One mole of an ideal monoatomic gas is taken round the cyclic process $MNOM$. The work done by the gas is (in $p_0 V_0$)

$A$ thermodynamic system is taken from an original state to an intermediate state by the linear process shown in the figure. Its volume is then reduced to the original value from $E$ to $F$ by an isobaric process. Calculate the total work done (in $J$) by the gas from $D$ to $E$ to $F$.

In the reported figure,the heat energy absorbed by a system in going through a cyclic process is $......\,\pi \text{ J}$.

The following figure shows two processes $A$ and $B$ for a gas. If $\Delta Q_A$ and $\Delta Q_B$ are the amounts of heat absorbed by the system in the two cases,and $\Delta U_A$ and $\Delta U_B$ are the changes in internal energies,respectively,then:

$A$ monoatomic gas is taken along path $AB$ as shown in the $V-P$ diagram. Calculate the change in internal energy of the system.

One mole of an ideal gas is taken from $A$ to $B$,from $B$ to $C$,and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure at $A$ is $P_{0}$ and volume is $V_{0}$. Then,the internal energy: