Temperature T = 37^{o}C = 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 (V _{2}/ V_{1})**

= – (1 mol) (8.314 J K

^{-1}mol

^{-1}) (310 K) ln (30 dm

^{3}/ 20dm

^{3})

= – 1045.02 J

From first law, ∆U= q + w

Since ∆U=0, q = – w = 1045.02 J