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    Thermodynamics NEET, Solutions To Question Papers Of Last Six Years

    By Safeer PP
    22 Sep'22  4 min read
    Thermodynamics NEET, Solutions To Question Papers Of Last Six Years
    Synopsis

    Analysis of the previous year's question papers is essential to understand the focus area for an exam. Here we analysed the chapter Thermodynamics from the past six years' NEET papers and listed the essential topics, and previous year's questions and solutions. 

    Thermodynamics NEET, Solutions To Question Papers Of Last Six Years
    Synopsis

    Analysis of the previous year's question papers is essential to understand the focus area for an exam. Here we analysed the chapter Thermodynamics from the past six years' NEET papers and listed the essential topics, and previous year's questions and solutions. 

    A good rank in the National Eligibility cum Entrance Test (NEET) for undergraduates is a dream of many students who wish to get an MBBS seat in colleges across India. To prepare for NEET, students should practice the previous year's questions and understand the important areas to be focused on for the NEET exam. Every question of NEET is important as it is a highly competitive exam. Here we analyse the Class 12 NCERT chapter Thermodynamics using past six-year papers and discuss the past six-year questions from Thermodynamics NEET Physics paper.

    Also Read| NEET: Is Physics Causing Nightmares? Doing This Can Make It Easy

    In the last six years of NEET papers a total of eight questions were asked from the chapter Thermodynamics. Thermodynamics belongs to the Class 11 NCERT Physics syllabus. NEET Thermodynamics is a small chapter that can be understood easily. Questions from Thermal Properties Of Matter are not included here. Only the NCERT Class 11 Physics, chapter 12 - Thermodynamics is included. Let us see the topics covered in the past six years of NEET papers and their weightage.

    Also Read| JEE Main, NEET: Logic Gate Made Easy

    Topics And Number Of Questions

    Topic

    Number Of Questions

    Weightage

    First Law Of Thermodynamics

    2

    25%

    Thermodynamic Processes

    4

    50%

    Carnot Engine

    2

    25%

    A total of eight questions are analysed out of which three are from thermodynamic processes. From the topic thermodynamic process, two questions are from adiabatic processes and one is a match to the following questions that cover all the important thermodynamic processes. Two questions were from the topics First Law Of Thermodynamics and Carnot Engine. Let us have a look at the previous year questions from Thermodynamics (2017-2022).

    Also Read| NEET Previous Year Papers’ Analysis: LCR Circuits, Concepts And Solutions

    Q 1 - NEET 2017

    A Carnot engine having an efficiency of 1/10as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is:

    Answer-

    For a Carnot cycle

    \beta =\frac{1}{\eta }-1

    \beta = Coefficient of performance

    \eta = Efficiency of Carnot engine

    \eta=\frac{1}{10}\:\:\:\:\:\:\:\:\:\:\:\:=> \beta =9

    => \beta=\frac{Q_{2}}{W}

    Q_{2}=\beta.W=9\times 10 J=90J

    Q 2 - 2017

    Thermodynamic processes are indicated in the following diagram.

    1663753499284

    Match the following

    Column-1 Column-2

    P. Process I

    a. Adiabatic


    Q. Process II

    b. Isobaric


    R. Process III

    c. Isochoric


    S. Process IV
    d. Isothermal


    Answer-

    P \rightarrow c, Q \rightarrow a, R \rightarrow d, S \rightarrow b

    Process I-volume is constant-isochoric

    Process IV-pressure is constant-Isobaric

    Process II is adiabatic and Process III is isothermal

    Q 3 - NEET 2018

    A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 105 Nm-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

    Answer-

    Heat\: Supplied= 54 cal = 54\times 4.2\: J= 226.8\: J

    Work\: Done =P\Delta V\: = 1.01\times 10^{5}\times 167.1\times 10^{-6}J

    = 168.8\times 10^{-1}= 16.92\: J

    Change in internal energy-

    \therefore \: \: \: \Delta U= 226.8-16.92J\simeq 209.88J

    Q 4 - NEET 2018

    The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

    Answer-

    \eta=1-\frac{T_{2}}{T_{1}}\times 100\%= (1-\frac{273}{373})\times 100\%

    =\frac{100}{373}*100\%=26.8\%

    Q 5 - 2018

    The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

    1663753498357

    Answer-work done =

    = P (\Delta V)= \frac{nR}{K}(K\Delta T)= n R\Delta T

    heat absorbed = nC_{p}\Delta T

    =n(\frac{5R}{2})\Delta T= \frac{5}{2}(nR\Delta T)

    ratio of work to heat

    \frac{nR\Delta T}{5/2 nR\Delta T}=2/5

    Q 6 - 2019

    In which processes, the heat is neither absorbed nor released by a system?

    Answer-

    In the adiabatic process heat is neither absorbed nor released by the system.

    Q 7 - 2021

    Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is:

    Answer-

    As the entire system is thermally insulated and as free expansion will be taking place the temperature of the gas remains the same. The process is adiabatic

    A similar type of question is asked in the NCERT Class 11 physics book. The question is given below-

    Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following

    (a) What is the final pressure of the gas in A and B ?

    (b) What is the change in internal energy of the gas?

    (c) What is the change in the temperature of the gas?

    (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

    Q 8- 2022

    An ideal gas follows a process described by PV^2 = C from (P_1, V_1, T_1) to (P_2, V_2, T_2) (C is a constant). Then

    1. if P1 > P2 then T2 > T1

    2. if V2 > V1 then T2 < T1

    3. if V2 > V1 then T2 > T1

    4. if P1 > P2 then V2 > V1

    PV^{2}= C

    \left [\frac{nRT}{V} \right ]V^{2}= C

    TV = constant

    so

    V_{2}>V_{1}then T_{2}<T_{1}

    All the concepts of NCERT Class 11 chapter Thermodynamics are essential for the NEET exam. Following the NCERT book and the previous year, question papers will give enough information to crack the NEET questions from Thermodynamics.

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