The density $\left(\rho \right)$ of an ideal gas varies with temperature $T$ as shown in figure. Then

From the diagram, the density $\left(\rho \right)$ of an ideal gas varies with temperature (T).

using the ideal gas equation,

$$\begin{gathered} PV=nRT \\ P=\frac{m}{v}\times \frac{I}{m}\times RT \\ =\rho \times \frac{RT}{m} \end{gathered}$$

Therefore, $\frac{P}{\rho }\times \frac{RT}{m}$

Here, m is mass , m is molar $\& n=\frac{m}{m} ,\rho =\frac{m}{v}$

where a changes takes place in $\rho$ , $T$ remains constant.

Then, $\frac{P}{\rho }$ is constant or we can also say, $PV$ is constant.

(A) we can understand, the product of PV at A is same as the product of PV at B.

(B) The  ${\rho }_B$ will be greater than ${\rho }_A.$

${\rho }_B>{\rho }_A$

Now, let us consider $k$ with both, then:

$k {\rho }_B > k {\rho }_A$

Here $k$ is constant.

Now, $P=kp$

Therefore, $P_B>P_A$

(C) 

From the above diagram, here Pv is every considered as a constant. It is very clear from the P-V graph that, The work done during AB will be negative.

(D) we know that,

$$\begin{gathered} ∆u=nc to ∆t \\ and T_A=T_B,∆T=0 \end{gathered}$$

Therefore, $∆u=0 in AB$

The correct answer is (A) ,(B), (C), (D).