$(A)$ $q=0$ $(B)$ $T_2=T_1$ $(C)$ $P_2 V_2=P_1 V_1$ $(D)$ $P_2 V_2^\gamma=P_1 V_1^\gamma$
- ✓$(A,B,C)$
- B$(A,B,D)$
- C$(A,C,D)$
- D$(B,C,D)$
$(A)$ $q=0$ $(B)$ $T_2=T_1$ $(C)$ $P_2 V_2=P_1 V_1$ $(D)$ $P_2 V_2^\gamma=P_1 V_1^\gamma$
$q =0 $
$p _{ ex }=0 \text {, so } w =0$
so $\Delta U=0$ (ideal gas)
Hence $\Delta T=0$
$\Rightarrow \Delta T=0 \quad \Rightarrow T_2=T_1 \quad \Rightarrow P_2 V_2=P_1 V_1$
The process is however adiabatic irriversible.
So we cannot apply $P _2 V _2^\gamma= P _1 V _1^\gamma$
Hence ans is $( A ),( B ),( C )$
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$[$ Given $: \sqrt{2}=1.41]$
$(i)\,Cl\xrightarrow{E.A.}Cl^-\,\,\,\,\,\,(ii)\,C{{l}^{-}}\xrightarrow{I.E.}Cl\,\,\,\,(iii)\,\,Cl\xrightarrow{I.E.}C{{l}^{+}}\,\,\,(iv)\,\,C{{l}^{+}}\xrightarrow{I.E.}Cl^{2+}$
( $Ba = 137,\,\,Cl = 35.5,\,\,S = 32,\,\,H = 1$ and $O = 16$ )