JEE Main · 2024 · Shift-IeasyTHERMO-006

Choose the correct option for free expansion of an ideal gas under adiabatic condition:

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Choose the correct option for free expansion of an ideal gas under adiabatic condition:

Options
  1. a

    q=0,ΔT0,w=0q=0,\, \Delta T \neq 0,\, w=0

  2. b

    q=0,ΔT<0,w0q=0,\, \Delta T<0,\, w \neq 0

  3. c

    q0,ΔT<0,w=0q \neq 0,\, \Delta T<0,\, w=0

  4. d

    q=0,ΔT=0,w=0q=0,\, \Delta T=0,\, w=0

Correct Answerd

q=0,ΔT=0,w=0q=0,\, \Delta T=0,\, w=0

Detailed Solution

🧠 Free expansion into vacuum — nothing pushes back, no work is done "Free expansion" means the gas expands into a vacuum: Pext=0P_{\text{ext}} = 0. "Adiabatic" means no heat exchange with surroundings.

🗺️ Each variable q=0q = 0: adiabatic, by definition. ✓

w=PextΔV=0ΔV=0w = -P_{\text{ext}} \cdot \Delta V = -0 \cdot \Delta V = 0: no opposing pressure. ✓

First Law: ΔU=q+w=0\Delta U = q + w = 0.

For ideal gas, ΔU=CvΔT\Delta U = C_v \Delta T. Since ΔU=0ΔT=0\Delta U = 0 \Rightarrow \Delta T = 0. ✓

Option (d) matches: q=0q = 0, ΔT=0\Delta T = 0, w=0w = 0.

⚠️ Trap: Option (b) tempts students who confuse free expansion with reversible adiabatic expansion. In reversible adiabatic expansion, w0w \neq 0 and ΔT<0\Delta T < 0. Free expansion has no opposing force, so w=0w = 0.

Answer: (d)\boxed{\text{Answer: (d)}}

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