The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100° C is (Given: Δ H text vap for water at 373 K =41 kJ / mol . R =8.314 JK -1 mol -1)

Question

The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100° C is (Given: Δ H text vap for water at 373 K =41 kJ / mol . R =8.314 JK -1 mol -1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

H 2 O (ℓ) leftharpoons H 2 O ( g ) 90 gm of H 2 O Δ H =Δ U +Δ n g RT ⇒ 5 moles of H 2 O 5 × 41000 J =Δ U +1 × 8.314 × 373 × 5 Δ U =189494.39 Joule

Q. The internal energy change (in J) when $90 g$ of water undergoes complete evaporation at $10 0^{\circ} C$ is

(Given : $Δ H_{vap}$ for water at $373 K = 41 k J / m o l$ $R = 8.314 J K^{- 1} m o l^{- 1})$

$H_{2} O (ℓ) ⇌ H_{2} O (g) 90 g m$ of $H_{2} O$

$Δ H = Δ U + Δ n_{g} RT$

$\Rightarrow 5$ moles of $H_{2} O$

$5 \times 41000 J = Δ U + 1 \times 8.314 \times 373 \times 5$

$Δ U = 189494.39$ Joule

Q. The internal energy change (in J) when 90g of water undergoes complete evaporation at 100∘C is (Given : ΔHvap ​ for water at 373K=41kJ/mol R=8.314JK−1mol−1)

H2​O(ℓ)⇌H2​O(g)90gm of H2​O ΔH=ΔU+Δng​RT ⇒5 moles of H2​O 5×41000J=ΔU+1×8.314×373×5 ΔU=189494.39 Joule

Q. The internal energy change (in J) when $90 g$ of water undergoes complete evaporation at $10 0^{\circ} C$ is

(Given : $Δ H_{vap}$ for water at $373 K = 41 k J / m o l$ $R = 8.314 J K^{- 1} m o l^{- 1})$

$H_{2} O (ℓ) ⇌ H_{2} O (g) 90 g m$ of $H_{2} O$

$Δ H = Δ U + Δ n_{g} RT$

$\Rightarrow 5$ moles of $H_{2} O$

$5 \times 41000 J = Δ U + 1 \times 8.314 \times 373 \times 5$

$Δ U = 189494.39$ Joule