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(D) For an ideal gas,$PV = nRT$. Since density $d = \frac{m}{V}$,we have $V = \frac{m}{d}$. Substituting this into the ideal gas equation,$P(\frac{m}{

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During the process $DA$,internal energy of the gas remains constant.

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(D) For an ideal gas,$PV = nRT$. Since density $d = \frac{m}{V}$,we have $V = \frac{m}{d}$. Substituting this into the ideal gas equation,$P(\frac{m}{d}) = nRT$,which gives $P = (\frac{nRT}{m})d$. Since $n, R, m$ are constant,$P \propto Td$. $1$. Process $AB$: The density $d$ is constant (isochoric process). Since $V$ is constant,the work done $W = \int P dV = 0$. Thus,options $A$ and $B$ are incorrect.$2$. Process $BC$: The graph shows $P$ increases while $d$ increases. Since $P = \frac{\rho RT}{M}$ (where $\rho$ is density),$T = \frac{PM}{\rho R}$. Along $BC$,the slope of the line $P$ vs $d$ is positive and passes through the origin,implying $T$ is constant (isothermal process). Thus,internal energy $U \propto T$ remains constant. Option $C$ is incorrect.$3$. Process $DA$: The graph shows $P$ decreases as $d$ decreases. Similar to $BC$,this line passes through the origin,meaning $T$ is constant. Therefore,the internal energy remains constant. Option $D$ is correct.

The pressure $(P)$ versus density $(d)$ graph of an ideal gas is shown in the figure. Choose the correct statement.

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