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(B) For an ideal gas,the equation of state is $PV = nRT$.Given that the pressure $P$ increases linearly with temperature $T$,we have $P = kT$,where $k

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$A, D, E$ Only

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(B) For an ideal gas,the equation of state is $PV = nRT$.Given that the pressure $P$ increases linearly with temperature $T$,we have $P = kT$,where $k$ is a constant.Substituting this into the ideal gas equation: $(kT)V = nRT$.This simplifies to $V = \frac{nR}{k} = 	ext{constant}$.Since the volume $V$ is constant,the process is isochoric.$1$. Work done $W = \int P dV = 0$ because $dV = 0$. Thus,statement $A$ is correct.$2$. According to the first law of thermodynamics,$Q = \Delta U + W$. Since $W = 0$,$Q = \Delta U$. Thus,statement $B$ is incorrect.$3$. Since the volume is constant,statement $C$ is incorrect.$4$. As the temperature increases,the internal energy $\Delta U = nC_v \Delta T$ increases. Thus,statement $D$ is correct.$5$. Since the volume is constant,the process is isochoric. Thus,statement $E$ is correct.Therefore,statements $A, D,$ and $E$ are correct.

Column-$I$	Column-$II$
$(a)$ Graph $(a)$	$(i)$ Adiabatic process
$(b)$ Graph $(b)$	$(ii)$ Isobaric process
	$(iii)$ Isochoric process

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