Three processes form a thermodynamic cycle as | vedclass

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(A) For process $2 	o 3$ (isochoric process):Volume is constant,so work done $W_{2 	o 3} = 0$.Given heat $Q_{2 	o 3} = -40\,J$ (heat leaves the sys

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$-40$

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(A) For process $2 	o 3$ (isochoric process):Volume is constant,so work done $W_{2 	o 3} = 0$.Given heat $Q_{2 	o 3} = -40\,J$ (heat leaves the system).From the first law of thermodynamics,$\Delta U = Q - W$.$\Delta U_{2 	o 3} = U_3 - U_2 = -40 - 0 = -40\,J$.  ...$(1)$For process $1 	o 2$ (isothermal process):Temperature is constant,so $T_1 = T_2$.For an ideal gas,internal energy $U$ depends only on temperature,so $U_1 = U_2$.  ...$(2)$From equations $(1)$ and $(2)$:$U_3 - U_1 = -40\,J$,which implies $U_1 - U_3 = 40\,J$.For process $3 	o 1$ (adiabatic process):$Q_{3 	o 1} = 0$.From the first law of thermodynamics,$\Delta U_{3 	o 1} = Q_{3 	o 1} - W_{3 	o 1}$.$U_1 - U_3 = 0 - W_{3 	o 1}$.$W_{3 	o 1} = -(U_1 - U_3) = -40\,J$.

List-$I$	List-$II$
$A$. Figure of merit $(k)$	$P$. in series
$B$. Current sensitivity $(S_I)$	$Q$. to get power loss minimized
$C$. Deflection is made half by using resistance	$R$. in shunt/parallel
$D$. High resistance box connected	$S$. current per unit deflection
	$T$. deflection per unit current

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