What is the change in internal energy if a system does $140 \ kJ$ of work on the surroundings and $40 \ kJ$ of heat is added to the system (in $kJ$)?

Based on the first law of thermodynamics,which one of the following is correct?

In a process,the work done by the system is equal to the change in its internal energy. Then the process will be ...........

For water at $100^{\circ} C$ and $1 \, bar$,$\Delta_{vap} H - \Delta_{vap} U = ...... \times 10^{2} \, J \, mol^{-1}$. (Round off to the Nearest Integer) $[Use : R = 8.31 \, J \, mol^{-1} \, K^{-1}]$ [Assume volume of $H_{2}O_{(\ell)}$ is much smaller than volume of $H_{2}O_{(g)}$. Assume $H_{2}O_{(g)}$ treated as an ideal gas]

The correct relationship is

$A$ piston filled with $0.04 \ mol$ of an ideal gas expands reversibly from $50.0 \ mL$ to $375 \ mL$ at a constant temperature of $37.0 \ ^oC$. As it does so,it absorbs $208 \ J$ of heat. The values of $q$ and $w$ for the process will be
$(R = 8.314 \ J/mol \ K) \ (\ln 7.5 = 2.01)$