An ideal gas undergoes a process in which its internal energy...

From the first law of thermodynamics:

$\Delta Q = \Delta U + \Delta W$

It is given that the decrease in internal energy is equal to the work done by the system:

$-\Delta U = \Delta W \Rightarrow \Delta U + \Delta W = 0$

Substituting this into the first law equation gives $\Delta Q = 0$. Since there is no heat exchange, the process must be adiabatic.

Furthermore, the problem states that the internal energy decreases ($\Delta U < 0$). For an ideal gas, internal energy is a function of temperature alone ($U = nC_vT$).

Since $\Delta U < 0$, the temperature of the gas must decrease.