18.0 g of water completely vaporises at $100^{\circ} C$ and 1 bar pressure and the enthalpy change in the process is 40.79 kJ $m o l^{- 1}$ . What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

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Solution

Given that, quantity of water =18.0 g, pressure =1 bar
As we know that, 18.0 $g H_{2} O$ = 1 mole $H_{2} O$
Enthalpy change for vaporising 1 mole of $H_{2} O$ = 40.79 kJ $m o 1^{- 1}$
$∴$ Enthalpy change for vaporising 2 moles of $H_{2} O$ = 2 $\times$ 40.79 kJ= 81.358kJ
Standard enthalpy of vaporisation at $100^{\circ} C$ and 1 bar pressure, $Δ_{v a p} H^{\circ} = + 40.79 k J m o l^{- 1}$

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18.0 g of water completely vaporises at 100∘C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol−1. What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

18.0 g of water completely vaporises at $100^{\circ} C$ and 1 bar pressure and the enthalpy change in the process is 40.79 kJ $m o l^{- 1}$ . What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?