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Three moles of an ideal gas undergo a cyclic process shown in figure. The work done by the gas during the process is [Take ln ( 2 ) = 0 . 693 ]

Options

  1. A2.36 R T 0
  2. B0.58   R T 0
  3. C1.16 R T 0
  4. D– 0.58   R T 0

Correct answer

C. 1.16 R T 0

Step-by-step solution

If V 0 be volume of the gas at C , P 0 V 0 = 3 R T 0 W = 3 R ( 2 T 0 ) l n 2 – P 0 V 0 = 6 R T 0 l n 2 – 3 R T 0 = 3 R T 0 [ 2 × 0.693 – 1 ] = 3 R T 0 [ 1.386 – 1 ] = 1.16 R T 0

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