An ideal gas undergoes a reversible isothermal expansion from state...

$T =$ constant

$\Delta H = \Delta U = 0$

$H =$ constant.

$\Delta S \ne 0$ Entropy will increase $PV =$ constant.

and during adiabatic reversible expansion.

$q = 0$

$\Delta U = w = - ve$

so $T$ will decrease

$\Delta H = {nc}_{p}\Delta T = - ve$

$H$ will decrease

$\Delta S = 0$ (isoentropic process)

$S =$constant. and ${pv}^{r} =$ constant.

Keeping in mind correct graph are 1,2 and 4.