Calculate w, q and ∆U when 0.75 mol of an ideal gas expands isothermally and slowly at 27°C from a volume of 15 L to 25 L. 

For isothermal reversible expansion of an ideal gas,  w = −nRT ln

V₂/V₁ = −2.303nRT log V₂/V₁  Putting n = 0.75 mol,

V₁ = 15L, V₂ = 25L, T = 27 + 273 = 300K  and  R = 8.314 JK⁻¹ mol⁻¹, we get  w = −2.303 × 0.75 × 8.314 × 300 log 25/15

= − 955.5 J  (−ve sign represents work of expansion)

For isothermal expansion of an ideal gas,  ΔU = 0  

∴  ΔU = q + w  gives  q = − w = + 955.5 J.