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Heat and Thermodynamics question

2024 · 1 Feb · Shift 2 · Q70
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Heat and Thermodynamics question

2024 · 1 Feb · Shift 2 · Q70

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A diatomic gas (γ=1.4)(\gamma=1.4)(γ=1.4) does 200 J200 \mathrm{~J}200 J of work when it is expanded isobarically. The heat given to the gas in the process is :
  1. A
    800 J800 \mathrm{~J}800 J
  2. B
    600 J600 \mathrm{~J}600 J
  3. C
    700 J700 \mathrm{~J}700 J
  4. D
    850 J850 \mathrm{~J}850 J
View written solutionFree

Correct answer: C

  1. Given

    • Diatomic gas with γ=1.4\gamma = 1.4γ=1.4
    • Isobaric expansion
    • Work done by gas: W=200 JW = 200\,\text{J}W=200J
  2. Use the relation for an isobaric process

    For an ideal gas in an isobaric process, W=nRΔTW = nR\Delta TW=nRΔT

    Heat supplied is Q=nCpΔTQ = nC_p\Delta TQ=nCp​ΔT

    Therefore, QW=CpR\frac{Q}{W} = \frac{C_p}{R}WQ​=RCp​​

  3. Find CpC_pCp​ using γ\gammaγ

    We know γ=CpCv=1.4\gamma = \frac{C_p}{C_v} = 1.4γ=Cv​Cp​​=1.4 and Cp−Cv=RC_p - C_v = RCp​−Cv​=R

    Using the standard relation, Cp=γRγ−1C_p = \frac{\gamma R}{\gamma - 1}Cp​=γ−1γR​

    So, CpR=γγ−1=1.40.4=3.5\frac{C_p}{R} = \frac{\gamma}{\gamma - 1} = \frac{1.4}{0.4} = 3.5RCp​​=γ−1γ​=0.41.4​=3.5

  4. Calculate heat supplied

    Q=W⋅CpR=200×3.5=700 JQ = W\cdot \frac{C_p}{R} = 200 \times 3.5 = 700\,\text{J}Q=W⋅RCp​​=200×3.5=700J

  5. Match with the options

    Q=700 JQ = 700\,\text{J}Q=700J

    So the correct option is C.

  6. Comparison with stored answer

    Stored correct answer: C

    Our derived answer: C

    Hence, they agree.

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