The percent increase in the vapour pressure of water when the temperature increases by 1 °c from 14°c to 15°c will be 5.1% increase.
Using Combined gas law ,
P₁V₁/T₁ = P₂V₂/T₂
Where ,
P₁ = 1 atm , V₁ and V₂ = 1 L , T₁ = 287.15 K (14°c) and T₂ = 288.15 K (15°c)
P₁V₁/T₁ = P₂V₂/T₂
V₁/T₁= (P₂/P₁)V₂/T₂
(P₂V₁) = (V₁/T₁1)×(T₂/V₂)
(P₂V₁) = (V₁/V₂)×(T₂/T₁)
(P₂/P₁) = (1L/1L)×(288.15K/287.15K)
(P₂/P₁) = 1.051×(1.051 - 1.00)
= 0.051 atm(0.051 atm/1.00 atm)
= 0.051
The vapour pressure at a liquid's normal boiling point is the same as the ordinary atmospheric pressure, which is 1 atmosphere, 760 Torr, 101.325 kPa, or 14.69595 psi.
The equilibrium pressure between molecules moving from a liquid into a gaseous phase and molecules moving from a gaseous phase into a liquid phase in a closed container is known as the vapour pressure of a liquid.
The equilibrium pressure between molecules moving from a liquid into a gaseous phase and molecules moving from a gaseous phase into a liquid phase in a closed container is known as the vapour pressure of a liquid.
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