For water Δ text vap H =41 kJ mol -1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is kJ mol -1 [Use: R =8.3 J mol -1 K -1 ]

Question

For water Δ text vap H =41 kJ mol -1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is kJ mol -1 [Use: R =8.3 J mol -1 K -1 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given equation is H 2 O (ℓ) longrightarrow H 2 O ( g ): Δ H =41 ( kJ / mol ) ⇒ From the relation: Δ H =Δ U +Δ n g RT ⇒ 41 ( kJ / mol )=Δ U +(1) × (8.3/1000) × 373 ⇒ DU =41-3.0959 =38 kJ / mol

Q. For water $Δ_{vap} H = 41 k J m o l^{- 1}$ at $373 K$ and $1$ bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___ $k J m o l^{- 1}$
[Use : $R = 8.3 J m o l^{- 1} K^{- 1}$ ]

Given equation is
$H_{2} O (ℓ) ⟶ H_{2} O (g) : Δ H = 41 \frac{k J}{m o l}$
$\Rightarrow$ From the relation : $Δ H = Δ U + Δ n_{g} RT$
$\Rightarrow 41 \frac{k J}{m o l} = Δ U + (1) \times \frac{8.3}{1000} \times 373$
$\Rightarrow D U = 41 - 3.0959$
$= 38 k J / m o l$

Q. For water Δvap ​H=41kJmol−1 at 373K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___kJmol−1 [Use : R=8.3Jmol−1K−1 ]

Given equation is H2​O(ℓ)⟶H2​O(g):ΔH=41molkJ​ ⇒ From the relation : ΔH=ΔU+Δng​RT ⇒41molkJ​=ΔU+(1)×10008.3​×373 ⇒DU=41−3.0959 =38kJ/mol

Q. For water $Δ_{vap} H = 41 k J m o l^{- 1}$ at $373 K$ and $1$ bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___ $k J m o l^{- 1}$
[Use : $R = 8.3 J m o l^{- 1} K^{- 1}$ ]

Given equation is
$H_{2} O (ℓ) ⟶ H_{2} O (g) : Δ H = 41 \frac{k J}{m o l}$
$\Rightarrow$ From the relation : $Δ H = Δ U + Δ n_{g} RT$
$\Rightarrow 41 \frac{k J}{m o l} = Δ U + (1) \times \frac{8.3}{1000} \times 373$
$\Rightarrow D U = 41 - 3.0959$
$= 38 k J / m o l$