Question

On a humid day in summer, the mole fraction of gaseous ${{H}_{2}}O$ (water vapour) in the air at 25 $^{0}C$ can be as high as 0.0287. Assuming a total pressure of 0.977 atm. What is the partial pressure of dry air ?

Answer 1

An attempt to this question can be made by understanding the term partial pressure. Try to formulate a relation between partial pressure, total pressure and mole fraction. Based on this you can substitute values and thus find the partial pressure of dry air.

Complete step-by-step solution:
Partial pressure can be defined as the pressure applied by the particular gas in a container consisting of more than 1 gas. The total pressure of an ideal gas mixture is equal to the sum of the partial pressure of constituent gases. Partial pressure in terms of calculation is defined as the product of mole fraction of the gas and the total pressure of the mixture.
The formula is given below:
$\text{p = (X) x (}{{\text{p}}_{T}})$
Where,
p is the partial pressure of the gas
X is the mole fraction of gas
${{\text{p}}_{T}}$ is the total pressure of the mixture
We will now substitute the values to obtain the partial pressure of water vapor.
${{\text{p}}_{{{H}_{2}}O}}\text{ = }{{\text{X}}_{{{H}_{2}}O}}{{\text{p}}_{Total}}$
= 0.0287 x 0.977 = 0.028 atm
The partial pressure of dry air will be the difference of total pressure and the partial pressure of water vapor.
${{p}_{dry\text{ }air}}\text{ = 0}\text{.977 - 0}\text{.028}$ = 0.949 atm.

Therefore, the partial pressure of dry air is 0.949 atm and the correct answer is option (B).

Note: It is important to know that the above equation holds good for ideal gases only. In case of real gases, the equation ceases to exist.