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:

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

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

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

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

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Solution

The correct option is C $△ H_{A C} < △ U_{B C} a n d q_{A C} = △ U_{B C}$
$△ H_{A C} + △ H_{C B} + △ H_{B A} = 0$
$△ H_{B A} = 0$ (Temperature is constant)
$△ H_{A C} = △ H_{B C}$
We know that $q_{p} = △ H$ (In path BC, P = constant)
Hence $q_{B C} = △ H_{B C}$
From Eq. (1)
$q_{B C} = △ H_{A C}$
(B) $q_{B C} = - P_{2} (V_{1} - V_{2}) = P_{2} (V_{2} - V_{1})$
(C) $△ H_{A C} = n C_{p} (T_{1} - T_{2}) = n C_{p} (T_{2} - T_{1})$
$△ U_{C A} = n C_{v} (T_{1} - T_{2}) = - n C_{v} (T_{2} - T_{1})$
As, $C_{p} > C_{v}$
So, $△ H_{A C} < △ U_{C A}$

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