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