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The figure shows the p-V plot of an ideal gas taken through a cycle A B C D A . The part A B C is a semi-circle and C D A is half of an ellipse. Then,

Options

  1. Athe process during the path A B is isothermal
  2. Bheat flows out of the gas during the path B C D
  3. Cwork done during the path A B C is zero
  4. Dpositive work is done by the gas in the cycle A B C D A .

Correct answer

B. heat flows out of the gas during the path B C D

Step-by-step solution

(A) p-V graph is not rectangular hyperbola. Therefore, process A-B is not isothermal. (B) In process B C D , product of p V (therefore temperature and intemal energy) is decreasing. Further, volume is decreasing. Hence, work done is also negative. Hence, Q will be negative or heat will flow out of the gas. (C) W_ A B C = positive (D) For clockwise cycle on p-V diagram with P on y -axis, net work done is positive.

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