Correct Answer - A
Process (AB) is an isothermal process with (T) = constant and `p_(B) gt p_(A)`.
(p - V) graph : `p prop 1/V "i.e." p - V` graph is a rectangular hyperbola with `(p_B gt p_A)`
and `(V_B lt V_A)`
(V - T) graph : (T - constant). Therefore, (V - T) graph is a straight line parallel to (V - axis) with `(V_B lt V_A)`.
(p - T) graph : `rho = (pM)/(RT)`
or `rho prop p`
As (T) is constant. Therefore, (`rho - T`) graph is a straight the line parallel to (`rho` - axis) with `(p_(A) "as" p_(B) gt p_(A)`.
Process (BC) is an isobaric process with (p = constant)
and `(T_C gt T_B)`
(p - V) graph : As (p) is constant. therefore, (p - V) graph is a straight line parallel to (V - axis) with `(V_C gt V_B)` ( because (V prop T) is an isobaric process)
(V - T) graph : In isobaric process (V `prop` T, i.e V - T) graph is a straight line passing through origin, with `(T_C gt T_B)`
and `(V_C gt V_B)`
(p - T) graph : `(rho prop 1/T` (when (P) = constant) i.e. (rho - T) graph is a hyperbola with `(T_C gt T_B)`
and `(rho _C ltrho _B)`
There is no need of discussing (C - D) and (D - A) processes. As they are opposite to (AB) and (BC) respectively. The corresponding three graphs are shown above.
