An ideal gas of 0.5 mol expands isothermally; its final volume V₂ increases from 0.1 m³ at Q=0 to 0.3 m³ when heat supplied is 1500 J (as shown in the V₂–Q graph). If its temperature is 120A K, find A. Use ln9=2 and R=25/3 J mol⁻¹ K⁻¹.
Physics · Mixed Chapter
JEE Mains

Question

An ideal gas of 0.5 mol expands isothermally; its final volume V₂ increases from 0.1 m³ at Q=0 to 0.3 m³ when heat supplied is 1500 J (as shown in the V₂–Q graph). If its temperature is 120A K, find A. Use ln9=2 and R=25/3 J mol⁻¹ K⁻¹.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Iₛothermal expansion of an ideal gas obeys Q = nRT ln(V₂/V₁). Step 2 — Read the given graph: at Q = 0, initial volume V₁ = 0.1 m³; at Q = 1500 J, final volume V₂ = 0.3 m³. Step 3 — Hence V₂/V₁ = 3 and Q = nRT ln3. Given ln9 = 2, we have ln3 = (ln9)/2 = 1. Step 4 — With n = 0.5 mol and R = 25/3 J mol⁻¹ K⁻¹: 1500 = 0.5 × (25/3) × T × 1. Step 5 — T = 1500 × 6/25 = 360 K. From T = 120A K, A = 360/120 = 3. Answer: 3.