In water saturated air the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is :

Question

In water saturated air the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is : (A) 1.18 atm (B) 1.76 atm (C) 1.176 atm (D) 0.98 atm

Tardigrade Admin · Accepted Answer

XH2O = 0.02 ∴ Xgas = 0.98 Ptotal =1.2 atm partial pressure of dry-air = PT × mole fraction of dry-air Partial pressure of dry-air = 1.2 atm × 0.98 = 1 .176 atm.

Q. In water saturated air the mole fraction of water vapour is $0.02$ . If the total pressure of the saturated air is $1.2 a t m$ , the partial pressure of dry air is :

1.18 atm

1.76 atm

1.176 atm

0.98 atm

$X_{H_{2} O} = 0.02$
$∴ X_{g a s} = 0.98$
$P_{t o t a l} = 1.2 a t m$
partial pressure of dry-air $= P_{T} \times$ mole fraction of dry-air
Partial pressure of dry-air $= 1.2 a t m \times 0.98$
$= 1.176 a t m$ .

Q. In water saturated air the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2atm, the partial pressure of dry air is :