120 g of a liquid X (molar mass = 40 g mol ⁻¹ ) completely vaporises at 400 K and 1 bar pressure. The molar enthalpy of vaporisation of X under these conditions is 33.32 kJ mol ⁻¹ . The change in internal energy for the vaporisation of 120 g of liquid X is ________ kJ . (Given: R = 8.3 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

Number of moles of liquid X , n = 120 g 40 g mol ⁻¹ = 3 moles Total enthalpy change for 3 moles, H = n _ vap H H = 3 mol 33.32 kJ mol ⁻¹ = 99.96 kJ The vaporisation process for 3 moles is: 3 X (l) 3 X (g) Change in gaseous moles, n_g = 3 - 0 = 3 Using the first law of thermodynamics: H = U + n_g RT U = H - n_g RT U = 99.96 kJ - (3 8.3 J K ⁻¹ mol ⁻¹ 400 K 10⁻³ kJ J ⁻¹ ) U = 99.96 kJ - 9.96 kJ = 90 kJ Answer: 90