Answer:
Option A,C,D
Explanation:
(a) Since, change of state ($p_{1},V_{1},T_{1}$) to ($p_{2}, V_{2},T_{2}$) is isothermal therefore $T_{1}=T_{2}$
(b) Since, change of state ($p_{1},V_{1},T_{1}$) to ($p_{3}, V_{3},T_{3}$) is an adiabatic expansion it brings about cooling of gas , therefore $T_{3} <T_{1}$
(c) Work-done is the area under the curve of p-V diagram' As obvious from the given diagram, magnitude
of area under the isothermal curve adiabatic curve , hence
$w_{isothermal}$ > $W_{adiabatic}$
Note: Here only magnitudes of work is being considered otherwise both works have negative sign.
(d)$\triangle U=nC_{v}\triangle T$
In isothermal Process, $\triangle U=0, $ as $\triangle T=0$ in adiabatic process
$\triangle U=nC_{v}$ ($T_{3}-T_{1})<0$ as $T_{3}<T_{1}$
$\Rightarrow$ $\triangle U_{isothermal}$ > $\triangle U_{adiabatic}$
