Question

A reversible cyclic process for an ideal gas is shown. 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?

• A. $q_{AC}=\Delta U_{BC}$ and $w_{AB}=P_2(V_2-V_1)$
• B. $w_{BC}=P_2(V_2-V_1)$ and $q_{BC}=\Delta H_{AC}$
• C. $\Delta H_{CA}<\Delta U_{CA}$ and $q_{AC}=\Delta U_{BC}$
• D. $q_{BC}=\Delta H_{AC}$ and $\Delta H_{CA}>\Delta U_{CA}$

Answer 1

Answer: (B), (C)

The correct options are
B $w_{BC}=P_2(V_2-V_1)$ and $q_{BC}=\Delta H_{AC}$
C $\Delta H_{CA}<\Delta U_{CA}$ and $q_{AC}=\Delta U_{BC}$

AC $\to$ Isochoric
AB $\to$ Isothermal
BC $\to$ Isobaric

$\left(A\right)q_{AC}=\Delta U_{BC}=n{C}_v(T_2-T_1)$
$W_{AB}=nR{T}_{1}ln\left(\frac{V_2}{V_1}\right)$

$\therefore$ A is wrong.

$\left(B\right)q_{BC}=\Delta H_{AC}=n{C}_P(T_2-T_1)$
$W_{BC}=-P_2(V_1-V_2)$

$\therefore$ B is correct.

$\left(C\right)n{C}_P(T_1-T_2)<n{C}_v(T_1-T_2)$

$\therefore$ C is correct.

$\left(D\right)\Delta H_{CA}<\Delta U_{CA}$
$\therefore$ D is wrong.