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

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27\,^{\circ}C$. If the work done during the process is $3\, kJ$,then the final temperature of the gas is $(C_v = 20\, J/K)$ .....$K$

For the $1^{st}$ law of thermodynamics,select the correct option.

Assertion : For an isothermal reversible process $Q = -W$,i.e.,work done by the system equals the heat absorbed by the system.
Reason : Enthalpy change $(\Delta H)$ is zero for an isothermal process.

$\Delta_{vap} H^{\ominus}$ for water is $+40.49 \ kJ \ mol^{-1}$ at $1 \ bar$ and $100^{\circ} C$. The change in internal energy for this vaporization under the same conditions is . . . . . . . . . . $kJ \ mol^{-1}$. (Integer answer) (Given $R = 8.3 \ J \ K^{-1} \ mol^{-1}$)

What is the difference between $\Delta H$ and $\Delta U$ for the reaction given below at $298 \ K$ (in $kJ$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$
$2 \ C_6H_{6(\ell)} + 15 \ O_{2(g)} \rightarrow 12 \ CO_{2(g)} + 6 \ H_2O_{(\ell)}$