The value for Δ U for the reversible isothermal evaporation of 90 g water at 100 °C will be (Δ Hevap of water = 40.8 kJ mol-1, R = 8.314 J K-1 mol-1)

Question

The value for Δ U for the reversible isothermal evaporation of 90 g water at 100 °C will be (Δ Hevap of water = 40.8 kJ mol-1, R = 8.314 J K-1 mol-1) (A) 4800 kJ (B) 188.494 kJ (C) 40.8 kJ (D) 125.03 kJ

Tardigrade Admin · Accepted Answer

Δ H for 18 g water = 40.8 kJ For 90 g water = (40.8/18)×90 = 204 kJ n for 90 g water = 90/18 = 5 Δ ng = 5 - 0 = 5 Δ H = Δ U+Δ ngRT Δ U = 204000-(5×8.314×373) = 188494 J or 188.494 kJ

Q. The value for $Δ U$ for the reversible isothermal evaporation of $90 g$ water at $100^{°} C$ will be ( $Δ H_{e v a p}$ of water $= 40.8 k J m o l^{- 1}$ , $R = 8.314 J K^{- 1} m o l^{- 1}$ )

$4800 k J$

$188.494 k J$

$40.8 k J$

$125.03 k J$

$Δ H$ for $18 g$ water $= 40.8 k J$
For $90 g$ water $= \frac{40.8}{18} \times 90 = 204 k J$
$n$ for $90 g$ water $= 90/18 = 5$
$Δ n_{g} = 5 - 0 = 5$
$Δ H = Δ U + Δ n_{g} RT$
$Δ U = 204000 - (5 \times 8.314 \times 373)$
$= 188494 J$ or $188.494 k J$

Q. The value for ΔU for the reversible isothermal evaporation of 90g water at 100°C will be (ΔHevap​ of water =40.8kJmol−1, R=8.314JK−1mol−1)

4800kJ

188.494kJ

40.8kJ

125.03kJ