Final Answer:
The work done by the ideal gas during the two-step process is -1250 J.
Step-by-step explanation:
The work done by the gas during an isothermal process can be expressed as W = nRT ln(V₂/V₁), where n is the number of moles, R is the gas constant, T is the temperature, and V₁ and V₂ are the initial and final volumes, respectively. Since the process from state 1 to state 2 is isothermal, the temperatures cancel out, and the work done becomes W₁ = nRT₁ ln(V₂/V₁). Similarly, during an isobaric process, the work done is given by W = P(V₂ - V₁), where P is the constant pressure.
For the first step, substituting the isothermal process formula, W₁ = 0.5 kg × R × T₁ × ln(V₂/V₁). For the second step, substituting the isobaric process formula, W₂ = P × (V₃ - V₂). Adding the two work values gives the total work done, W_total = W₁ + W₂.
Given that T₁ is the same for both processes, we can calculate W_total using the provided values. For the final numerical answer, W_total = -1250 J, where the negative sign indicates work done on the gas.