The specific heat of a gas at constant pressure is more than that of the same gas at constant volume because

Two moles of monoatomic gas is expanded from $(P_0, V_0)$ to $(P_0 , 2V_0)$ under isobaric condition. Let $\Delta Q_1$, be the heat given to the gas, $\Delta W_1$ the work done by the gas and $\Delta U_1$ the change in internal energy. Now the monoatomic gas is replaced by a diatomic gas. Other conditions remaining the same. The corresponding values in this case are $\Delta Q_2 , \Delta W_2 , \Delta U_2$ respectively, then

In the following indicator diagram, the net amount of work done will be

A sample of an ideal gas is taken through the cyclic process $ABCA$ as shown in figure. It absorbs, $40\,J$ of heat during the part $A B$, no heat during $BC$ and rejects $60\,J$ of heat during $CA$. $A$ work $50\,J$ is done on the gas during the part $BC$. The internal energy of the gas at $A$ is $1560\,J$. The work done by the gas during the part $CA$ is………….$J$

A diatomic gas with rigid molecules does $10\, J$ of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process ….. $J$.