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JEE Main201912 Apr 2019Morning ShiftPhysicsThermodynamicsActual

A sample of an ideal gas is taken through the cyclic process a b c a as shown in the figure. The change in the internal energy of the gas along the path c a is - 180 J . The gas absorbs 250 J of heat along the path a b and 60 J along the path b c . The work done by the gas along the path a b c is:

Options

  1. A130 J
  2. B100 J
  3. C120 J
  4. D140 J

Correct answer

A. 130 J

Step-by-step solution

In a cyclic process, Δ U = 0 From 1st law of thermodynamics Q = Δ U + W Q = W Q a b + Q b c + Q c a = W a b c + W c a W a b c = Q a b + Q b c + Q c a - W c a W a b c = Q a b + Q b c + U c a W a b c = 250 + 60 - 180 = 130 J

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