$A$ Carnot cycle for a gas is represented in the pressure-volume $(P-V)$ diagram shown below. Consider the following statements:
$I.$ The area of $ABCD$ is the work done by the gas.
$II.$ The area of $ABCD$ is the total heat absorbed.
$III.$ The change in internal energy of the cycle is zero.
Which of the following is correct?

The graph for $y = 1 - e^{-x}$ is represented by (for $x > 0$):

The amount of work,which can be obtained by supplying $200 \, cal$ of heat,is

The figure shows the $P-V$ plot of an ideal gas taken through a cycle $ABCDA$. The part $ABC$ is a semi-circle and $CDA$ is half of an ellipse. Then,
$(A)$ the process during the path $A \rightarrow B$ is isothermal
$(B)$ heat flows out of the gas during the path $B \rightarrow C \rightarrow D$
$(C)$ work done during the path $A \rightarrow B \rightarrow C$ is zero
$(D)$ positive work is done by the gas in the cycle $ABCDA$

Heat is supplied to a diatomic gas at constant pressure. The ratio of $\Delta Q : \Delta U : \Delta W$ is

Two cylinders $A$ and $B$ fitted with pistons contain an equal amount of an ideal diatomic gas at temperature $T$ $K$. The piston of cylinder $A$ is free to move,while that of $B$ is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in $A$ is $dT_{A}$,then the rise in temperature of the gas in cylinder $B$ is (where $\gamma = \frac{C_{P}}{C_{V}}$):