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