A reversible cyclic process for an ideal gas is shown below. Here, P, V, and T are pressure, volume and temperature, respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.
The correct option(s) is (are)
Text Solution
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(b,c) AC ⇒ isochoric process
AB ⇒ isothermal process
BC ⇒ isobaric process
⇒ q AC = Δ U AC = nC v,m (T 2 – T 1 ) = Δ U BC
⇒ W AB = –nRT 1 
⇒ W BC = –P 2 (V 1 – V 2 ) = P 2 (V 2 – V 1 )
⇒ q BC = Δ H BC = nC P,m (T 2 – T 1 ) = Δ H AC
⇒ Δ H CA = nC P,m (T 1 – T 2 )
⇒ Δ U CA = nC V,m (T 1 – T 2 )
Δ H CA < Δ U CA since both are negative (T 1 < T 2 )
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