According to the First Law of Thermodynamics, the relationship between heat ($Q$), internal energy ($\\Delta U$), and work ($W$) is given by $Q = \\Delta U + W$ (using the Physics sign convention where $W$ is work done by the gas).

1. Isothermal Process: Since the process is isothermal (constant temperature) and involves an ideal gas, the internal energy change is zero ($\\Delta U = 0$).
2. First Law Application: Substituting $\\Delta U = 0$ into the First Law equation gives $Q = W$.
3. Calculation: The problem states the gas does $-150\text{ J}$ of work against its surroundings. Thus, $W = -150\text{ J}$. $Q = -150\text{ J}$ The negative sign for heat ($Q$) indicates that heat is released or removed from the system.

Therefore, $150\text{ J}$ of heat has been removed from the gas.