One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature of 27°C and does W joule of work. If the gas undergoes the same magnitude of expansion adiabatically from 27°C doing the same amount of work W, its final temperature will be close to °C. (Take logₑ 2=0.693.)

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(4) -56. Isothermal leg: W=nRT ln 2=8.314×300×0.693=1728 J. Adiabatic leg must do the same work, and there W=(nR(T₁-T₂))/(γ-1). Diatomic, so γ=7/5 and γ-1=0.4. 1728=(8.314(300-T₂))/(0.4), so 300-T₂=83.2 and T₂=216.8 K. 216.8-273=-56.2°C, closest to -56°C.

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