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

Temperature T = 37^oC = 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