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A cyclic process for 1 mol of an ideal gas is shown in the V - T diagram. The work done in AB , BC and CA respectively are

Options

  1. A0, RT 1 ln V ⁡ 1 V ⁡ 2 , R ⁡ T ⁡ 1 - T ⁡ 2
  2. BR ⁡ ,  T ⁡ 1 - T ⁡ 2 R ⁡ , RT 1 ln V ⁡ 1 V ⁡ 2
  3. C0, RT 2 ln V ⁡ 2 V ⁡ 1 , ⁡ RT 1 V ⁡ 1 V ⁡ 1 - V ⁡ 2
  4. D0, RT 2 ln V ⁡ 1 V ⁡ 2 , ⁡ R ⁡ T ⁡ 1 - T

Correct answer

C. 0, RT 2 ln V ⁡ 2 V ⁡ 1 , ⁡ RT 1 V ⁡ 1 V ⁡ 1 - V ⁡ 2

Step-by-step solution

During AB process is isochoric. ∴     Δ V   =   0     ∴     W   =   0 During BC , the process is isothermal. ∴   Δ T = 0 ∴ W ⁡ = RT 2 ln V ⁡ 2 V ⁡ 1 During CA , the process is isobaric. So, the pressure is constant. ∴     W   =   P ( V 1 – V 2 ) But P V 1 =   R T 1 ∴ P ⁡ = RT 1 V ⁡ 1 = RT 2 V ⁡ 2 ∴ W ⁡ = RT 1 V ⁡ 1 V ⁡ 1

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