One mole of diatomic ideal gas at 27°C undergoes isothermal expansion until its volume doubles, then adiabatic compression back to initial volume. Determine total work done (J). Take 2¹⁄⁵ = 1.149 or as specified in the PDF.
Physics · Mixed Chapter
JEE Main Practice

Question

One mole of diatomic ideal gas at 27°C undergoes isothermal expansion until its volume doubles, then adiabatic compression back to initial volume. Determine total work done (J). Take 2¹⁄⁵ = 1.149 or as specified in the PDF.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The isothermal expansion at T₁=300 K yields W₁ = RT₁ ln2. Step 2: For a diatomic gas γ = 7/5. During adiabatic compression from 2V to V: T₂ = T₁×2²⁄⁵ = 300×2²⁄⁵. Step 3: Work by gas in adiabatic compression W₂ = Cᵥ(T₁−T₂) = (5R/2)×300(1−2²⁄⁵). Step 4: Wtotal = R×300[ln2 + (5/2)(1−2²⁄⁵)] ≈ −265 J, close to printed −263.77 J depending on the PDF’s stated 2²⁄⁵ approximation.