$H_{2}O(l)$ [image:5_image21.emf] $H_{2}O(g),\ \mathrm{\Delta}H_{vap} = 10\ kcal\ mol^{- 1}$. If pressure is increased

Question

Dr. Abubacker · Accepted Answer

(c)
$H_{2}O \rightleftharpoons H_{2}O(g)$. This is endothermic process, taking place with increase in pressure. If pressure is increased, equilibrium is displaced in backward side (Le-Chatelier) hence, steam is liquefied. To boil the liquid again, boiling point increases

Q. $H_{2}O(l)$

$H_{2}O(g),\ \mathrm{\Delta}H_{vap} = 10\ kcal\ mol^{- 1}$. If pressure is increased

Explanation

Reviewed by Dr. Abubacker

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Q. $H_{2}O(l)$ $H_{2}O(g),\ \mathrm{\Delta}H_{vap} = 10\ kcal\ mol^{- 1}$. If pressure is increased

Explanation

Reviewed by Dr. Abubacker

Want more practice?

Q. $H_{2}O(l)$

$H_{2}O(g),\ \mathrm{\Delta}H_{vap} = 10\ kcal\ mol^{- 1}$. If pressure is increased