When heat is provided to an ideal gas during an isothermal process,

For an isothermal process, the temperature of the ideal gas remains constant, so $\Delta T = 0$.

The change in internal energy of an ideal gas depends only on temperature and is given by $\Delta U = nC_v\Delta T$. Since $\Delta T = 0$, $\Delta U = 0$.

According to the first law of thermodynamics, $Q = \Delta U + W$.

Substituting $\Delta U = 0$, we get $Q = W$.

Since heat is provided to the gas, $Q > 0$. Therefore, $W > 0$, which means the gas performs positive work.