solution:
Enthalpy change of vapourisation for
![\[1\text{ }mole\text{ }=\text{ }40.79\text{ }kJ\text{ }mol1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1da8136a9ea2ca4dc78216be57999d46_l3.png "Rendered by QuickLaTeX.com")
enthalpy change of vapourisation for
![\[2\text{ }moles\text{ }of\text{ }water\text{ }=\text{ }\left( 40.79\text{ }\times \text{ }2 \right)\text{ }=\text{ }81.58kJ\text{ }mol1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cfce431c79beac8a69fd231e24881a47_l3.png "Rendered by QuickLaTeX.com")
for water
![\[Hvapourisation,\text{ }\text{ }will\text{ }be\text{ }equivalent\text{ }to\text{ }=40.79\text{ }kJ\text{ }mol1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d03291c8a5510a69c9825a9a9422921b_l3.png "Rendered by QuickLaTeX.com")
18.0 g of water totally vapourises 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 vapourising two moles of water under similar conditions? What is the standard enthalpy of vapourisation for water?
18.0 g of water totally vapourises 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 vapourising two moles of water under similar conditions? What is the standard enthalpy of vapourisation for water?