RE-NEET 2026 Chemistry Laws of Thermodynamics MCQ Question
Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure. w₁, w₂, w₃ and w₄ represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; ΔU₂ and ΔU₄ are changes in the internal energy for the processes 2 and 4, respectively. [use R = 2 cal K⁻¹ mol⁻¹] The correct option is
w₂ + w₄ = ΔU₂ - ΔU₄
w₁ + w₂ = 2T₁ ln V₂/V₁
w₁ + w₂ + w₃ + w₄ = 0
w₁ + w₃ = -2T₁ ln V₂/V₁ - 2T₂ ln V₄/V₃
Correct Answer
Detailed Explanation
In this problem, we analyze the work done in a cyclic process involving an ideal gas. The correct answer is option D:
This equation arises from the first law of thermodynamics, which states that the change in internal energy () equals the heat added to the system minus the work done by the system. Since processes 2 and 4 are adiabatic, no heat is exchanged, leading to the relationship between work and internal energy changes.
The other options fail because they either misrepresent the relationships between work and internal energy or do not account for the adiabatic nature of the processes correctly.
Found an issue with this question?