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Thermodynamics question

2025 · 22 Jan · Shift 1 · Q10
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  5. /2025 · 22 Jan · Shift 1 · Q10

Thermodynamics question

2025 · 22 Jan · Shift 1 · Q10

JEE MainChemistryThermodynamicsMCQ+4 / −1
A liquid when kept inside a thermally insulated closed vessel at 25∘C25^{\circ} \mathrm{C}25∘C was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ?
  1. A
    ΔU=0,q0\Delta \mathrm{U}=0, \mathrm{q}0ΔU=0,q0
  2. B
    ΔU>0,q=0,w>0\Delta \mathrm{U}\gt 0, \mathrm{q}=0, \mathrm{w}\gt 0ΔU>0,q=0,w>0
  3. C
    ΔU=0,q=0,w=0\Delta \mathrm{U}=0, \mathrm{q}=0, \mathrm{w}=0ΔU=0,q=0,w=0
  4. D
    ΔU0\Delta \mathrm{U}0ΔU0
View written solutionFree

Correct answer: B

  1. Given conditions

    • The vessel is thermally insulated  this means no heat exchange with surroundings.
    • The vessel is closed  matter cannot enter or leave.
    • The liquid is mechanically stirred from outside  work is done on the system.

  2. Determine heat, qqq Since the vessel is thermally insulated, q=0q = 0q=0

  3. Determine work, www Mechanical stirring means shaft work is supplied to the liquid from outside. Therefore, work is done on the system, so internal energy increases.

    Using the chemistry sign convention:

    • work done on the system is taken as positive. Hence, w>0w > 0w>0

  4. Apply first law of thermodynamics ΔU=q+w\Delta U = q + wΔU=q+w Since q=0q=0q=0 and w>0w>0w>0, ΔU=0+w>0\Delta U = 0 + w > 0ΔU=0+w>0 So, ΔU>0\Delta U > 0ΔU>0

  5. Check options

    • A: ΔU=0,q<0,w<0\Delta U = 0, q<0, w<0ΔU=0,q<0,w<0  incorrect
    • B: ΔU>0,q=0,w>0\Delta U > 0, q = 0, w > 0ΔU>0,q=0,w>0  correct
    • C: ΔU=0,q=0,w=0\Delta U = 0, q = 0, w = 0ΔU=0,q=0,w=0  incorrect
    • D: indicates negative/incorrect signs  incorrect

Therefore, the correct option is: B\boxed{B}B​

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