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Thermodynamics formulas include Entropy, Heat capacity, Isothermal Processes, Adiabatic processes, among many others.
- The study of the transfer of heat and other aspects of energy from one substance to another is called Thermodynamics.
- How matter changes when heat and energy are applied to a particular substance or object is the study of Thermodynamics.
- Any substance comprises atoms and molecules.
- A thermodynamic system is a system with defined boundaries and surroundings and the transfer of temperature(T), Pressure(P), and Volume(V) between atoms and molecules.
Key Terms: Thermodynamics, Isothermal Process, Adiabatic Process, Heat Capacity, Quasi-Static Process, Energy, Heat, Temperature, Work
What is Thermodynamics?
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Thermodynamics is one of the many branches of Physics that deals with the transfer of work, temperature, heat, and energy from one substance to another or from one form to another.
- Some thermodynamic systems that can be found in our everyday life are, Air conditioners, Refrigerators, washing machines, and all other heating and cooling systems, which function on the principles of Thermodynamics.
- The laws of thermodynamics defines how a system’s energy changes, alongside whether the system can perform efficient work on its surroundings.
- It simply explains how thermal energy is converted to or from different forms of energy, and how matter has been affected due to it.
There are numerous Thermodynamic formulas, including Enthalpy of Vaporization formula, Entropy formula, Adiabatic process, among many others.
Enthalpy of Vaporization Formula
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Enthalpy of Vaporization is the quantity of energy required to turn a liquid substance into its gaseous form. It is also known as Heat of Vaporization or Heat of Evaporation. It is denoted by ΔHvap. The formula for Heat of Vaporization is,
ΔHvap = ΔUvap + pΔV
Where,
- ΔUvap= increased internal energy and
- pΔV= work done under ambient pressure
The SI unit for the heat of vaporization is J or Cal
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Entropy Formula
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Entropy can be defined as the measure of the randomness and disturbance in the system.
- Entropy is highest if the boundaries of the system are not defined.
- Solid substances or closed-space substances will have less entropy whereas gaseous substances will have high entropy.
- The entropy of a substance depends on two factors- Heat and Temperature.
- Entropy changes according to the temperature applied.
- A lower temperature is equal to More entropy whereas a Higher Temperature applied is equal to Less Entropy.
- Hence we can say that Entropy is inversely proportional to the temperature applied to the system. So to put it in a formula,
| ΔS = \(\frac{q_{(rev)}}{T}\) |
where, q(rev) is the heat applied on the system.
The SI unit for entropy is J/K
Boltzmann’s Entropy Formula
Boltzmann’s Entropy formula talks about the relationship between the entropy of the system and the atoms and molecules arranged in the system. It is formulated as:-
S = kb log W
‘kb ‘is Boltzmann’s constant which is equal to 1.380649 x 10-23 J/K and ‘log’ is the natural logarithm function and W is the microscopic configurations. ‘W’ can also be written as Ω (Omega).
Read More: Boltzmann Equation: Statement, Equation, Applications and Examples
Law of Thermodynamics
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There are two Laws of Thermodynamics. All the thermodynamics systems and processes obey both the law of thermodynamics.
First Law of Thermodynamics
The first law of Thermodynamics states that ‘the energy can neither be created nor be destroyed, it can only be changed from one form to another’. It states the relationship between work, heat, and internal energy between molecules of the system.

First law of thermodynamics
The work of a thermodynamics system can be calculated as negative external pressure applied on the thermodynamic system into the change of volume of a thermodynamic system, W= -p ΔV
The first law of thermodynamics can be represented as:
| ΔU = Q - W |
- ΔU is the net sum of the internal change of energy of the system.
- Q is the sum of the heat between the system.
- And work is the quantity of energy exchanged between thermodynamics.
The first law of Thermodynamics is also called as ‘Law of Conservation of Energy
Second Law of Thermodynamics
The Second Law of Thermodynamics states that the ‘Entropy of an isolated system will never be zero’. It can also be represented mathematically as
| ΔSuniv = ΔSsys + ΔSsurr ≥ 0 |
Here, S represents the entropy.
The second Law of Thermodynamics is also called as Law of Increased Entropy.
Heat Capacity Formula
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The heat capacity formula can be denoted as the product of mass, specific heat and the change in the temperature. It can be shown as, Q = mc Δ T.
Here,
- m is the mass of the substance
- T is the change in the temperature of the substance and
- C is the Specific Heat Capacity of the substance.
Specific Heat Capacity Formula
Specific heat is defined as the heat required to change the temperature of a substance by 1 °C. The substance should not be of changing volume.
Hence the specific heat capacity formula can be associated with the energy required to change the temperature of a substance by 1 unit respective of 1kg mass volume of the substance. The formula of Specific heat Capacity is:
| Specific Heat Capacity = \(\frac{Energy Required}{Mass * ΔT}\) |
Hence,
C = \(\frac{Q}{mΔt}\)
- Here Q is the heat energy applied
- m is the mass of a substance in kg
- and t is the temperature change
The changes in temperature can be calculated as
Δ T = (Tf – Ti)
Where Tf is the final temperature and Ti is the initial temperature in K.
The SI unit for the Specific Heat Capacity formula is J/Kg.K or J/Kg.oC
Thermodynamic Processes
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The many Thermodynamic Process include:
Isothermal Process
Isothermal Processes are those processes where the temperature of the system remains constant. The equal amount of transfer of heat between the inside and outside of the system maintains the state equilibrium and temperature constant. The pressure and volume of the system can change or vary the in Isothermal process.
Quasi-Static Process
Changes in the system happen very slowly in Quasi-Static Process. The changes in temperature, pressure, or volume are slower. Physically these processes do not show any changes for a long time and chemically they maintain a state of thermodynamic equilibrium.
Reversible Process
As the name suggests, Reversible Processes are processes in which change in the substance and its surroundings can be restored to its initial state by retracing the steps and following a reverse path.
Adiabatic Process
Adiabatic Processes are processes where there is no transfer of Heat. No exchange of heat takes place during the compression or decompression of the system. Such a process can be either reversible or irreversible. The
Adiabatic Process formula is:
| PV\(\gamma\)= constant |
Here,
- P is the pressure applied to the system
- V is the volume of the system and
- \(\gamma\) is the Adiabatic Index.
Adiabatic Index can be calculated as the ratio of heat capacity at constant value Cp to the heat capacity at constant volume Cv.
Work in an Adiabatic Process:
Work done in an Adiabatic Process depends on the pressure and volume of the substance. Similar to the Adiabatic process formula which depends on Pressure and Volume. The work done in an Adiabatic process can be calculated as
W = ∫ Pdv
- Where W is equal to the work done
- P is the pressure applied on the substance whereas
- V is the volume of the substance
Read Also: Adiabatic Process Derivation: Formula, Examples & Equation
Mayer’s Formula
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Mayer’s Formula explains the relationship between molar specific heat of a gas at constant pressure and volume.
Consider one mole of gas. Let us assign ‘P’ as Pressure, ‘V’ as Volume, and ‘T’ as Temperature.
Let there be the change in temperature from T to dT at constant volume applied on the gas.
Considering First Law of Thermodynamics-
ΔU=ΔQ -ΔW
ΔQ= ΔU+ ΔQ
= ΔU + PΔV
Since we are considering Volume as constant, ΔV=0
Hence ΔQ= ΔU
Cv = (ΔQ/ΔT)v
Cv = (ΔU/ ΔT) v
Cv = (ΔU/ ΔT) v
Now, considering Pressure as constant,
ΔQ= ΔU+PΔV
Cp= (ΔQ/ ΔT) p
Cp= (ΔQ/ ΔT)p + P(ΔV/ΔT)p
Cp= (ΔQ/ ΔT)p + P(ΔV/ΔT)p
Now for an ideal molar gas,pressure,volume will be equal to gas constant and temperature
PV=RT
ΔV/ΔT=R/P
Cp=(ΔU/ ΔT)p + P*R/P
Cp= Cv + R
| Cp = - Cv = R |
This is known as Mayer’s formula and the SI unit for Cp,Cv and R is joule/mole oC
Read Also:
| Concepts Related to Adiabatic Process | ||
|---|---|---|
| Reversible and Irreversible Processes | Zeroth law of Thermodynamics | MCQ on Thermodynamics |
| Carnot engine | Thermodynamics Important Questions | Kelvin Planck Statement |
Things to Remember
- All the processes which occur in nature are guided by Thermodynamics laws.
- Adiabatic Processes are processes where there is no transfer of Heat.
- Boltzmann’s Entropy formula talks about the relationship between the entropy of the system and the atoms and molecules arranged in the system.
- Mayer’s Formula can be represented as, Cp = - Cv = R.
- Specific heat Capacity can be represented as, Specific Heat Capacity, = Energy Required / Mass ΔT.
- Enthalpy of Vaporization can be shown as, ΔHvap=ΔUvap+ pΔV.
Previous Year Questions
- An ideal monoatomic gas is confined in a cylinder by a spring-loaded piston… [JEE Main 2014]
- A Carnot freezer takes heat from water at 0oC… [JEE Main 2016]
- The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is …?
- A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 Nm-2) requires 54 cal...[NEET 2018]
- Thermodynamic processes are indicated in the following diagram...[NEET 2017]
- A carnot engine having an efficiency of 1/10 as heat engine, is used as a refrigerator...[NEET 2017]
- A carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should...[NEET 2006]
- A gas can be taken from A to B via two different processes… [JEE Main 2019]
- A gas is compressed from a volume of 2m3… [JEE Main 2014]
Sample Questions
Ques. Under what conditions, the adiabatic process takes place? (1 mark)
a) Q = 0
b) T = 0
c) P = 0
d) W = 0
Ans. a) Q = 0
Explanation: When heat transfer is 0, the adiabatic process can be seen to take place.
Ques. Can we calculate heat capacity using the Specific heat capacity of the substance? (1 mark)
Ans. Q = mcΔT
- M = mass of the substance
- C = Specific Heat Capacity of the substance.
- T= change in the temperature of the substance
Ques. What is the first law of Thermodynamics? (1 mark)
Ans. The first law of thermodynamics says that energy can neither be created nor be destroyed but can be transformed from one form to another
Ques. What is the change in internal energy for an isolated system at constant volume? (2 marks)
Ans. For an isolated system, there is no transfer of heat, energy, or work, hence w=0 and q = 0.Using the first law of thermodynamics:
Δ U = q + w
= 0 + 0 = 0
ΔU = 0
Ques. The difference between Cp and Cv can be derived using the Mayer’s formula. Calculate the difference between Cp and Cv for five moles of an ideal gas. (2 marks)
Ans. For 1 mole of an ideal gas
Cp – Cv = R
For 5 moles of gas, the relation is Cp – Cv = 5R = 10 × 4.184 J
Cp – Cv = 20.925 J.
Ques. The enthalpy of vapourization of Co2 is 15.326 kJ/mol Calculate the heat required for the vapourization of 185g of Co2 at constant pressure. (Molar mass of CCl4 = 44.01 g/mol. (2 marks)
Ans. qp = ΔH = 15.326 kJ/mol
The heat required for vapourization of 185g of Co2 is
=(185/44.01 g/mol ) x 15.326 kJ/mol
= 64.41 kJ
Ques. It has been found that 121.4J is needed to heat 30g of ethanol from 14 0C to 18 0C. Calculate (a) the specific heat capacity, and (b) the molar heat capacity of ethanol. (3 marks)
Ans. (a) c=Q/mΔt
=121.4/30 x (18-14)
=121.4/30 x 4
=1.001J/g.0C
1 0C is equal to 1k, the specific heat capacity of ethanol = 1.001J/g.0C
(b) Molar heat capacity, Cm = specific heat x molar mass.
Cm (ethanol) = 1.001 x 46
= 46.046 Jmol-1 0c-1
Ques. What are the conditions when heat absorbed by the system under certain specific conditions is independent of the path? Explain in detail. (3 marks)
Ans. 1) At constant volume
By first law of thermodynamics:
q = ΔU + (–w)
(–w) = pΔV
Hence q = ΔU + pΔV
Since volume is constant, ΔV = 0
qV = ΔU + 0
qV = ΔU = change in internal energy
2) At constant pressure
qp = ΔU + pΔV
But, ΔU + pΔV = ΔH
qp = ΔH = change in enthalpy.
Ques. Determine the quantity out of ΔrG and ΔrGӨ that is going to be zero at equilibrium? (2 marks)
Ans. ΔrG is always going to be zero.
ΔrGӨ is zero for K = 1 since ΔrGӨ = – RT ln K
ΔrGӨ is going to be non-zero for other values of K.
Ques. What is the change in internal energy in case of an isolated system at constant volume? (2 marks)
Ans. There is going to be no energy transfer because heat or work in an isolated system,
Hence, w = 0 and q = 0.
As per the first law of thermodynamics-
Δ U = q + w = 0 + 0 = 0
ΔU = 0
Ques. Determine the relationship between ΔH and ΔU for ideal gases. (5 marks)
Ans. To indicate the volume change
VA =volume of gaseous reactants
VB =volume of gaseous products
nA=moles of reactant
nB =moles of product
At constant pressure and temperature,
pVA = nART and pVB = nBRT
Subtracting both the equations,
pVB − pVA = (nB − nA) RT
pΔV = Δ ngRT
Δng=nB−nA
Now substituting the value of pΔV, we get
ΔH = ΔU + ΔngRT
Heat change at constant pressure,ΔH = qp
Heat change at constant temperature, ΔH = qV
Hence ,for gaseous system
qP = qV + ΔngRT
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