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Thermodynamics question

2019 · 8 Apr · Shift 1 · Q10
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Thermodynamics question

2019 · 8 Apr · Shift 1 · Q10

JEE MainChemistryThermodynamicsMCQ+4 / −1
Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)
  1. A
    Adiabatic process : Δ\DeltaΔ U= – w
  2. B
    Cyclic process : q = –w
  3. C
    Isochoric process : Δ\DeltaΔ U= q
  4. D
    Isothermal process : q = – w
View written solutionFree

Correct answer: A

  1. First law of thermodynamics

    Using the chemistry sign convention: ΔU=q+w\Delta U = q + wΔU=q+w where:

    • qqq = heat absorbed by the system
    • www = work done on the system

    Also, non-expansion work is zero.

  2. Check each process one by one

    Option A: Adiabatic process : ΔU=−w\Delta U = -wΔU=−w

    For an adiabatic process, q=0q = 0q=0 So from the first law, ΔU=q+w=0+w=w\Delta U = q + w = 0 + w = wΔU=q+w=0+w=w Hence the correct relation is: ΔU=w\Delta U = wΔU=w not ΔU=−w\Delta U = -wΔU=−w

    Therefore, Option A is incorrect.

    Option B: Cyclic process : q=−wq = -wq=−w

    For a cyclic process, the system returns to its initial state, so ΔU=0\Delta U = 0ΔU=0 From the first law, 0=q+w0 = q + w0=q+w q=−wq = -wq=−w So Option B is correct.

    Option C: Isochoric process : ΔU=q\Delta U = qΔU=q

    In an isochoric process, volume is constant, so expansion work is zero: w=0w = 0w=0 Therefore, ΔU=q+0=q\Delta U = q + 0 = qΔU=q+0=q So Option C is correct.

    Option D: Isothermal process : q=−wq = -wq=−w

    For an ideal gas in an isothermal process, temperature is constant. Since internal energy of an ideal gas depends only on temperature, ΔU=0\Delta U = 0ΔU=0 From the first law, 0=q+w0 = q + w0=q+w q=−wq = -wq=−w So Option D is correct.

  3. Conclusion

    The equation that does not correctly represent the first law is: A\boxed{\text{A}}A​

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