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 .... $J$.

$A$ gas is compressed at a constant pressure of $50 \,N/m^2$ from a volume of $10 \,m^3$ to a volume of $4 \,m^3$. Energy of $100 \,J$ is then added to the gas by heating. Its internal energy is

When a system is taken from state $a$ to state $c$ along a path $abc$,it is found that $Q = 80 \ cal$ and $W = 35 \ cal$. Along path $adc$,$Q = 65 \ cal$. The work done $W$ along path $adc$ is: (in $cal$)

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 $\,J$)

$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes $I$ and $II$ respectively, then the relation between $\Delta U_1$ and $\Delta U_2$ is

$A$ system is given $300$ calories of heat and it does $600$ joules of work. How much does the internal energy of the system change in this process? $(1 \text{ calorie} = 4.18 \text{ joules})$