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# NEET Study Notes for Thermodynamics, Simple Definitions with Important Formulas and Previous Year Questions Thermodynamics is a branch of Physics which deals with processes involving heat, work and internal energy. The system is defined as the region of space or quantity of matter under investigation or study. Everything external to that system is called environment or surroundings.

• The topic of Thermodynamics carries a weightage of around 8-10% in NEET that means 2-3 questions or around 8- 12 marks can be asked from this section.
• Some of the important sub-topics of Thermodynamics include- Thermal equilibrium, first law of thermodynamics, second law of thermodynamics, etc. Check  NEET Physics Syllabus

Thermodynamics also deals with macroscopic behaviour instead of microscopic behaviour of this system and understanding of such topics is very important to solve questions from this section. Given below in the article are quick notes and important questions on these topics with tips and tricks to help you prepare for NEET 2020

## Thermodynamic Processes

The process by which one or more parameters of the thermodynamic system undergo changes is called thermodynamic processes. It includes:

1. Isothermal Process: In this change in pressure and volume takes place at a constant rate and the total amount of heat in the system does not remain constant.
2. Isobaric Process: In this process, change in volume and temperature of a gas takes place at constant level.
3. Isochoric Process: In this process, change in pressure and temperature takes place in such a way that volume of the system remains constant.
4. Adiabatic Process: In this process, change in volume, pressure and temperature takes place without any heat entering or leaving the system.

#### Sample Question

Question: “CT “, given below is the molar heat capacity. Find its value at constant temperature?

(A) 0

(B) 0 < CT < Cv

(C) Cv < CT < Cp

(D) CT =

Solution: (A)

First Law of Thermodynamics

## First Law of Thermodynamics

First law of Thermodynamics states that if a quantity of heat is supplied to a system capable of doing work, then the quantity of heat absorbed by the system is equal to the sum of increase in the internal energy of the system, i.e,

dQ= dU+ dW

Application of first law of dynamics-

1. Cooling caused in Adiabitic Process: dT=PdV/Cv
2. Melting: dU= mLf
3. Boiling: Du=mLv – P(Vf -Vi)
4. Mayer’s Formula: Cp – Cv = R

#### Sample Question

Question: Read the processes given below and identify which of these violate the First law of Thermodynamics?

(A) W > 0, Q < 0, and ?Eint > 0

(B) W > 0, Q < 0, and ?Eint < 0

(C) W < 0, Q > 0, and ?Eint < 0

(D) W > 0, Q > 0, and ?Eint < 0

Solution: (D)

Second Law of Thermodynamics

## Second Law of Thermodynamics

It states that state of entropy of the entire universe is always increasing with time and the changes in the universe can never be negative.

1. Clausius Statement: It states that heat cannot flow from a cold body to a hot body without performance of work by some external agency.
2. Kelvin’s Statement: According to Kelvin’s statement, it is Impossible to obtain a continuous supply of energy by cooling a body below the coldest of its surroundings.
3. Planck’s Statement: It says that it is impossible to extract heat from a single body and convert the whole of it into work.

#### Sample Question

Question: How much heat is delivered to a higher temperature reservoir, if a heat pump uses 300 J or work to remove 400 J of heat from the low temperature reservoir?

1. 500 J
2. 100 J
3. 700 J
4. 1500 J

Solution: (C)

What is Carnot Engine?

## Carnot Engine

Carnot Engine is a theoretical concept which gives the estimate of the maximum possible efficiency that a heat engine during the conversion of heat into work can possess.

Efficiency of Carnot Engine: Efficiency η of an engine is defined as the ratio between useful heat (heat converted into work) to the total heat supplied to the engine.

η = W / Q1 = [Q1- Q2]/ Q1 = 1-[Q2/Q1] = 1- T2/T1

Carnot’s cycle:

Given below are steps involved in Carnot’s cycle:

1. Isothermal Expansion: W1= RT1 loge[V2/V1]
2. Adiabatic Expansion: W2= R/y-1 [T1-T2]
3. Isothermal Expansion: W3= RT2 logeV3/V4
4. Adiabatic Compression: W4= R/γ-1 [T1-T2]
5. Total work done: W = W1+ W2+ W3+ W4 = R (T1-T2) loge (V2/V1)

#### Sample Question

Question: If a Carnot engine having an efficiency of 40% takes in heat from a source maintained at a temperature of 500k, then, what will be the intake temperature from the same source, if the engine’s efficiency now is 60%?

1. 750 K
2. 1200 K
3. 1000K
4. 600 K

Solution: (A)

Question: How much heat energy is extracted from a high temperature reservoir per cycle if efficiency is 33% and work output is 24J per cycle?

1. 10J
2. 72 J
3. 28 J
4. 16 J

Solution: (B)

### Some Important definitions to remember

SystemPart of the universe under investigation.
Open SystemA system which can exchange both energy and matter with its surroundings.
Closed SystemA system which permits passage of energy but not mass, across its boundary.
Isolated systemA system which can neither exchange energy nor matter with its surrounding.
SurroundingsPart of the universe other than system, which can interact with it.
BoundaryAnything which separates system from surrounding.
State variablesThe variables which are required to be defined in order to define state of any system i.e. pressure, volume, mass, temperature, surface area, etc.
State FunctionsProperty of system which depend only on the state of the system and not on the path. Example: Pressure, volume, temperature, internal energy, enthalpy, entropy etc.
Intensive propertiesProperties of a system which do not depend on mass of the system i.e. Temperature, pressure, density, concentration,
Extensive propertiesProperties of a system which depend on mass of the system i.e. Volume, energy, enthalpy, entropy etc.
ProcessPath along which state of a system changes.

Important Formulas

## Important Formulas

1. Work:
1. dW = Fdx= Pdv
2. W= P(Vf-Vi)
2. ­Difference between heat Capacities (Mayer’s Formula) :
3. Specific Heat capacity at constant pressure: Cp = MCP
4. Specific Heat capacity at constant value: Cv=MCv
5. First Law of Thermodynamics= dQ= dU+ dW
6. 1 joule= 4.186 cal
7. Equation of State for ideal gas : PV= RT
1. Cp – Cv = R/J
2. For 1 g of gas, cp – cv = r/J
3. Adiabatic gas Constant, y= Cp /Cv = cp/ cv
8. Relation of CV with energy: CV= 1/m (dU/dT)
9. Adiabatic Gas Equation: PVY = constant
10. Slope for PV diagram:
1. For isobaric process: zero
2. For isochoric process: infinite
11. Enthalpy: H= U+PV

### Important Topics on Thermodynamics

Thermodynamics is a important and vast topic when it comes to NEET. It is important to give due attention to this topic. Not only this, there are many sub topics in Thermodynamics which also require a lot of time and attention. These are given below:

1. Zeroth Law of Thermodynamics
2. First Law of Thermodynamics
3. Second Law of Thermodynamics
4. Thermal Equilibrium
5. Concept of Temperature
6. Heat, Work and Energy
7. Enthalpy
8. Cornet Engine and Cornet Cycle
9. Thermodynamic Processes
10. Reversible Processes and irreversible processes
11. Specific Heat Capacity of Gases
12. Comparison of slopes of Isothermal and adiabatic Curves
13. Change in internal energy

Previous Year Solved Questions

### Previous Year Solved Sample Questions

Question: In the figure given below, in which path between initial state “I” and final state “f”, the work done on gas is greatest?

(A)- A

(B)- B

(C ) – C

(D)- D

Solution: (D)

Question: A Carnot freezer takes heat from water at 0oC and rejects it at the the temperature of 27oC. Given: the latent heat of ice is 336×103 J kg−1. Find the energy consumed by the freezer if 5 kg of water at 0oC is converted into ice at 0oC by the freezer.

(A) 1.67 ×× 105 J

(B) 1.68 ×× 106 J

(C ) 1.51 ×× 105 J

(D) 1.71 ×× 107 J

Solution: (A)

Question: What is the efficiency of Carnot energy which discharges 3 J of heat into low temperature reservoir for every 2 J of work output.

1. 1/3
2. 2/5
3. 3/5

(D) 2/3

Solution: (A)

Question: Find, for which of the given processes the would the entropy change be zero?

1. Isobaric

(B) Isothermal

(D) Constant volume

(E) None of these

Solution: (C )

Question: 200 g water is heated from 40oC to 60oC. Find the change in its internal energy where heat of water = 4184 J/kg/K. (The small change in expansion of water can be ignored.)

(A) 8.4 kJ

(B) 4.2 kJ

(C )16.7 kJ

(D) 167.4 kJ

Solution: (C )

### Some tips & tricks to solve questions

Some tips and tricks to solve the NEET paper are given below:

1. Understand the Concepts Thoroughly:

Understanding the concepts to solve thermodynamics and NEET paper is prerequisite since there will be no direct questions and each question will require a different application and approach. So, be clear and thorough with your concepts.

1. Take Practice Test/ Mock Tests/Previous Year Questions:

Taking practice test, mock tests or sectional tests is a main process to score good marks in any exam. Similar is the case with NEET exam. Solving these tests and questions will prepare you about the level of questions asked and will also help you in increasing efficiency and time management for the exam day.

1. Make a Proper Time table:

Following a proper time table is very important. It helps you to give proper time to studies and other activities. Also remember, taking breaks in between is equally important.

1. Do Revision

Take time out and revise all the important topics so that they are on your tips on the exam day.

For Exam Day,

1. Attempt easy questions first

On the day of exam, start with attempting easy questions first and then move to moderate question and if time permits then only attempt difficult questions. Also, do not spend more time on one question. If you are unable to solve it, leave that question and move to the next one.

1. Do Not panic

Stay calm and focussed and do not panic if you can’t solve one question. Breathe in, relax and solve next one, and come back to it later on, if you have time.

Read questions carefully twice and then look at the options before solving a question. Since the paper would be MCQ, many a times, the question can be solved through elimination process and does not require the complete solving and application procedure.

Study Plan for NEET Physics

## Study Plan for NEET Physics

Since almost 30 days are left for NEET, most of the students would be done with their syllabus and would be doing revision only. But the trick here is that, your revision also needs to be planned. Given below are some things that you can do in next 30 days to prepare well for NEET:

1. Make a checklist of all the important formulas and read and revise them regularly when you wake up in the morning and at night before going to bed.
2. Start giving mocks at the same time in the day, when you have exam on the final day. This will help your body and mind to get ready for the exam day.
3. After attempting mocks, analyise the paper and see what different approach you could have taken to solve that question in less time. Also solve unsolved questions.
4. Instead of mock tests, you can also try solving previous year question papers. Also remember to solve this year’s sample paper.
5. Take mock test on one day and keep next day for revision. Remember, revision and analysis of mocks are important.
6. Do not try learning new topics now, instead just give time for revision.
7. Do not attempt any mock tests, in the last week before exam. Just go through your notes and look into difficult questions.
8. Do not study on last 2-3 days before the exam. Just revise the formulas and do anything that helps you relax- watch movies, web shows, anything that you like.

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