$A$ gas absorbs $18 \ J$ of heat and work done on the gas is $12 \ J$. Then the change in internal energy of the gas is: (in $J$)

$A$ gas is at a constant pressure of $4 \times 10^5 \,N/m^2$. When a heat energy of $2000 \,J$ is supplied to the gas, its volume changes by $3 \times 10^{-3} \,m^3$. What is the increase in its internal energy (in $\,J$)?

When a system is taken from thermodynamic state $i$ to $f$ along the path $iaf$ (see figure),it is found that the heat $Q$ absorbed by the system is $50 \ cal$ and the work $W$ done by the system is $20 \ cal$. Along the path $ibf$,$Q = 36 \ cal$. What is the work $W$ done along the path $ibf$ (in $cal$)?

Consider $1 \,kg$ of liquid water undergoing a change in phase to water vapour at $100^{\circ} C$. At $100^{\circ} C$,the vapour pressure is $1.01 \times 10^5 \,N m^{-2}$ and the latent heat of vaporization is $22.6 \times 10^5 \,J kg^{-1}$. The density of liquid water is $10^3 \,kg m^{-3}$ and that of vapour is $\frac{1}{1.8} \,kg m^{-3}$. The change in internal energy in this phase change is nearly ............ $J kg^{-1}$.

$A$ unit mass of a liquid with volume $V_1$ is completely changed into a gas of volume $V_2$ at a constant external pressure $P$ and temperature $T$. If the latent heat of evaporation for the given mass is $L$,then the increase in the internal energy of the system is:

When a system is taken from state $i$ to a state $f$ along path $iaf$,$Q = 50 \, J$ and $W = 20 \, J$. Along path $ibf$,$Q = 35 \, J$. If $W = -13 \, J$ for the curved return path $fi$,$Q$ for this path is ...... $J$.