One mole (18.0 g) of ice at -10 degC is converted to steam at 100 degC at 1 atm. Data: c_p(ice) = 2.09, c_p(water) = 4.18 J g^-1K^-1; Delta_fusH = 6.01 kJ mol^-1 at 0 degC; Delta_vapH = 40.7 kJ mol^-1 at 100 degC. Treat steam as an ideal gas, neglect the volume of condensed phases and use T = 373 K and R = 8.314 J mol^-1K^-1 for the PV term. Calculate Delta U of the whole process in kJ (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3td014

One mole ( $18.0 g$ ) of ice at $- 10^{\circ} C$ is converted to steam at $100^{\circ} C$ at $1 atm$ . Data: $c_{p} (ice) = 2.09$ , $c_{p} (water) = 4.18 J g^{- 1} K^{- 1}$ ; $Δ_{f u s} H = 6.01 kJ mo l^{- 1}$ at $0^{\circ} C$ ; $Δ_{v a p} H = 40.7 kJ mo l^{- 1}$ at $100^{\circ} C$ . Treat steam as an ideal gas, neglect the volume of condensed phases and use $T = 373 K$ and $R = 8.314 J mo l^{- 1} K^{- 1}$ for the $P V$ term. Calculate $Δ U$ of the whole process in $kJ$ (two decimals).

Numerical answer type, answer in kJ, tolerance 0.02.

Δ H = q_{p} = 18 (2.09) (10) + 6010 + 18 (4.18) (100) + 40700 = 54610.2 J

. Only the vaporisation produces gas, so

Δ (P V) = RT = (8.314) (373) = 3101.1 J

Δ U = Δ H - Δ (P V) = (54610.2 - 3101.1) /1000 = 51.509 kJ