Calculate q, w, ∆U and ∆H for the reversible isothermal expansion of one mole of an ideal gas at 37oC from a volume of 20dm3 to a volume of 30 dm3

Temperature T = 37oC = 37 + 273 = 310 K
Since the process is Isothermal,

Therefore, ∆U=0 and ∆H = 0 (as for an isothermal expansion of an ideal gas ∆U=0 and ∆H = 0)

As work done in reversible isothermal expansion is given by:

w= -nRT ln (V2/ V1)
= – (1 mol) (8.314 J K-1 mol-1) (310 K) ln (30 dm3 / 20dm3)
= – 1045.02 J

From first law, ∆U= q + w
Since ∆U=0, q = – w = 1045.02 J

Category: First Law of Thermodynamics

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