An ideal monatomic gas of n moles follows the cycle WXYZW shown in the V-T diagram. Successive processes are adiabatic and isobaric. Volumes at W, X and Y are 64 cm³, 125 cm³ and 250 cm³. Given that nRT at W is 1 J, find the heat absorbed along XY in joules. Image
Physics · Circular Motion
JEE Advanced 2025 Paper 2

Question

An ideal monatomic gas of n moles follows the cycle WXYZW shown in the V-T diagram. Successive processes are adiabatic and isobaric. Volumes at W, X and Y are 64 cm³, 125 cm³ and 250 cm³. Given that nRT at W is 1 J, find the heat absorbed along XY in joules. Image

Physics · Circular Motion

JEE Advanced 2025 Paper 2

G1 Physics · Accepted Answer

For a monatomic gas, γ = 5/3. Along the adiabatic path WX, T V²ᐟ³ = constant. Therefore T at X / T at W = (64/125)²ᐟ³ = (4/5)² = 16/25. Given nRT at W = 1 J, nRT at X = 16/25 = 0.64 J. XY is isobaric, so T ∝ V: T at Y / T at X = 250/125 = 2. Thus nRT at Y = 1.28 J and nRΔT = 1.28 − 0.64 = 0.64 J. For a monatomic gas, Cₚ = 5R/2. Hence Qₓᵧ = (5/2)nRΔT = (5/2)(0.64) = 1.6 J.