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Heat and Thermodynamics question

2022 · 27 Jun · Shift 2 · Q65
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Heat and Thermodynamics question

2022 · 27 Jun · Shift 2 · Q65

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
A diatomic gas (γ\gammaγ = 1.4) does 400J of work when it is expanded isobarically. The heat given to the gas in the process is ‾\underline{\hspace{2cm}}​ J.
Numerical answer
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Correct answer: 1400

  1. Given:

    • Diatomic gas with γ=1.4\gamma = 1.4γ=1.4
    • Isobaric work done by gas: W=400 JW = 400\,\text{J}W=400J
    • We need heat supplied: QQQ

  2. Use first law of thermodynamics: Q=ΔU+WQ = \Delta U + WQ=ΔU+W

  3. For an ideal gas: γ=CpCv\gamma = \frac{C_p}{C_v}γ=Cv​Cp​​ and Cp−Cv=RC_p - C_v = RCp​−Cv​=R

    Hence, CpCv=1.4=75\frac{C_p}{C_v} = 1.4 = \frac{7}{5}Cv​Cp​​=1.4=57​

    So the molar heat capacities are in the ratio Cv:R=Rγ−1:RC_v : R = \frac{R}{\gamma-1} : RCv​:R=γ−1R​:R

    Since Cv=Rγ−1=R0.4=2.5RC_v = \frac{R}{\gamma - 1} = \frac{R}{0.4} = 2.5RCv​=γ−1R​=0.4R​=2.5R and Cp=Cv+R=3.5RC_p = C_v + R = 3.5RCp​=Cv​+R=3.5R

  4. For an isobaric process: W=nRΔTW = nR\Delta TW=nRΔT Q=nCpΔTQ = nC_p\Delta TQ=nCp​ΔT

    Therefore, QW=nCpΔTnRΔT=CpR=3.5\frac{Q}{W} = \frac{nC_p\Delta T}{nR\Delta T} = \frac{C_p}{R} = 3.5WQ​=nRΔTnCp​ΔT​=RCp​​=3.5

  5. So, Q=3.5W=3.5×400=1400 JQ = 3.5W = 3.5 \times 400 = 1400\,\text{J}Q=3.5W=3.5×400=1400J

  6. Final answer: Q=1400 JQ = 1400\,\text{J}Q=1400J

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