Assuming that water vapour is an ideal gas, the internal energy (ΔU) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (Given: Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol−1 and R = 8.3 J mol−1K−1) will be

Question

Assuming that water vapour is an ideal gas, the internal energy (ΔU) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (Given: Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol−1 and R = 8.3 J mol−1K−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

H2O(ℓ) xrightarrow textvaporisation H2O(g) Δ ng=1-0=1 Δ H=Δ U+Δ ngRT Δ U=Δ H-Δ ngRT =41-8.3×10-3×373 =37.9 kJ mol-1 Hence, (3) is correct.

Q. Assuming that water vapour is an ideal gas, the internal energy $(Δ U)$ when 1 mol of water is vapourised at 1 bar pressure and $10 0^{°} C$ , (Given: Molar enthalpy of vapourization of water at 1 bar and $373 K = 41 k J m o l^{- 1}$ and $R = 8.3 J m o l^{- 1} K^{- 1}$ ) will be

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

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

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

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

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

Q. Assuming that water vapour is an ideal gas, the internal energy (ΔU) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (Given: Molar enthalpy of vapourization of water at 1 bar and 373K=41kJmol−1 and R=8.3Jmol−1K−1) will be

4.100kJmol−1

3.7904kJmol−1

37.904kJmol−1

41.00kJmol−1