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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.
(image)
The correct option ($s$) is (are)
$(A)$ $q_{A C}=\Delta U_{B C}$ and $W_{A B}=P_2\left(V_2-V_1\right)$
$(B)$ $\mathrm{W}_{\mathrm{BC}}=\mathrm{P}_2\left(\mathrm{~V}_2-\mathrm{V}_1\right)$ and $\mathrm{q}_{\mathrm{BC}}=\mathrm{H}_{\mathrm{AC}}$
$(C)$ $\Delta \mathrm{H}_{\mathrm{CA}}<\Delta \mathrm{U}_{\mathrm{CA}}$ and $\mathrm{q}_{\mathrm{AC}}=\Delta \mathrm{U}_{\mathrm{BC}}$
$(D)$ $\mathrm{q}_{\mathrm{BC}}=\Delta \mathrm{H}_{\mathrm{AC}}$ and $\Delta \mathrm{H}_{\mathrm{CA}}>\Delta \mathrm{U}_{\mathrm{CA}}$
$A,B,C$
$B,C$
$A,C$
$A,B$
Solution
(image)
$\mathrm{AB}=\text { Isothermal process, } \Delta \mathrm{E} \text { or } \Delta \mathrm{U}=0$
$\mathrm{AC}=\text { Isochoric process, } \theta_{\mathrm{v}}=\Delta \mathrm{U}$
$\mathrm{BC}=\text { Isobaric process, } \theta_{\mathrm{P}}=\Delta \mathrm{H}$
