Corresponding to p-T graph as shown in figure, select the correct option(s)

 

 In process  ab from the given graph, we can see that

$\Rightarrow p\propto T \text{ or } V=\text{ constant } \rho =\text{ constant }$

p and T both are increasing.

Therefore, U is also increasing (as $U\propto T$ )

Now,

$p·$V graph is straight line parallel to p-axis as V is constant.

In process bc from the given graph, we can see that

$$\begin{gathered} \Rightarrow T =\text{ constant } \Rightarrow U =\text{ constant } \Rightarrow pV =\text{ constant } \text{ or } \\ p\propto \frac{1}{V}\left(isothermal\right) \end{gathered}$$

p is increasing. Therefore, V will decrease.

Hence, $\rho$ will increase.

Now,

($p·V$ graph is a rectangular hyperbola $\left( \right.$ as$\left.P\propto \frac{1}{V}\right).$

In process cd

$$\begin{gathered} p=\text{ constant } \\ V\propto T \end{gathered}$$

Temperature is decreasing. So, volume will also decrease. But density will increase.

Now,

p-V graph is a straight line parallel to V-axis because p is constant.

Process da  is just inverse of process bc. So, this process will complete the cycle following the steps discussed in process bc.