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(D) For path $ab$ (isochoric process): The heat absorbed is given as $Q_{ab} = 7000 \; J$. Since it is an isochoric process,$W_{ab} = 0$,so $\Delta U_

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$9800$

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(D) For path $ab$ (isochoric process): The heat absorbed is given as $Q_{ab} = 7000 \; J$. Since it is an isochoric process,$W_{ab} = 0$,so $\Delta U_{ab} = Q_{ab} = 7000 \; J$.Using $\Delta U = \mu C_V \Delta T$,where $C_V = \frac{5}{2}R$ for a diatomic gas like carbon monoxide:$7000 = \mu 	imes \frac{5}{2}R 	imes (1000 - 300) = \mu 	imes \frac{5}{2}R 	imes 700$.$\mu R = \frac{7000 	imes 2}{5 	imes 700} = 4 \; J/K$.For the complete cycle $abc$,the change in internal energy is zero: $\Delta U_{ab} + \Delta U_{bc} + \Delta U_{ca} = 0$.Since $bc$ is isothermal,$\Delta U_{bc} = 0$. Thus,$\Delta U_{ca} = -\Delta U_{ab} = -7000 \; J$.For path $ca$ (isobaric process at pressure $P_1$): The work done is $W_{ca} = P_1(V_1 - V_2) = \mu R(T_a - T_c)$.Since $T_a = 300 \; K$ and $T_c = T_b = 1000 \; K$ (as $bc$ is isothermal),$W_{ca} = 4 	imes (300 - 1000) = 4 	imes (-700) = -2800 \; J$.Using the first law of thermodynamics for path $ca$: $Q_{ca} = \Delta U_{ca} + W_{ca} = -7000 + (-2800) = -9800 \; J$.The magnitude of heat rejected is $9800 \; J$.

List-$I$	List-$II$
$A$. Pressure varies inversely with volume of an ideal gas.	$I$. Adiabatic process
$B$. Heat absorbed goes partly to increase internal energy and partly to do work.	$II$. Isochoric process
$C$. Heat is neither absorbed nor released by a system.	$III$. Isothermal process
$D$. No work is done on or by a gas.	$IV$. Isobaric process

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