Final answer:
The isobaric process requires more work from the gas, with a ratio of 3:1.
Step-by-step explanation:
In the case of an ideal gas expanding quasi-statically to three times its original volume, the work done by the gas will be greater in the isobaric process compared to the isothermal process.
In an isobaric process, the pressure remains constant while the volume increases. The work done is given by the equation W = PΔV, where ΔV is the change in volume and P is the constant pressure.
In an isothermal process, the temperature remains constant while the volume increases. The work done is given by the equation W = nRT ln(Vf/Vi), where n is the number of moles, R is the gas constant, T is the temperature, and Vf and Vi are the final and initial volumes, respectively.
Since the work done in the isobaric process is directly proportional to the change in volume, while the work done in the isothermal process is proportional to the natural logarithm of the volume ratio, the work done in the isobaric process will be greater than the work done in the isothermal process.
The ratio of the work done in these processes will be 3:1, with the isobaric process requiring more work from the gas.