$A$ total of $48 \,J$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} C$. The work done by the gas is : (Given,$R=8.3 \,J \,K^{-1} \,mol^{-1}$.) (in $\,J$)

In changing the state of a thermodynamic system from state $A$ to state $B$, the heat supplied to the system is $Q$ and the work done by the system is $W$. The change in its internal energy is

$1.00 \ kg$ of liquid water at $100^{\circ} C$ undergoes a phase change into steam at $100^{\circ} C$ at $1.0 \ atm$ (take it to be $1.00 \times 10^5 \ Pa$). The initial volume of the liquid water was $1.00 \times 10^{-3} \ m^3$ which is changed to $2.001 \ m^3$ of steam. Find the change in the internal energy of the system. [Use heat of vaporization $\simeq 2000 \ kJ \ kg^{-1}$] (in $kJ$)

The latent heat of vaporization of water is $2240 \, J/g$. If the work done in the process of vaporization of $1 \, g$ is $168 \, J$,then the increase in internal energy is .... $J$.

State the first law of thermodynamics and write its limitations.

The $p-V$ diagram of a system undergoing thermodynamic changes is as shown in the figure. The work done by the system in going from $A \rightarrow B \rightarrow C$ is $30 \,J$. If $68 \,J$ of heat is given to the system,then the change in the internal energy of the system between $A$ and $C$ is (in $\,J$)