A swimmer coming out from a pool is covered with a film of water weighing about 18 g. The internal energy of vaporization at 100°C is (Δ textvapH for water at 373 K = 40.66 kJ mol-1)

Question

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. The internal energy of vaporization at 100°C is (Δ textvapH for water at 373 K = 40.66 kJ mol-1) (A) 87.56 kJ mol-1 (B) 35.56 kJ mol-1 (C) 37.56 kJ mol-1 (D) 40.5 kJ mol-1

Tardigrade Admin · Accepted Answer

Δ textvap U=Δ textvapH-Δ ngRT H2O(l) arrow H2O(g) Δ ng=1-0=1 assuming steam behaving as an ideal gas, then Δ textvap U=40.66 -1 × 8.314× 10-3× 373 =37.56 kJ mol-1

Q. A swimmer coming out from a pool is covered with a film of water weighing about $18 g$ . The internal energy of vaporization at $10 0^{\circ} C$ is $(Δ_{vap} H$ for water at $373 K$ $= 40.66 k J m o l^{- 1})$

$87.56 k J m o l^{- 1}$

$35.56 k J m o l^{- 1}$

$37.56 k J m o l^{- 1}$

$40.5 k J m o l^{- 1}$

$Δ_{vap} U = Δ_{vap} H - Δ n_{g} RT$
$H_{2} O_{(l)} \to H_{2} O_{(g)}$
$Δ n_{g} = 1 - 0 = 1$
assuming steam behaving as an ideal gas, then
$Δ_{vap} U = 40.66 - 1 \times 8.314 \times 1 0^{- 3} \times 373$
$= 37.56 k J m o l^{- 1}$

Q. A swimmer coming out from a pool is covered with a film of water weighing about 18g. The internal energy of vaporization at 100∘C is (Δvap​H for water at 373K =40.66kJmol−1)

87.56kJmol−1

35.56kJmol−1

37.56kJmol−1

40.5kJmol−1