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GATE 2020 CH – Question 50

Thermodynamics · Laws of thermodynamics, open and closed systems, entropy and chemical potential · 2 marks · Multiple choice

One mole of ideal gas moves reversibly from L to N by two paths on the P–V diagram. Path I is straight LN. Path II follows horizontal LM, then vertical MN. Coordinates are L: (0.02 m³/mol,100 kPa), M: (0.03,100), N: (0.03,200). If net heats into the gas are Q₁,Q₂, find Q₁ − Q₂ in J.

Diagram for GATE 2020 CH question 50
  1. 250
  2. 500
  3. 1000
  4. 1500

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Correct answer: (B) 500

Explanation

Both paths have the same initial and final states, so their internal-energy changes are equal. From Q = ΔU + W, the heat difference equals the work difference. Path I work is the trapezoid area $[(100+200)/2](0.03-0.02)=1.5$ kJ. Path II work is $100(0.01)=1.0$ kJ; the constant-volume segment does no work. Thus Q₁ − Q₂ = 0.5 kJ = 500 J, B.

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