Question

Pure water was allowed to come to equilibrium with its vapour in two different vessels of volume V and 2V at same temperature and having vapour pressure P1 and P2 respectively. Assuming that the only gas present inside the vessels to be H2O (g), the correct relation between P1 and P2 is


P1 > P2


P1 < P2


P1 = P2


P1 = 2P2

Answer 1

Answer:

sorry friend I not know the answer of this question

Answer 2

Answer:

The vapour pressure of pure water ( P₁ ) is equal to the vapour pressure of the vapours/gaseous state of water ( P₂ ).

Hence, option c) P₁ = P₂ is correct.

Explanation:

Given,

The volume of pure water = V

The volume of vapours = 2V

The vapour pressure of pure water = P₁

The vapour pressure of vapours/gaseous state of water = P₂

The pure water is in equilibrium with the vapour state.

The reaction: $H_{2}O(l) \rightleftharpoons H_{2}O (g)$

The temperature remains constant.

The relation between P₁ and P₂ =?

Vapour pressure: When the thermodynamic equilibrium state is reached in a closed container, the pressure is applied between molecules leaving the liquid phase and going into the gaseous phase and the molecules leaving the gaseous phase and entering the liquid phase.

As we know,

• The vapour pressure is independent of volume, i.e., vapour pressure does not depend upon the volume of the liquid or gaseous phase.
• The vapour pressure depends upon the temperature.
• The vapour pressure increases as the temperature increases and vice versa.

And in the given case, the temperature is the same.

Therefore, the vapour pressure is also the same.

Hence, P₁ = P₂.