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JEE Main 2014 (19 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2014 (19 Apr, Shift 1)

Q229JEE Main 2014 · 19 APR · SHIFT 1Easy

Thermodynamic Processes

A gas is compressed from a volume of 2 m3 to a volume of 1 m3 at a constant pressure of 100 N/m2. Then it is heated at constant volume by supplying 150 J of energy. As a result, the internal energy of the gas:

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Answer: (a)

  1. Step 1 (isobaric compression): W1=PΔV=100×(1−2)=−100W_1 = P\Delta V = 100 \times (1 - 2) = -100 J (work done on the gas is 100 J).
  2. Step 2 (isochoric): W2=0W_2 = 0, Q=150Q = 150 J.
  3. Taking no heat exchange in step 1: ΔU=Q−W=150−(−100)=250\Delta U = Q - W = 150 - (-100) = 250 J.
  4. So the answer is (a) Increases by 250 J.
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