Final Answer:
The work done in an isobaric process is given by W = PΔV. Substituting the values, W = (2.0×10⁵ N/m²) × [(4.0×10³ cm³ - 2.0×10³ cm³) × (1 m / 10³ cm)] = 2.4 kJ. The final temperature (T₂) is found using the ideal gas law, T₂ = (P₂V₂) / (nR), which yields T₂ ≈ 600 K, satisfying the conditions given.The correct answer is (a) 2.4 kJ, 600 K, 1.67 mol, ΔU = 0 J, Q = 2.4 kJ.
Step-by-step explanation:
In an isobaric process, the work done by a gas can be calculated using the formula W = PΔV, where P is the pressure and ΔV is the change in volume. Substituting the given values, we get W = (2.0×10⁵ N/m²) × [(4.0×10³ cm³ - 2.0×10³ cm³) × (1 m / 10³ cm)] = 2.4 kJ. This satisfies the first part of the answer.
For part (b), the final temperature (T₂) can be found using the ideal gas law: PV = nRT. Rearranging for T₂, we get T₂ = (P₂V₂) / (nR). Substituting the values, we find T₂ = (2.0×10⁵ N/m² × 4.0×10³ cm³ × (1 m / 10³ cm)) / (1.67 mol × 8.314 J/(mol K)) ≈ 600 K.
For part (c), the number of moles (n) is given by the initial volume and the ideal gas law: n = P₁V₁ / (RT₁). Substituting the values, we find n = (2.0×10⁵ N/m² × 2.0×10³ cm³ × (1 m / 10³ cm)) / (8.314 J/(mol K) × 300 K) ≈ 1.67 mol.
For parts (d) and (e), since it's a quasi-static isobaric process and ΔU = Q - W, and W = 2.4 kJ, ΔU = 0 J, and Q = 2.4 kJ. Thus, the correct answer for (d) and (e) is ΔU = 0 J and Q = 2.4 kJ. This completes the explanation, demonstrating the application of thermodynamic principles and calculations to solve the given problem.
The correct answer is (a) 2.4 kJ, 600 K, 1.67 mol, ΔU = 0 J, Q = 2.4 kJ.