thermodynamics,

• \(\Delta Q=\Delta W+\Delta U\)
• \(\Delta{Q}=\Delta{W}(\Delta{U}=0)\)

Adiabatic Process

In an adiabatic process, the system is insulated from the surroundings, and the heat absorbed or released is zero. Therefore, \(Q=0\). Mathematically  an adiabatic process is represented as

\(
P V^\gamma=\text { const } \dots(12.13)
\)

where \(\gamma\) is the ratio of specific heats (ordinary or molar) at constant pressure and at constant volume.
\(
\gamma=\frac{C_p}{C_v}
\)
Thus if an ideal gas undergoes a change in its state adiabatically from \(\left(P_1, V_1\right)\) to \(\left(P_2, V_2\right)\) :
\(
P_1 V_1^\gamma=P_2 V_2^\gamma \dots(12.14)
\)
Figure \(12.8\) shows the \(P\) – \(V\) curves of an ideal gas for two adiabatic processes connecting two isotherms.

We can calculate, as before, the work done in an adiabatic change of an ideal gas from the state \(\left(P_1, V_1, T_1\right)\) to the state \(\left(P_2, V_2, T_2\right)\).
\(
\begin{gathered}
W=\int_{V_1}^{V_2} P \mathrm{~d} V \\
=\text { constant } \times \int_{V_1}^{V_2} \frac{\mathrm{d} V}{V^\gamma}=\text { constant } \times \frac{V^{-\gamma+1}}{1-\gamma} \mid V_2 \\
\frac{\text { constant }}{(1-\gamma)} \times\left[\frac{1}{V_2^{\gamma-1}}-\frac{1}{V_1^{\gamma-1}}\right] \dots(12.15)
\end{gathered}
\)
From Eq. (12.14), the constant is \(P_1 V_1^\gamma\) or \(P_2 V_2^\gamma\)
\(
\begin{aligned}
W & =\frac{1}{1-\gamma}\left[\frac{P_2 V_2^\gamma}{V_2^{\gamma-1}}-\frac{P_1 V_1^\gamma}{V_1^{\gamma-1}}\right] \\
& =\frac{1}{1-\gamma}\left[P_2 V_2-P_1 V_1\right]=\frac{\mu R\left(T_1-T_2\right)}{\gamma-1} \dots(12.16)
\end{aligned}
\)
As expected, if work is done by the gas in an adiabatic process \((W>0)\), from Eq. (12.16), \(T_2<T_1\). On the other hand, if work is done on the gas \((W<0)\), we get \(T_2>T_1\) 1.e., the temperature of the gas rises.

Key Points
According to the 1st law of the thermodynamic process

• \(\Delta{Q}=\Delta{W} + \Delta{U}\)
• \(\Delta{U}=-\Delta{W}\)
  So if work done is negative internal energy increases and vice versa.

Isochoric process

In an isochoric process, \(V\) is constant. No work is done on or by the gas. A volume change is zero, so the work done is zero.

• The volume of the system = Constant
• Change in volume \(\Delta V=0\)
• If the change in volume \(\Delta{V}=0\), then the work done is zero.

According to the first law of thermodynamic law

• \(\Delta{Q}=\Delta{W}+\Delta{U}\)
•  If \(\Delta{W}=0\), then \(\Delta{Q}=\Delta{U}\)

Isobaric process

In an isobaric process, pressure P is fixed. Work done by the gas is
\(
W=P\left(V_2-V_1\right)=\mu R\left(T_2-T_1\right) \dots(12.17)
\)
So if volume increases, work done is positive, else negative.

Since temperature changes, so does internal energy. The heat absorbed goes partly to increase internal energy and partly to do work. The change in temperature for a given amount of heat is determined by the specific heat of the gas at constant pressure.

Key Points

• Pressure \(=\) Constant during this process
• \(\Delta{W}={P} \Delta {V}\)
  So if volume increases, work done is positive, else negative.

According to the thermodynamic first law of thermodynamics

• \(\Delta Q=\Delta W+\Delta U\)
• \(\Delta {Q}={P} \Delta {V}+\Delta{U}\)

Cyclic process

In a cyclic process, the system returns to its initial state. Since internal energy is a state variable, \(\Delta U=0\) for a cyclic process. From first law of thermodynamics, the total heat absorbed equals the work done by the system. Therefore, \(\Delta Q=\Delta W\)

The definitions of these special processes are summarised in Table. 12.2 below.