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Paragraph: In the following P-V diagram, two adiabatics cut two isothermals at T₁ and T ₂ .2 moles of an ideal diatomic gas is taken through the cyclic process ABCDA. Question: If Q₁ is heat absorbed and Q₂ is heat rejected in one cycle then-

Options

  1. AQ ₁ ~T ₁= Q ₂ ~T ₂
  2. BQ ₁ ~T ₁^2= Q ₂ ~T ₂^2
  3. CQ ₁ ~T ₂^2= Q ₂ ~T ₁^2
  4. DQ ₁ ~T ₂= Q ₂ ~T ₁

Correct answer

D. Q ₁ ~T ₂= Q ₂ ~T ₁

Step-by-step solution

Q₁=n R T₁ ( V_B V_A ) ; Q₂=n R T₂ ( V_D V_C ) But V_D V_C = V_A V_B so Q₂=n R T₂ ( V_B V_A ) Q₁ Q₂ = T₁ T₂

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