$A$ thermodynamic system is changed from state $(P_1, V_1)$ to $(P_2, V_2)$ by two different processes. The quantity which will remain the same is

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 ............ $J kg^{-1}$.

In a given process,for an ideal gas,$\Delta W = 0$ and $\Delta Q < 0$. Then,for the gas,

$A$ perfect gas goes from state $A$ to another state $B$ by absorbing $8 \times 10^5 \ J$ of heat and doing $6.5 \times 10^5 \ J$ of external work. It is now transferred between the same two states in another process in which it absorbs $10^5 \ J$ of heat. Then in the second process:

The increase in internal energy of $2$ moles of gas when its temperature is raised from $340 \ K$ to $342 \ K$ is ........ $cal$. Given $C_v = 4.96 \ cal/mole \ K$.

$A$ system is taken from state $A$ to state $B$ along two different paths. The heat absorbed and work done by the system along these two paths are $Q_1, Q_2$ and $W_1, W_2$ respectively,then