Thermodynamics in Physics

Thermodynamics is the field of physics concerned with the interplay among heat, work, temperature, and energy and encompasses principles governing the behaviour of macroscopic systems. Thermodynamics is foundational in understanding and designing various systems across engineering, Physics, Chemistry, and Biology, providing essential principles for energy conversion, heat transfer and system equilibrium studies.

1.0Thermodynamics System and Important Terms

A thermodynamic system can be represented in pressure (P), volume (V), and temperature (T). A system is a specified portion of matter consisting of one or more substances on which the effects of thermodynamic variables such as temperature, volume, and pressure are to be studied. For example, a gas enclosed in a cylinder fitted with a piston is a system.

• Heterogeneous System- The heterogeneous System is not uniform throughout. For example, a system consisting of two or more immiscible liquids is said to be heterogeneous.
• Homogeneous System—A system is considered homogeneous if it is completely uniform throughout, e.g., a pure solid or liquid.
• Surroundings-Anything outside the System that exchanges energy with the System and tends to change its properties is called its surroundings.
• Universe-The system and its surroundings are together known as the universe.
• Boundary-Real or imaginary wall that separates the system from the surroundings

Types of Systems

(1) Open System - Any interaction is possible.

(2) Closed System - Impenetrable by matter, but other  interactions can occur.

(3) Semipermeable - Penetrable by some species, but not by others; rest interactions possible.

(4) Insulated System-Thermal interactions are impossible, but nonthermal interactions can occur.

(5) Rigid System - Boundary cannot be mechanically deformed.

(6) Isolated System - No interactions can occur.

• Intensive-Independent of the size (or mass) of the system. Example:- pressure, temperature, etc.
• Extensive-depends on the size (or mass) of the system. Example:-volume, number of moles, etc.

Thermodynamic variables of the system :

• Composition (m)             
• Temperature (T)                
•  Volume (V)   
• Pressure (P)                    
•  Mass

2.0Thermodynamic State

• The state of a system can be described completely by composition, temperature, volume, and pressure. If a system is homogeneous and has definite mass and composition, then the remaining three variables, namely temperature, pressure, and volume, can describe the state of the system. These variables are interrelated by the equation PV = µRT. The thermodynamic state of the system is its condition as identified by two independent thermodynamic variables (P, V or P, T or V, T).

3.0Process

• When a system changes from one equilibrium state to another, the path of successive states through which the system passes is called a process.

• Cyclic Process-It is a thermodynamic process in which the system returns to its initial stage after undergoing a series of changes.
• Non–cyclic process-It is a process in which the system does not return to its initial stage.
• Quasi–static or equilibrium process-It A thermodynamic process that occurs very slowly, ensuring that at every moment, the temperature and pressure remain uniform throughout the system, is referred to as a quasi-static process..
• Reversible and Irreversible processes-A reversible process is one in which the changes in heat and work of direct process from initial to a final state are exactly retraced in the opposite sense in the reverse process, and the system and surroundings are left in their initial states. Reversibility is an ideal concept that can not be realized in practice. The process which is not reversible is the irreversible process. In nature the processes are irreversible.

4.0Heat and Work in Thermodynamic Processes

• Work and heat are both forms of energy. Work is ordered energy, whereas heat is disordered energy.
• Conversion of mechanical work into heat is entirely possible but the reverse is not true.
•    W=JQ (∴ J = Conversion Factor=Joule’s Constant=Mechanical Equivalent of       heat
• J=4.18 $\frac{\text{ Joule }}{\text{ calorie }}$
• 1 cal=4.18 ⋍ 4.2 Joule
• 1 cal > 1 joule

5.0Internal Energy

• The internal energy of a system arises from both the motion and configuration of its molecules. The energy associated with molecular motion is known as  internal kinetic energy (U_k) ,while the energy related to molecular configuration is referred to as internal potential energy (U_p).
• Internal energy in the absence of intermolecular forces is simply the function of temperature and state only, it is independent of the path followed. $\Delta$U = U_f – U_i

(∴ U_i = Internal energies in initial state and U_f = Internal energies in final state)

• For ideal gas it is a function of temperature, identity and moles.
• It depends mainly on the states of matter and temperature of the system.
• The Absolute value of internal energy is impossible to determine, so it is generally expressed as a difference between the two states of a system, or by applying zero value to a reference state. $\mathrm{dU}=\mathrm{d}{U}_k+\mathrm{d}{U}_p$

6.0Work done by Thermodynamic System

• If the gas expands against the piston, It exerts a force on it and displaces it through a distance, but it does work on the piston.

dW=PdV

$W={\int }_{V_i}^{V_f}dW={\int }_{V_i}^{V_f}PdV=$

$\Rightarrow$Work done by  gas is equal to the area under P-V graph

7.0Zeroth Law of Thermodynamics(ZLOT)

• According to the Zeroth Law of Thermodynamics, two objects (or systems) are in thermal equilibrium if they are separately in thermal equilibrium with a third object (such as a thermometer) and hence have the same temperature.
• ZLOT defines temperature only.
• If two systems are in thermal equilibrium with the third system separately, then they must also be in thermal equilibrium with each other.

8.0First Law of Thermodynamics(FLOT

• Suppose some heat is supplied to a system capable of doing external work. In that case, the heat absorbed by the system is equivalent to the combined effect of the increase in its internal energy and the external work it executes. $\Delta Q=\Delta U+\Delta \mathrm{W}$
• FLOT introduces the concept of internal energy.
• The first law of thermodynamics is based on the law of energy conservation.

9.0Sign Convention used in Physics Thermodynamics

                    $\Delta Q=\Delta U+\Delta \mathrm{W}$

10.0Work done in Various Thermodynamic Processes

Mathematical Method $\mathrm{W}={\int }_{V_i}^{V_f}PdV$

• Cyclic Process $\Rightarrow$ Work = Area of the closed curve in PV Graph
• Isochoric Process $\Rightarrow$Work = 0
• Isothermal Process $\Rightarrow$Work = $=\mu RT{\log }_e\left[\frac{V_f}{V_i}\right]=\mu RT{\log }_e\left[\frac{P_i}{P_f}\right]$
• Adiabatic Process $\Rightarrow$Work $=\mu R\frac{\left(T_f-T_i\right)}{1-\gamma }$
• Isobaric $\Rightarrow$Work = $\mathrm{P}\left(V_f-V_i\right)=\mu R\left(T_f-T_i\right)$
• Polytropic Process $\Rightarrow$ Work=$\frac{\mu R\left(T_f-T_i\right)}{1-x}$

Graphical Method

$\Delta$W = Area enclosed between PV curve and Volume axis

Work done $\Rightarrow W_{A\to B}$ = Area Shaded

 Note

• Work and heat are both path functions, it depends on the path.
• Internal energy is determined solely by the initial and final states of the system, making it a state function.

•   Order of Work

$\Delta W_c>\Delta W_b>\Delta W_a$

• Order of Internal energy

$\mathrm{d}{U}_a=\mathrm{d}{U}_b=\mathrm{d}{U}_c\text{ (Constant) }$

• Order of Heat

$\Delta Q_c>\Delta Q_b>\Delta Q_a$

11.0Cyclic Process

• W= Area enclosed between closed loop

In P-V graph direction

$(1)Clockwise(C.W.)=\Delta W=\text{ Positive }$

$(2)Anticlockwise(A.C.W)=\Delta W=\text{ Negative }$

In V-P graph direction

• $\text{ Clockwise(C.W.) }=\Delta W=\text{ Negative }$
• $\text{ Anticlockwise(A.C.W) }=\Delta W=\text{ Positive }$

12.0Different Thermodynamic Process

1. Isochoric Process

• Volume= Constant
• Equation of state $\Rightarrow \frac{P}{T}=\mathrm{constant}$
• Work done $\mathrm{w}={\int }_{V_i}^{V_f}PdV=0(\therefore \mathrm{dV}=0)$

 Form of First Law, $\mathrm{Q}=\Delta U=\mu C_v\Delta T$

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

2. Isobaric Process

• Pressure = constant
• Equation of state$\Rightarrow \frac{V}{T}=\mathrm{constant}(\mathrm{V}\propto \mathrm{T})$
• Work done $\mathrm{w}={\int }_{V_i}^{V_f}PdV=\mathrm{P}\left(V_f-V_i\right)$
• Form of First Law $\mathrm{Q}=\Delta U+\mathrm{P}\left(V_f-V_i\right)$              

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

• Slope of the PV curve ${\left(\frac{dP}{dV}\right)}_{\text{isobaric }}=0$

3. Isothermal Process

• Temperature= constant
• Equation of state PV=Constant
• Work done $\mathrm{w}=\mu RT{\log }_e\left[\frac{V_2}{V_1}\right]=2.303\mu RT{\log }_{10}\left[\frac{P_1}{P_2}\right]$
• Form of First Law $\Delta$U=0
• Slope of isothermal curve${\left[\frac{dP}{dV}\right]}_{\text{isothermal }}=-\frac{P}{V}$

4. Adiabatic Process

• No heat exchange between system and surroundings
• Work done $W=\frac{1}{\gamma -1}\left[P_1{V}_1-P_2{V}_2\right]=\frac{\mu R}{\gamma -1}\left(T_1-T_2\right)$
• Slope of adiabatic curve ${\left[\frac{dP}{dV}\right]}_{\text{adiabatic }}=\frac{-{\rm Y}P}{V}$

Note-Magnitude of slope of adiabatic curve is greater than slope of isotherm

5. Polytropic Process

• $\mathrm{P}{V}^x=\text{ Constant },\mathrm{C}=C_v+\frac{R}{1-x}$
• For isobaric $\mathrm{x}=0,\mathrm{C}=C_P=\frac{R}{\gamma -1}+\mathrm{R}=C_v+R$
• For isothermal x=1,C=∞
• For adiabatic x=,C=0

Drawbacks of FLOT

• No information about what part of heat converted into mechanical work.
• Does not give proper direction of heat flow.

13.0Second Law of Thermodynamics

1. Kelvin planck statement : It states that in a cyclic process total heat can not be converted into mechanical work.
2. Claussius statement : It is impossible to have net heat flow from a low temperature body to a high temperature body.

14.0Heat Engine

• Efficiency of Heat Engine, $(\eta )=\frac{\text{ Useful output }}{\text{ Input }}$

$\Rightarrow \eta =\frac{\text{ Net work }}{\text{ heat given }}$

$\Rightarrow \eta =\frac{w}{Q_1}=\frac{Q_1-Q_2}{Q_1}$

• Percentage Efficiency,

$\eta =\frac{T_1-T_2}{T_{}}\times 100\%$

• If T₁ = ∞ K or T₂ = 0 K then = 100% (That is practically impossible)
• Efficiency depends on temperature of source (T₁K) and temperature of sink (T₂ K)

15.0Carnot Cycle

Enclosed area represents work done by the engine.

• It is a hypothetical engine because the working substance is an ideal gas.
• Its efficiency is maximum but not 100%.
• This cyclic process has two isothermal and two adiabatic processes

16.0Refrigerators

• It takes heat from a place at lower temperature and gives it off to a place at relatively higher temperature.
•  Refrigerator is just opposite to heat engine
• Coefficient of Performance (COP) of refrigerator is,

$\mathrm{COP}(\beta )=\frac{\text{ Heat absorbed from Cold }}{W\text{ ork }}=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}$

$\mathrm{COP}(\beta )=\left(\frac{T_2}{T_1-T_2}\right)$

Relation between ( and )       

$\eta =\frac{1}{1+\beta }$                

$\beta =\frac{1}{n}-1$

17.0Sample Questions On Thermodynamics

Q-1. If Q amount of heat is given to a diatomic ideal gas in the process in which the gas performs the work $\frac{2Q}{3}$ on its surroundings. Find the molar heat capacity (in terms of R) for the process.

Solution:

Q=n C d T

d Q=d U+d W $\Rightarrow$ d U=d Q-d W $\Rightarrow$ d U=Q-$\frac{2}{3}$ Q=$\frac{Q}{3}$

$dU=\frac{Q}{3}\left(\therefore dU=n{C}_{V}dT\right)$

$n\times \frac{5}{2}R\times dT=\frac{Q}{3}$

$ndT=\frac{2Q}{15R}$

$Q=nCdT\Rightarrow (ndT)C$

$Q=(ndT)C$

$Q=\frac{2Q}{15R}\times C$

$C=\frac{15R}{2}$        

Q-2. A gas under constant pressure of 4.5×10⁵ Pa when subjected to 800 kJ of heat, changes the volume from 0.5m³ to 2.0 m³. Find the change in internal energy of the gas?

Solution:

$\Delta U=\Delta Q-\Delta W=\Delta Q-P\Delta V$

$\Delta U=800\times 10^3-4.5\times 10^5(2-0.5)=1000(800-675)$

$\Delta U=125\mathrm{kJ}$

Q-3. 1kg of water at 373 K is converted into steam at the same temperature. Volume of 1cm³ of water becomes 1671 cm³ on boiling. What is the change in the internal energy of the system given that the latent heat of vaporisation of water is 5.4 × 10⁵ cal kg^–1?

Solution:

Volume of 1 kg of water = 1000 cm³ = 10^–3 m³

Volume of 1 kg of steam = 10³ × 1671 cm³ = 1.671 m³ 

Change in volume $\Delta \mathrm{V}$= (1.671 – 10^–3) m³ = 1.670 m³, Pressure

P = 1 atm = 1.01 × 10⁵ N m^–2

In expansion work ,W=P

$\Delta V=1.01\times 10^5\times 1.67\mathrm{J}=\frac{1.686\times 15^5}{4.2}cal$ = 4.015 ×10⁴ cal

But $\Delta U=Q-W$  (First law of thermodynamics)

$\Delta U$ = (5.4 × 10⁵ – 0.4015 × 10⁵) cal = 4.9985 × 10⁵ cal

Q-4.Find the molar heat capacity (in terms of R) of a monatomic ideal gas undergoing the process PV^1/2 = constant?

Solution:

Use for polytropic process,

$\mathrm{C}={\mathrm{C}}_{\mathrm{v}}+\frac{R}{1-x}$

PV^x = constant, by comparison $x=\frac{1}{2}$

$\mathrm{C}=\frac{f}{2}\mathrm{R}+\frac{R}{1-x}$

$C=\frac{3}{2}R+\frac{R}{1-\frac{1}{2}}=\frac{7}{2}R$

( ∴ ${\mathrm{C}}_{\mathrm{v}}=\frac{f}{2}R$, degree of freedom for Monatomic gas is 3)

Q-5 A Carnot engine takes 10³ kilocalories of heat from a reservoir at 627°C and exhausts the remaining heat to a sink at 27°C. What will be the efficiency of the engine?

Solution:

Q₁ = 10³ × 10³ cal = 10⁶ × 4.2 J

T₁ = 900 K

T₂ = 300 K

$\eta =\frac{T_1-T_2}{T_1}\times 100=\frac{600}{900}\times 100=67\%$