calculate how muc heat must be supplied to evaporate 18g of water at 298 K

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Solution

Number of moles in 18 g H₂O(l) is = 18g/18 g mol^-1 = 1mol
Δ_vapU^o =Δ_vapH^o − pΔV = Δ_vapH^o −Δn_gRT
Δ_vapH^o – Δn_gRT = 40.66 kJ mol^-1 -(1)(8.314 JK^-1 mol^-1)(373k)(10^-3KJ J^-1)
Δ_vapU^o = 40.66 kJ mol^-1 – 3.10 kJ mol^-1
= 37.56kJ mol

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calculate how muc heat must be supplied to evaporate 18g of water at 298 K

Number of moles in 18 g H2O(l) is = 18g/18 g mol-1 = 1molΔvapUo =ΔvapHo − pΔV = ΔvapHo −ΔngRTΔvapHo – ΔngRT = 40.66 kJ mol-1 -(1)(8.314 JK-1 mol-1)(373k)(10-3KJ J-1)ΔvapUo = 40.66 kJ mol-1 – 3.10 kJ mol-1 = 37.56kJ mol

Number of moles in 18 g H₂O(l) is = 18g/18 g mol^-1 = 1mol
Δ_vapU^o =Δ_vapH^o − pΔV = Δ_vapH^o −Δn_gRT
Δ_vapH^o – Δn_gRT = 40.66 kJ mol^-1 -(1)(8.314 JK^-1 mol^-1)(373k)(10^-3KJ J^-1)
Δ_vapU^o = 40.66 kJ mol^-1 – 3.10 kJ mol^-1
= 37.56kJ mol