Chapter 1, Problem 1.9NP

(a)

Interpretation Introduction

Interpretation: Interpret the mole percentage of each of the gases in the 31.0 L sample of moist air at 60°C and 1 atm in which the partial pressure of water vapor is 0.131 atm.

Also,

• Mole percent of ${\text{N}}_{\text{2}}$ = 78.0%
• Mole percent of ${\text{O}}_{\text{2}}$ = 21.0%
• Mole percent of Ar = 1.00%

Concept Introduction: The ideal gas equation is the equation which gives the relations between P, V, n and T of gases. It can be written as:

  $\text{P × V = n × R × T}$

Here, P is pressure V is volume, n is number of moles, R is Universal gas constant and T is temperature.

The mole percent of any substance can be calculated in a mixture as follows:

  $\text{\% mole=}\frac{\text{Partial pressure of substance}}{\text{Total pressure of mixture}}\times 100\%$

Also, partial pressure of substance can be calculated as follows:

  $\text{Partial pressure of substance}=\text{mole fraction of substnace×total pressure of substance}$

(b)

Interpretation Introduction

Interpretation: The volume of the mixture needs to be determined.

Concept Introduction: The ideal gas equation is the equation which gives the relations between P, V, n and T of gases. It can be written as:

  $\text{P × V = n × R × T}$

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

(c)

Interpretation Introduction

Interpretation: The fraction of the water that will be condensed if the total pressure of the mixture is isothermally increased to 81.0 atm needs to be determined.

Concept Introduction: The partial pressure of a gas say A in mixture of gases can be calculated as follows:

  $P_A=X_A{P}_T$

Here, $P_A$ is partial pressure of gas A, $X_A$ is mole fraction of gas A and $P_T$ is total pressure of the gas in mixture.