An ideal gas undergoes a cyclic process as shown in the diagram. The net work done by the gas in the cycle is

Two moles of an ideal monoatomic gas undergo a cyclic process as shown in the figure. The temperatures in different states are given as $6 T_1 = 3 T_2 = 2 T_4 = T_3 = 2400 \text{ K}$. The work done by the gas during the complete cycle is $(R = \text{Universal gas constant})$ (in $R$)

Calculate the heat absorbed by the system in going through the process shown in the figure.

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:

Find the work done in the cyclic process $ABCA$ shown in the figure.

Using the given $P-V$ diagram, the work done by an ideal gas along the path $\text{ABCD}$ is $-$ (in $\text{P}_0 \text{V}_0$)