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**Hint:**Work done by an ideal gas is the area under the P-V curve. In the isothermal process, the temperature of the system remains constant. In the adiabatic process, there is no transferring of heat or mass between a thermodynamic system and its surroundings. In the isobaric and isochoric process, the pressure and volume remain constant respectively.

**Complete step by step solution:**

Since the work done by an ideal gas is the area under the P-V curve. So we will discuss the P-V curves of all given processes (Isothermal, Adiabatic, Isobaric, Isochoric).

Isothermal vs. Adiabatic process:

The slope of the adiabatic process is always steeper than the slope of the Isothermal process.

Therefore the isothermal curve lies above the adiabatic curve in the P-V diagram. Hence, work done in the isothermal process is greater than that in an adiabatic process

${W_1} > {W_2}$

Both ${W_1}$ and ${W_2}$ are > 0.

Isochoric Process:

Since \[V = \] constant, therefore the P-V graph will have a straight line parallel to the y-axis and will have no area under that curve.

Therefore, the work done is zero.

Also,

$W = \int_{V_1}^{V_2}PdV = 0$ , since $dV$ is zero

Therefore ${W_4} = 0$

Isobaric Process:

Since P=constant, therefore the P-V graph will have a straight line parallel to V(x-axis) and will have a maximum area under that curve.

Also,

$W = P({V_1} - {V_2})$

${W_3} > 0$

Since initial pressure remains constant max work will be done in this case.

Therefore combining all the cases,

We get ${W_3} > {W_1} > {W_2} > {W_4}$.

**Note:**In general if the volume of the gas decreases, the pressure of the gas increases. If the volume of the gas increases, the pressure decreases. These results support Boyle's law but in Isothermal, isochoric, and isobaric processes the temperature, volume, and pressure respectively are made constant.

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