21Thermodynamics

📚
THERMODYNAMICS
▢ Introduction:
❖ Definition: Thermodynamics deals with conversion of heat into mechanical energy

Table 1: Thermodynamic Variables

Feature
Intensive variables
Extensive variables
Dependence
Do not depend on size or amount of substance
Depend on size or amount of substance
Depend on
Nature of substance
Amount of system
Examples
Temperature, pressure, density, refractive index, surface tension, boiling point, melting point
Mass, volume, entropy, enthalpy, free energy
▢ Zeroth Law of Thermodynamics:
❖ Statement: If two bodies are separately in thermal equilibrium with a third body, then they are also in thermal equilibrium with each other
❖ Formula: If \(T_A=T_C\) and \(T_B=T_C\), then \(T_A=T_B\)
❖ Importance: Basis of temperature measurement
▢ Work:
❖ Definition: Work is done by gas during expansion and work is done on gas during compression

Table 1: Thermodynamic Work

Point
Description / Formula
Work during expansion
Work done by system = positive
Work during compression
Work done on system = negative
Depends on
Path, initial state and final state
PV diagram
Work done = area under PV curve
Formula
\(W=\int_{V_1}^{V_2}P\,dV\)
Cyclic clockwise process
Work done positive
Cyclic anticlockwise process
Work done negative
Net work in cycle
Area enclosed by PV diagram
Order of work done
\(W_{isobaric}>W_{isothermal}>W_{adiabatic}>W_{isochoric}\)
▢ Heat:

Table 1: Heat in Thermodynamics

Point
Description
Nature
Path dependent
Heat entering system
Positive
Heat leaving system
Negative
▢ Internal Energy:
❖ Symbol: \(U\)

Table 1: Internal Energy

Point
Description
Nature
State function
Depends on
Initial and final state only
Does not depend on
Path
Main dependence
Temperature
Cyclic process
\(\Delta U=0\)
Order
\(U_{gases}>U_{liquids}>U_{solids}\)
Ideal gas
Internal energy depends only on temperature
Real gas
Internal energy depends on temperature and volume
▢ First Law of Thermodynamics:
❖ Meaning: Law of conservation of energy
❖ Statement: Heat supplied to a system is used to increase internal energy and to do external work
❖ Formula: \(\Delta Q=\Delta U+\Delta W\)

Table 1: Sign Convention

Quantity
Positive
Negative
\(\Delta Q\)
Heat supplied to system
Heat extracted from system
\(\Delta U\)
Increase in internal energy
Decrease in internal energy
\(\Delta W\)
Expansion / work done by system
Compression / work done on system
▢ Thermodynamic Processes:
❖ Reversible vs Irreversible:

Table 1: Reversible and Irreversible Processes

Feature
Reversible process
Irreversible process
Definition
Changes in heat and work are exactly retraced in reverse direction
Path for heat and work is not retraced
Nature
Ideal process
Natural process
Occurrence
Almost impossible in nature
Processes in nature are irreversible
Examples
Ideal slow frictionless process
Pouring milk into tea, stirring, work against friction
❖ Isothermal Process:
◉ Definition: Temperature remains constant
◉ Condition: Change should be very slow
◉ Equation: \(PV=constant\)
◉ First Law: \(\Delta Q=\Delta W\)
◉ Internal Energy: \(\Delta U=0\)
◉ Work Done: \(W=nRT\ln\frac{V_2}{V_1}=nRT\ln\frac{P_1}{P_2}=\Delta Q\)
◉ Important Points:
  • •
    Obeys Boyle's law
  • •
    Work done depends on temperature and expansion ratio
  • •
    In cyclic and isothermal processes, change in internal energy is zero
❖ Adiabatic Process:
◉ Definition: No heat exchange with surroundings
◉ Conditions:
  • •
    Wall of container must be perfectly insulating
  • •
    Change must be sudden
◉ Heat Exchange: \(\Delta Q=0\)
◉ Equations:
  • •
    \(PV^\gamma=constant\)
  • •
    \(TV^{\gamma-1}=constant\)
  • •
    \(T^\gamma P^{1-\gamma}=constant\)
◉ First Law: \(0=\Delta U+\Delta W\Rightarrow \Delta U=-\Delta W\)
◉ Work Done: \(W=\frac{1}{1-\gamma}(P_2V_2-P_1V_1)=\frac{nR}{1-\gamma}(T_2-T_1)=-\Delta U\)
◉ Entropy: \(\Delta S=\frac{\Delta Q}{T}=0\)
◉ Important Points:
  • •
    Work done in adiabatic change depends only on change in temperature
  • •
    During adiabatic expansion, final pressure is less than isothermal expansion
  • •
    During adiabatic compression, final pressure is more than isothermal compression
❖ Isochoric / Isometric Process:
◉ Definition: Volume remains constant
◉ Condition: \(\Delta V=0\)
◉ Law: Obeys Gay-Lussac's law
◉ Work Done: \(W=0\)
◉ First Law: \(\Delta Q=\Delta U\)
◉ Heat: \(\Delta Q=nC_V\Delta T\)
❖ Isobaric Process:
◉ Definition: Pressure remains constant
◉ Condition: \(\Delta P=0\)
◉ Law: Obeys Charles law
◉ First Law: \(\Delta Q=\Delta U+\Delta W\)
◉ Formula: \(nC_P\Delta T=nC_V\Delta T+P\Delta V\)
◉ Work Done: \(W=P\Delta V=nR\Delta T\)
◉ Ratio: \(\Delta Q:\Delta U:\Delta W=(f+2):f:2\)
▢ Slopes and Elasticities:

Table 1: Slope of PV Curves

Process
Equation
Slope
Isothermal
\(PV=constant\)
\(\left(\frac{dP}{dV}\right)_{iso}=-\frac{P}{V}\)
Adiabatic
\(PV^\gamma=constant\)
\(\left(\frac{dP}{dV}\right)_{adi}=-\gamma\frac{P}{V}\)
Relation
—
Adiabatic slope = \(\gamma\times\) isothermal slope

Table 2: Elasticity of Gases

Process
Elasticity
Isothermal elasticity
\(E_{iso}=P\)
Adiabatic elasticity
\(E_{adi}=\gamma P\)
Isochoric elasticity
\(E_{choric}=\infty\)
Isobaric elasticity
\(E_{baric}=0\)
❖ Important Point: Adiabatic curve is steeper than isothermal curve
▢ Internal Energy in Processes:

Table 1: Change in Internal Energy

Process
\(\Delta U\)
Isobaric
Positive during heating
Isothermal
Zero
Adiabatic expansion
Negative
Cyclic
Zero
Order
\(\Delta U_{isobaric}>\Delta U_{isothermal}>\Delta U_{adiabatic}\)
▢ Second Law of Thermodynamics:

Table 1: Statements of Second Law

Statement
Meaning
Kelvin statement
It is impossible for a cyclic heat engine to convert whole heat extracted from a reservoir completely into work
Clausius statement
Heat cannot flow from colder body to hotter body by itself without external agency
❖ Important Points:
  • •
    Whole work can be converted into heat
  • •
    Whole heat cannot be converted into work
  • •
    Heat naturally flows from hot body to cold body
▢ Heat Engine:
❖ Definition: Device that continuously converts heat energy into mechanical work through cyclic process
❖ Working:
  • •
    Working substance takes heat \(Q_1\) from source
  • •
    Converts part of heat into work \(W\)
  • •
    Rejects remaining heat \(Q_2\) to sink
  • •
    Returns to initial state
  • •
    \(\Delta U=0\) for complete cycle

Table 1: Heat Engine Formulae

Quantity
Formula
Heat rejected
\(Q_2=Q_1-W\)
Work done
\(W=Q_1-Q_2\)
Efficiency
\(\eta=\frac{W}{Q_1}\times100\)
Efficiency
\(\eta=\frac{Q_1-Q_2}{Q_1}\times100\)
Efficiency
\(\eta=\left(1-\frac{Q_2}{Q_1}\right)\times100\)
Carnot efficiency
\(\eta=\left(1-\frac{T_2}{T_1}\right)\times100\)
❖ Efficiency Increase:
  • •
    Increase temperature of source
  • •
    Decrease temperature of sink
▢ Carnot Engine:
❖ Definition: Ideal reversible heat engine working on Carnot cycle
❖ Parts:
  • •
    Hot reservoir / source
  • •
    Cylinder with insulating wall
  • •
    Perfectly conducting base
  • •
    Perfect gas as working substance
  • •
    Frictionless insulated piston
  • •
    Non-conducting stand
  • •
    Sink
❖ Carnot Cycle:

Table 1: Four Strokes of Carnot Cycle

Step
Process
Path
1
Isothermal expansion
AB
2
Adiabatic expansion
BC
3
Isothermal compression
CD
4
Adiabatic compression
DA
❖ Efficiency:

Table 1: Carnot Engine Efficiency

Quantity
Formula / Point
Efficiency
\(\eta=\frac{W}{Q_1}\times100\)
Efficiency
\(\eta=\left(1-\frac{Q_2}{Q_1}\right)\times100\)
Temperature form
\(\eta=\left(1-\frac{T_2}{T_1}\right)\times100\)
Heat-temperature relation
\(\frac{Q_1}{Q_2}=\frac{T_1}{T_2}\)
Between steam point and ice point
\(\eta=26.81\%\)
❖ Carnot Theorem: No irreversible engine can have efficiency greater than Carnot reversible engine working between same hot and cold reservoirs
❖ Important Points:
  • •
    Efficiency depends only on source temperature \(T_1\) and sink temperature \(T_2\)
  • •
    Efficiency is independent of nature of working substance
  • •
    Carnot engine is most efficient engine
  • •
    Efficiency cannot be 100% because sink temperature can never be 0 K
  • •
    All reversible heat engines working between same hot and cold reservoirs have same efficiency
  • •
    Change in entropy of working substance in Carnot cycle is zero because it returns to initial state
  • •
    Carnot cycle contains only two isothermal and two adiabatic processes
▢ Types of Heat Engines:
❖ External vs Internal Combustion:

Table 1: Combustion Engines

Feature
External combustion engine
Internal combustion engine
Fuel burning
Fuel burnt outside main cylinder
Fuel burnt inside main cylinder
Examples
Steam engine
Petrol engine, diesel engine
❖ Steam Engine:

Table 1: Steam Engine

Point
Description
Working substance
Steam
Cycle
Rankine cycle
Use
Powerful engine; drags long trains
Nature
Less efficient and heavy
Efficiency
10% to 20%
❖ Petrol Engine:

Table 1: Petrol Engine

Point
Description
Cycle
Otto cycle
Working substance
Air 98% + petrol 2%
Use
Light vehicles, scooters, cars, aeroplanes
Efficiency
40% to 50%
Engine type
Four-stroke engine
◉ Four Strokes:
  1. Charging stroke
  2. Compression stroke → adiabatic compression
  3. Working / power stroke → adiabatic expansion
  4. Exhaust stroke
❖ Diesel Engine:

Table 1: Diesel Engine

Point
Description
Discovered by
Rudolf Diesel
Working substance
Air 98% + diesel 2%
Efficiency
55% to 70%
Nature
Very heavy engine
Engine type
Four-stroke engine
Spark
Does not require spark
◉ Four Strokes:
  1. Charging stroke
  2. Compression stroke → adiabatic compression
  3. Working / power stroke → adiabatic expansion
  4. Exhaust stroke
❖ Petrol and Diesel Engine Efficiency:
◉ Formula: \(\eta=1-\left(\frac{V_2}{V_1}\right)^{\gamma-1}\)
◉ Compression Ratio Form: \(\eta=1-\left(\frac{1}{\rho}\right)^{\gamma-1}\)
◉ Symbols:
  • •
    \(\gamma=\frac{C_p}{C_v}\) of air
  • •
    \(\rho=\frac{V_1}{V_2}\) = compression ratio
◉ Important Points:
  • •
    Petrol and diesel engines have air as working substance
  • •
    Petrol and diesel oil are used for ignition
  • •
    Useful work is done in third stroke
  • •
    Third stroke is called power stroke
▢ Refrigerator / Heat Pump:
❖ Definition: Heat engine working in reverse direction
❖ Working:
  • •
    Working substance takes heat \(Q_2\) from lower temperature body \(T_2\)
  • •
    External work \(W\) is supplied
  • •
    Heat \(Q_1=Q_2+W\) is rejected to hot body/surrounding at \(T_1\)
❖ Refrigerant: Freon, ammonia, etc.

Table 1: Refrigerator Formulae

Quantity
Formula
Heat rejected
\(Q_1=Q_2+W\)
Work input
\(W=Q_1-Q_2\)
Coefficient of performance
\(\beta=\frac{Q_2}{W}\)
COP
\(\beta=\frac{Q_2}{Q_1-Q_2}\)
Temperature form
\(\beta=\frac{1}{\frac{T_1}{T_2}-1}\)
❖ Open Refrigerator Door: If refrigerator door is kept open inside a room, room temperature increases
▢ Read and Digest:

Table 1: Important Thermodynamics Points

Fact
Point
Milk poured into tea and stirred
Irreversible process
Work done against friction
Irreversible process
Isobaric process ratio
\(\Delta Q:\Delta U:\Delta W=(f+2):f:2\)
Cyclic process
\(\Delta U=0\)
Isothermal process
\(\Delta U=0\)
Isothermal work
Depends on temperature and expansion ratio
Adiabatic work
Depends only on change in temperature
Ideal gas internal energy
Depends only on temperature
Real gas internal energy
Depends on temperature and volume
Increase Carnot efficiency
Increase \(T_1\) or decrease \(T_2\)
Reversible engines between same reservoirs
Same efficiency
Carnot engine
Most efficient engine
Four-stroke heat engine
Power obtained only in third stroke
Same compression volume range
Final pressure in adiabatic compression > isothermal compression
Carnot cycle
Two isothermal + two adiabatic processes
▢ High-Yield Recall:

Table 1: Thermodynamics One-Liners

Fact
Answer
Thermodynamics
Heat ↔ mechanical energy
Intensive variables
Temperature, pressure, density
Extensive variables
Mass, volume, entropy, enthalpy
Zeroth law
Basis of thermal equilibrium
Work done by gas
Positive during expansion
Work done on gas
Negative during compression
Work in PV diagram
\(W=\int P\,dV\)
Clockwise cyclic process
Positive work
Anticlockwise cyclic process
Negative work
Heat entering system
Positive
Heat leaving system
Negative
Internal energy
State function
First law
\(\Delta Q=\Delta U+\Delta W\)
Isothermal process
\(T=constant,\ \Delta U=0\)
Isothermal equation
\(PV=constant\)
Isothermal work
\(W=nRT\ln\frac{V_2}{V_1}\)
Adiabatic process
\(\Delta Q=0\)
Adiabatic equation
\(PV^\gamma=constant\)
Adiabatic work
\(W=\frac{P_2V_2-P_1V_1}{1-\gamma}\)
Isochoric process
\(V=constant,\ W=0\)
Isochoric first law
\(\Delta Q=\Delta U\)
Isobaric process
\(P=constant\)
Isobaric work
\(W=P\Delta V\)
Adiabatic slope
\(\gamma\times\) isothermal slope
Isothermal elasticity
\(P\)
Adiabatic elasticity
\(\gamma P\)
Kelvin statement
Whole heat cannot be converted into work
Clausius statement
Heat cannot flow cold → hot by itself
Heat engine work
\(W=Q_1-Q_2\)
Heat engine efficiency
\(\eta=\left(1-\frac{Q_2}{Q_1}\right)\times100\)
Carnot efficiency
\(\eta=\left(1-\frac{T_2}{T_1}\right)\times100\)
Carnot cycle
2 isothermal + 2 adiabatic
Carnot engine between ice and steam point
26.81%
Steam engine cycle
Rankine cycle
Petrol engine cycle
Otto cycle
Diesel engine
No spark required
Power stroke
Third stroke
Petrol/diesel efficiency
\(\eta=1-\left(\frac{1}{\rho}\right)^{\gamma-1}\)
Refrigerator
Reverse heat engine
Refrigerator COP
\(\beta=\frac{Q_2}{Q_1-Q_2}\)
Refrigerator COP temperature form
\(\beta=\frac{1}{T_1/T_2-1}\)
Q1.
An inventor claims to have made an engine which consumes 1g of fuel per second (of calorific value 2 K cal/gm) and delivers 10 KW of power. Mark the correct statement
📅BP 2010
Q2.
Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and volume V. The mass of gas in A is mA, and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The change in pressure in A and B are found to be ΔP and 1.5ΔP respectively. Then:
📅BP 2009
Q3.
The pressure and volume are changing but the temperature is constant in the process:
📅IOM 2012
Q4.
A refrigerator has to transfer an average of 263J of heat per second from -10°C to 25°C. The average power consumed by the refrigerator is
📅IOM 2010/2009
Q5.
If a gas is allowed to expand adiabatically against external pressure:
📅MOE 2009
Q6.
Efficiency of Carnot engine working between 27°C and 127°C is
📅MOE 2011
Q7.
In a Carnot engine, the temperature of the heat sink is 27°C and that of the source is 327°C. The efficiency is:
📅MOE 2012
Q8.
One mole of an ideal gas with γ = 1.4 is adiabatically compressed so that its temperature rises from 27°C to 35°C. The change in internal energy of the gas is
📅MOE 2012
Q9.
The efficiency of a Carnot engine is 1/5. On reducing the temperature of sink by 45°C, efficiency becomes 1/3. The initial temperature of the sink was
📅MOE 2014
Q10.
The efficiency of a Carnot engine is 20%. On reducing the temperature of sink by 45°C, efficiency becomes 33.3%. The initial temperature of the sink was
📅MOE 2014
Q11.
In an isothermal condition
📅KU 2014
Q12.
If 1500 cal of heat is supplied to a system and 1000 J of work is done, what is the increase in internal energy?
📅IOM 2014
Q13.
What happens in adiabatic process?
📅KU 2013
Q14.
A Carnot engine kept at temperature at 800K and 400K, the output of cycle is 800J. Then the energy supplied by the source is
📅KU 2013
Q15.
In an isothermal process
📅KU 2009
Q16.
During adiabatic compression of 5 moles of gas, 250 J work was done, the change in internal energy will be:
📅IE 2012
Q17.
The equation of adiabatic process is
📅IE 2012
Q18.
A Carnot engine with efficiency η=10% works same as heat engine, it is made to work with refrigerator having work done =10J. The heat transferred is
📅TE 2013
Q19.
If a gas is allowed to expand adiabatically against external pressure
📅MOE 2009
Q20.
A Carnot engine takes in 3000 kcal of heat from a reservoir at 627°C and gives it to a sink at 27°C. The work done by the engine is
📅MOE 2010
Q21.
The specific heat capacity of an ideal gas under isothermal condition is
📅IOM 1997
Q22.
Which of the following is not correct?
📅IOM 1997
Q23.
Find out the work done from the graph:
📅Graph-based question
Q24.
If one mole of an ideal gas at STP is heated through 1K, the work done by the gas in heat unit will be
📅MOE Curriculum
Q25.
A Carnot engine takes 300 calories of heat from a source at 500K and rejects 150 calories of heat to the sink. The temperature of the sink is
📅MOE 2065
Q26.
The maximum efficiency of an engine operating between 30°C and 300°C is
📅MOE 2061
Q27.
An inflated tyre of a bicycle bursts. Which of the following relation between pressure P and temperature T holds good if γ is the ratio of the specific heats of air?
📅MOE 2000
Q28.
When a gas undergoes adiabatic expansion, its internal energy:
📅KU 2008
Q29.
Internal energy of an ideal gas depends on
📅KU 2008
Q30.
A Carnot engine has the same efficiency between 800K and 500K and xK to 600K. The value of x is
📅IE 2004
Q31.
Two steam engines A and B, A working between temperature 650K and 700K and another B working between temperature 300K and 350K. Then
📅Bangladesh 2009