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?
![\[1\text{ }mole\text{ }=\text{ }40.79\text{ }kJ\text{ }mol1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1da8136a9ea2ca4dc78216be57999d46_l3.png "Rendered by QuickLaTeX.com")
![\[2\text{ }moles\text{ }of\text{ }water\text{ }=\text{ }\left( 40.79\text{ }\times \text{ }2 \right)\text{ }=\text{ }81.58kJ\text{ }mol1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cfce431c79beac8a69fd231e24881a47_l3.png "Rendered by QuickLaTeX.com")
![\[Hvapourisation,\text{ }\text{ }will\text{ }be\text{ }equivalent\text{ }to\text{ }=40.79\text{ }kJ\text{ }mol1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d03291c8a5510a69c9825a9a9422921b_l3.png "Rendered by QuickLaTeX.com")
