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Standard XII

Chemistry

Enthalpy

90g of water ...

Question

$90 g$ of water spilled out from a vessel in the room on the floor. Assuming that water vapour behaving as an ideal gas, Calculate the internal energy change when the spilled water undergoes complete evaporation at $100^{o} C$ . (Given the molar enthalpy of vaporisation of water at 1 bar and $373 k = 41 k J / m o l^{- 1}$ ).

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Solution

$△ U = ?$
$H_{2} O (l) ⟶ H_{2} O (g)$

[ $41 \times 10^{3}$ J/mol]
$∴$ For 18g $⟶ 41 \times 10^{3}$
For 90g $⟶ ? = \frac{90 \times 41 \times 10^{3}}{18}$
$= 205$ KJ/mol

$△ H = △ U + △ n R T$
$\Rightarrow 205 \times 10^{3} = △ U + 1 \times 8.314 \times 373$
$\Rightarrow △ U = 205000 - 3101.12$
$= 201.8$ KJ

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△U=?H2O(l)⟶H2O(g) [41×103 J/mol]∴ For 18g⟶41×103For 90g⟶?=90×41×10318=205 KJ/mol△H=△U+△nRT ⇒205×103=△U+1×8.314×373⇒△U=205000−3101.12=201.8 KJ