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

Question

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, 0, RT 1 ln V ⁡ 1 V ⁡ 2 , R ⁡ T ⁡ 1 - T ⁡ 2, R ⁡ , T ⁡ 1 - T ⁡ 2 R ⁡ , RT 1 ln V ⁡ 1 V ⁡ 2, 0, RT 2 ln V ⁡ 2 V ⁡ 1 , ⁡ RT 1 V ⁡ 1 V ⁡ 1 - V ⁡ 2, 0, RT 2 ln V ⁡ 1 V ⁡ 2 , ⁡ R ⁡ T ⁡ 1 - T ⁡ 2

MARKS App · Accepted Answer

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 - V ⁡ 2 = RT 2 V ⁡ 2 V ⁡ 1 - V ⁡ 2

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