Assuming that water vapour is an ideal gas, the internal energy change (Δ U ) when 1 mole of water is vaporised at 1 bar pressure and 100° C, (given: molar enthalpy of vaporization of water 41 kJ mol -1 at 1 bar and 373 K and R =8.3 J mol -1 K -1 ) will be :

Question

Assuming that water vapour is an ideal gas, the internal energy change (Δ U ) when 1 mole of water is vaporised at 1 bar pressure and 100° C, (given: molar enthalpy of vaporization of water 41 kJ mol -1 at 1 bar and 373 K and R =8.3 J mol -1 K -1 ) will be : (A) 4.100 kJ mol -1 (B) 3.7904 kJ mol -1 (C) 37.904 kJ mol -1 (D) 41.00 kJ mol -1

Tardigrade Admin · Accepted Answer

H 2 O (ℓ) xrightarrow text vaporisation H 2 O (g) Δ n g =1-0=1 Δ H =Δ U +Δ n 9 RT Δ U =Δ H -Δ n g RT =41-8.3 × 10-3 × 373 =37.9 kJ mol -1

$4.100 k J m o l^{- 1}$

$3.7904 k J m o l^{- 1}$

$37.904 k J m o l^{- 1}$

$41.00 k J m o l^{- 1}$

$H_{2} O_{(ℓ)} vaporisation H_{2} O_{(g)}$
$Δ n_{g} = 1 - 0 = 1$
$Δ H = Δ U + Δ n_{9} RT$
$Δ U = Δ H - Δ n_{g} RT$
$= 41 - 8.3 \times 1 0^{- 3} \times 373$
$= 37.9 k J m o l^{- 1}$

Q. Assuming that water vapour is an ideal gas, the internal energy change (ΔU) when 1 mole of water is vaporised at 1 bar pressure and 100∘C, (given : molar enthalpy of vaporization of water 41kJ mol−1 at 1 bar and 373K and R=8.3Jmol−1K−1 ) will be :

4.100kJmol−1

3.7904kJmol−1

37.904kJmol−1

41.00kJmol−1

H2​O(ℓ)​ vaporisation ​H2​O(g)​ Δng​=1−0=1 ΔH=ΔU+Δn9​RT ΔU=ΔH−Δng​RT =41−8.3×10−3×373 =37.9kJmol−1

$4.100 k J m o l^{- 1}$

$3.7904 k J m o l^{- 1}$

$37.904 k J m o l^{- 1}$

$41.00 k J m o l^{- 1}$

$H_{2} O_{(ℓ)} vaporisation H_{2} O_{(g)}$
$Δ n_{g} = 1 - 0 = 1$
$Δ H = Δ U + Δ n_{9} RT$
$Δ U = Δ H - Δ n_{g} RT$
$= 41 - 8.3 \times 1 0^{- 3} \times 373$
$= 37.9 k J m o l^{- 1}$