The heat supplied to a gas at a constant pressure of $5 \times 10^5 \ Pa$ is $1000 \ kJ$. If the volume of gas changes from $1 \ m^3$ to $2.5 \ m^3$,then the change in internal energy of the gas is (in $kJ$)

$A$ thermally insulated rigid container contains an ideal gas heated by a filament of resistance $100 \,\Omega$ through a current of $1 \,A$ for $5 \,min$. The change in internal energy is...... $kJ$.

$1.00 \ kg$ 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 internal energy of an ideal gas is given by $U = 1.5 PV$. It expands from $10 \ cm^3$ to $20 \ cm^3$ against a constant pressure of $2 \times 10^5 \ Pa$. Heat absorbed by the gas in the process is (in $J$)

$1 \,g$ of water at $100^{\circ} C$ is completely converted into steam at $100^{\circ} C$. $1 \,g$ of steam occupies a volume of $1650 \,cc$. (Neglect the volume of $1 \,g$ of water at $100^{\circ} C$). At the pressure of $10^5 \,N/m^2$, the latent heat of steam is $540 \,cal/g$ $(1 \,calorie = 4.2 \,joule)$. The increase in the internal energy (in joule) is:

If $Q$,$E$,and $W$ denote respectively the heat added,change in internal energy,and the work done by a closed cycle process,then