A swimmer coming out from a pool is covered with a film of water weigh

1.	A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporization at 100°C.∆vapHΘ for water at 373 K = 40.66 kJ mol-1
Answer» Given that 18gH₂O(l) \(\overset{vaporisation}{\rightarrow}\) 18gH₂O (g) No. of moles in 18 g H₂O(l) =\(\frac{18g}{18g\,mol^{-1}}\) ∆_vapU^Θ = ∆_vapH^Θ − p∆V = ∆_vapH^Θ − ∆n_gRT (Assuming steam behaves as an ideal gas) = 40.66 kJ mol⁻¹ − 1 × 8.314 JK⁻¹ mol⁻¹ × 373K × 103 kJ J⁻¹ = 40.66 kJ mol⁻¹ − 3.10 kJ mol⁻¹ = 37.56 kJ mol⁻¹

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporization at 100°C.∆vapHΘ for water at 373 K = 40.66 kJ mol-1

Answer»

Given that

18gH₂O(l) \(\overset{vaporisation}{\rightarrow}\) 18gH₂O (g)

No. of moles in 18 g H₂O(l) =\(\frac{18g}{18g\,mol^{-1}}\)

∆_vapU^Θ = ∆_vapH^Θ − p∆V

= ∆_vapH^Θ − ∆n_gRT

(Assuming steam behaves as an ideal gas)

= 40.66 kJ mol⁻¹ − 1 × 8.314 JK⁻¹ mol⁻¹ × 373K × 103 kJ J⁻¹

= 40.66 kJ mol⁻¹ − 3.10 kJ mol⁻¹

= 37.56 kJ mol⁻¹

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A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporization at 100°C.∆vapHΘ for water at 373 K = 40.66 kJ mol-1

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