Process in Thermodynamics

In thermodynamics, a process refers to the sequence of changes a system undergoes as it moves from one state to another. These changes involve the transfer of energy and matter. Understanding these processes is crucial in thermodynamics for analyzing how systems exchange energy, perform work, and respond to various changes. A thermodynamic process transforms a system from one equilibrium state to another. This transformation involves changes in the system's properties, such as pressure, volume, and temperature. Understanding thermodynamic processes is essential in engineering, chemistry, and physics. It helps in designing engines and refrigerators and understanding natural phenomena. This theory forms the basis for analyzing and understanding how energy and matter interact in various systems, providing critical insights for practical and theoretical applications.

1.0Definition of Thermodynamics Processes

A thermodynamic process is a series of transformations that a thermodynamic system undergoes as it changes from one equilibrium state to another. This process involves alterations in the system's properties, such as pressure, volume, and temperature. It is characterized by how energy is transferred to or from the system, typically heat and work. The critical aspects of the thermodynamics process are:

1. Change in State: A thermodynamic process involves a change in the state of a system, which is defined by its thermodynamic variables like temperature, pressure, and volume. State functions such as internal energy and entropy describe the initial and final states.
2. Energy Transfer: During the process, work and heat transfer energy between the system and its surroundings. Understanding how this energy is transferred and how it affects the system is crucial for analyzing the process.

2.0Types of Processes

Isometric or Isochoric Process

• The isochoric process is a thermodynamic process that takes place at a constant volume of the system,but pressure and temperature vary with the change in the system's state.
• Equation of state : $\frac{P}{T}$ = constant (P and T are variable, V is constant)
• Work done-In this process volume remains constant so dV=0 

$W={\int }_{V_1}^{V_2}PdV=0$

• The Form of First Law- Q=$\Delta$U, it means whole of the heat is utilized for change in internal energy of the system.

$\mathrm{Q}=\Delta U=\mu C_V\Delta T$

• The Slope of the P-V curve- $\frac{dP}{dV}=\infty$

Isobaric Process

• An isobaric process is a thermodynamic process that occurs under constant pressure, but the volume and temperature vary with the change in the system's state.
• Equation of state-  $\frac{V}{T}=\text{ Constant or }V\propto T$
• Work Done- In this process pressure remains constant (∴dP=0)

$\mathrm{W}={\int }_{V_i}^{V_f}PdV=P\left(V_f-V_i\right)$

• Fraction of heat given which-

1. Is converted into Internal Energy $\frac{\Delta U}{\Delta Q}=\frac{1}{\gamma }$
2. Does work against external Pressure $\frac{\Delta W}{\Delta Q}=1-\frac{1}{\gamma }$

• Form First Law of Thermodynamics

$\mathrm{Q}=\Delta U+P\left(V_f-V_i\right)$

$C_p\left(T_f-T_i\right)=\mu C_v\left(T_f-T_i\right)+P\left(V_f-V_i\right)$

It is clear that heat supplied to the system is utilized for:      

(i) Increasing internal energy and

(ii) Work done against the surrounding atmosphere. 

• The Slope of the PV Curve

${\left(\frac{dP}{dV}\right)}_{\text{isobaric }}=0$

Isothermal Process

• In this process, the pressure and volume of the system change, but the temperature remains constant. Heat exchange between the system and the surroundings is allowed in an isothermal process. The isothermal process is carried out by either supplying heat to the substance or extracting heat from it. A process has to be extremely slow to be isothermal.
• Equation of State - P V = Constant (\mu R T) T is Constant.
• Work Done - Consider µ moles of an ideal gas, enclosed in a cylinder, at absolute temperature T, fitted with a frictionless piston. Suppose that gas undergoes an isothermal expansion from the initial state (P₁, V₁) to the final state $\left(P_2,V_2\right)$

$\mathrm{W}={\int }_{V_1}^{V_2}PdV(\therefore \mathrm{PV}=\mu RT\text{ then }\mathrm{P}=\frac{\mu RT}{V}$

$W={\int }_{V_1}^{V_2}\frac{\mu RT}{V}dV=\frac{\mu RT}{V}{\int }_{V_1}^{V_2}\frac{dV}{V}=\mu RT\left[{\log }_e{V}_2-{\log }_e{V}_1\right]=\mu RT\log e_e\left[\frac{V_2}{V_1}\right]$

$\Rightarrow \mathrm{W}=2.303\mu RT{\log }_{10}\left[\frac{P_1}{P_2}\right]$

$\left(\because P_1{V}_1=P_2{V}_2\right)$

• Form of First Law

There is no change in temperature and internal energy of the system depends on temperature only.

$\Delta$U=0

$Q=2.303\mu RT{\log }_{10}\left[\frac{V_2}{V_1}\right]$

$\Rightarrow$ It is clear that the whole of the heat energy supplied to the system is utilized by the system for external work.There is no change in the system's internal energy.

• Slope of Isothermal Curve

For an Isothermal Process, PV = Constant

Differentiating we get

$\Rightarrow PdV+VdP=0$

$\Rightarrow VdP=-PdV$

$\Rightarrow \frac{dP}{dV}=-\frac{P}{V}$

Slope of Isothermal Curve,

${\left[\frac{dP}{dV}\right]}_{\text{isothermal }}=\frac{-P}{V}$

For a given system

• The product of the pressure and volume of a given mass of a perfect gas remains constant in an isothermal process.
• Boyle’s Law is obeyed in an isothermal process.
• A graph depicting the relationship between pressure and volume for a given mass of gas at a constant temperature is called an isothermal curve or isotherm.
• Two isotherms for a given gas at two different temperatures T₁ and T₂ are shown in figure

• The curves drawn for the same gas at different temperatures are mutually parallel and do not cut each other.
• If two isotherms intersect each other at a single point we get the same value of P and V at intersection point. PV = µRT₁ for temperature T₁ and PV = µRT₂ for temperature T₂.It means T₁ = T₂ which is not possible.
• An ideal gas is enclosed in a conducting cylinder fitted with a conducting piston.
• Let the gas expand very–very slowly. This will cause a prolonged cooling of the gas, but heat from the surroundings will be conducted into the cylinder. Therefore, the gas's temperature will remain unchanged. If the gas is compressed very slowly, heat will be produced, but this heat will be conducted to the surroundings, and the gas's temperature will remain constant.
• The temperature of a substance stays the same during the melting process, which is isothermal.
• Boiling is an isothermal process; when a liquid boils, its temperature remains constant.
• If sudden changes are executed in a vessel of infinite conductivity, they will be isothermal.

Adiabatic Process

It is a thermodynamic process in which the system's pressure, volume, and temperature change, but there is no exchange of heat between the system and its surroundings. A sudden and quick process will be adiabatic since there is no sufficient time available for the exchange of heat.

• Equation of State

$\mathrm{P}{V}^{\gamma }=\text{ Constant },\mathrm{T}{V}^{\gamma -1}=\text{ Constant },T^{\gamma }{P}^{1-\gamma }=\text{ Constant }$

• Work done - Let initial state of the system is (P₁, V₁, T₁) and after adiabatic change the final state of system is (P₂, V₂, T₂) then we can write,

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }=K(\text{ Here }K\text{ is Constant })$

$\mathrm{W}={\int }_{V_1}^{V_2}\mathrm{P}\mathrm{dV}=\mathrm{K}{\int }_{V_1}^{V_2}{V}^{-\gamma }dV=K{\left(\frac{V^{-\gamma +1}}{\gamma +1}\right)}_{V_1}^{V_2}=\frac{K}{(-\gamma +1)}\left[V_2^{-\gamma +1}-V_1^{-\gamma +1}\right]$

$\left(\because P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }\right)$

$W=\frac{1}{\gamma -1}\left[P_1{V}_1{V}_1^{-\gamma }\cdot V_1-P_2{V}_2^{\gamma }{V}_2^{-\gamma }\cdot V_2\right]=\frac{1}{\gamma -1}\left[P_1{V}_1-P_2{V}_2\right]$

$W=\frac{\mu R}{\gamma -1}\left[P_1{V}_1-P_2{V}_2\right]$

$\mathrm{W}=\frac{\mu R}{\gamma -1}\left[T_1-T_2\right]$

From the  First Law : dU = – δW 

• This means the work done by an ideal gas during adiabatic expansion (or compression) is at the cost of a change in internal energy proportional to the gas's fall (or rise) in temperature.
• If the gas expands adiabatically, the gas does the work. So, W_adia is positive.
• The gas cools during adiabatic expansion and T₁ > T₂.
• If the gas is compressed adiabatically, work is done on the gas. So, W_adia is negative.
• The gas heats up during adiabatic compression and T₁ < T₂.

The Slope of the Adiabatic Curve

For an Adiabatic Process, $P{V}^{\gamma }=\text{ Constant }$

Differentiating both sides

$\mathrm{yP}{V}^{\gamma -1}\mathrm{dV}+V^{\gamma }\mathrm{dP}=0$

$V^{\gamma }\mathrm{dp}=-\mathrm{yp}{V}^{\gamma -1}\mathrm{v}$

$\frac{dP}{dV}=-\frac{\gamma P{V}^{\gamma -1}}{V^{\gamma }}=-\mathrm{Y}\frac{P}{V}$

The Slope of the  adiabatic curve, ${\lfloor \frac{dP}{dV}\rfloor }_{\text{Adiabatic }}=\frac{-\gamma P}{V}$

• The Magnitude of the  slope of adiabatic curve is greater than the slope of Isotherm

${\left[\frac{dP}{dV}\right]}_{\text{adiabatic }}=\gamma \frac{P}{V}=\gamma {\left[\frac{dP}{dV}\right]}_{\text{isothermal }}\Rightarrow \frac{\text{ slope of adiabatic changes }}{\text{ slope of isothermal changes }}=\gamma$

Since $\gamma$ is always greater than one so an adiabatic curve is steeper than an isothermal curve.

Examples of Adiabatic Process

• If the gas is suddenly compressed by pushing the piston downward in a thermally insulated cylinder with a non-conducting piston, heat is generated. This heat cannot escape from the cylinder, increasing the gas's temperature.
• If a gas is suddenly expanded by moving the piston outwards, its temperature will decrease.
• Bursting of a cycle tube.
• Propagation of sound waves in a gaseous medium.
• In diesel engines, burning of diesel without a spark plug is done due to adiabatic compression of diesel vapour and air mixture.

3.0Reversible Processes

• A reversible process is one in which a change can be exactly reversed by altering the conditions. In this type of process, every change that occurs during the forward direction is precisely undone in the reverse direction.
• If the system does some work in the direct process, an equal amount of work must be done on it in the reverse process.
• If the system absorbs some heat in the direct process, it must release equal heat to the surroundings in the reverse process.
• At the end of the reversible process, both the system and surroundings must return to their initial states.

Necessary Conditions for a Reversible Process:

1. The process must be quasi-static and carried out infinitesimally slowly. This ensures that the system remains in thermal and mechanical equilibrium with its surroundings.
2. The dissipative forces, such as viscosity, friction, etc., should be absent.

4.0Irreversible Process

• Any process that cannot be retraced precisely in the reverse direction is called irreversible. Most of the processes occurring in nature are irreversible.
• An irreversible process in thermodynamics is a transformation where the system and its surroundings cannot be returned to their original states without leaving some net changes. Departures from equilibrium characterize this process and involve dissipative effects that result in the loss of valuable energy.

Conditions for an Irreversible Process:

• Presence of Dissipative Effects--The process involves the dissipation of energy in the form of heat, friction, or turbulence. Examples include frictional heating, unrestrained expansion, or mixing of substances. These effects lead to a loss of helpful work and increased entropy.
• Non Ideal Conditions-Real-world factors such as friction, viscosity, electrical resistance, and chemical reaction kinetics contribute to irreversibility. For example, due to non-ideal behaviour and frictional losses, the process will be irreversible in a natural gas undergoing expansion.

Examples of Reversible and Irreversible Processes

1. Examples of Reversible Processes 

• An infinitesimally slow compression and expansion of an ideal gas at constant temperature.

2. Example of Irreversible Process

• Diffusion of Gases
• Sudden expansion or contraction of gas
• Rusting of Iron

5.0Sample Questions on Process in Thermodynamics

Q-1. For a monoatomic gas, if the work done at constant pressure is W, What is the quantity of heat required at constant volume to achieve the same temperature increase?

Solution: 

For Monoatomic Gas $\left(C_V=\frac{3}{2}R,C_P=\frac{5}{2}R\right)$

$\Delta Q=\Delta U+\Delta W$

$C_P\Delta T=C_V\Delta T+W$

$\frac{5}{2}R\Delta T=\frac{3}{2}R\Delta T+W$

W= R$\Delta$T

$\Rightarrow$ For Isochoric Process, V=Constant

$\Delta Q=\Delta U+\Delta W$

$\Delta Q=\Delta U+P\Delta V$

$\Delta Q=\Delta U$

$\Delta Q=C_V\Delta T=\frac{3}{2}R\Delta T$

$\Delta Q=\frac{3}{2}W(\because \mathrm{W}=\mathrm{R}\Delta T)$

Q-2. A gas expands so that  its pressure and volume comply with the condition PV²=Constant. Will the gas cools or get heated on expansion?

Solution:

For a perfect gas, $\frac{PV}{T}=\mathrm{Constant}(k)$

$\left(\because \mathrm{P}{V}^2=\mathrm{Constant}\left({\mathrm{K}}^{\prime }\right)\right)$

$\mathrm{P}=\frac{K^{\prime }}{V^2}$

$\frac{K^{\prime }}{V^2}\cdot \frac{V}{T}=K\text{ or }\mathrm{VT}=\frac{K^{\prime }}{K}=\text{ Constant }$

Hence $\mathrm{V}\propto \frac{1}{T}$

$\Rightarrow$ This expansion of gas will result in the decrease of temperature.

Q-3. What happens to the change in internal energy of a gas during

1. Isothermal Expansion
2. Adiabatic Expansion

Solution:

1. Isothermal Expansion- Temperature remains constant during an isothermal change, as internal energy is a function of temperature only, so it will remain constant during an isothermal change.

$\Delta T=0$

$\Delta U=C_V\Delta T=0$

2. Adiabatic Expansion - For an adiabatic change 

$\Delta Q=0$

So from First Law of Thermodynamics

$\Delta Q=\Delta U+\Delta W$

$\Delta W=-\Delta U$

During Expansion, Work is done by a gas i.e. W is positive. So U must be        Negative. Hence Internal Energy of a gas decreases during an adiabatic expansion.