Q1(b). Δ U for heating one mole of water from 25 to 30°C

Reading of the data: the two densities printed (0.9970 and 0.9956 g cm⁻³) are those of water at 25°C and 30°C; water has density 1.000 g cm⁻³ at 0°C and 0.958 g cm⁻³ at 100°C, so the two values cannot belong to 0 and 100°C. They are used for the initial and the final state. The heat capacity is not given; the standard value C_(p,m)=75.3 J K⁻¹ mol⁻¹ is assumed. The question asks for the change Δ U.

1. Heat at constant pressure: q_P=Δ H=nC_(p,m)Δ T=1.00×75.3×5.0 Δ H=376.5 J
2. Molar volumes (M=18.015 g mol⁻¹): V₁=(18.015)/(0.9970)=18.069cm³, V₂=(18.015)/(0.9956)=18.095cm³, Δ V=+0.0254cm³
3. Expansion work against the constant 1.00 atm: w=-pΔ V=-(101 325 Pa)(0.0254×10⁻⁶m³)=-0.0026 J
4. First law: Δ U=q+w=Δ H-pΔ V=376.5-0.0026 Δ U=376.5 J (≈0.38 kJ)
5. The work term is 0.0026 J, about one part in 10⁵ of Δ H. Liquids and solids hardly expand, so for a condensed phase Δ U≈Δ H and C_V≈ Cₚ to within a few per cent.
6. Read literally (densities at 0 and 100°C), the volume change over 5 K is even smaller (0.0013 cm³), and Δ U is still 376.5 J. The answer does not depend on the reading.

Application. for gases the difference is not negligible: Δ H-Δ U=Δ n_gRT. That is why bomb-calorimeter values (Δ U) of combustion must be corrected to Δ H, while for reactions of solids and liquids alone no correction is needed.