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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)

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$q_{A C} = Δ U_{B C} a n d w_{A B} = P_{2} (V_{2} - V_{1})$

$w_{B C} = P_{2} (V_{2} - V_{1}) a n d q_{B C} = Δ H_{A C}$

$Δ H_{C A} < Δ U_{C A} a n d q_{A C} = Δ U_{B C}$

$q_{B C} = Δ H_{A C} a n d Δ H_{C A} > Δ U_{C A}$

answer is B, C.

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Detailed Solution

AC ® Isochoric

AB ® Isothermal

BC ® Isobaric

$# q_{AC} = U_{BC} = nCv (T_{2} - T_{1})$

$\begin{array}{l} W_{AB} = {nRT}_{\underline{\underline{1}} \ln} \frac{V_{2}}{V_{1}} \end{array} A (w r o n g)$

$\begin{array}{ll} # q_{BC} = H_{AC} = nCp (T_{2} - T_{1}) \\ W_{BC} = - P_{2} (V_{1} - V_{2}) B (correct) \\ # {nC}_{2} (T_{1} - T_{2}) < nCv (T_{1} - T_{2}) C (correct) \\ Δ \underline{\underline{H_{CA}}} < {ΔU}_{CA} \\ # D (wrong) \end{array}$

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