Chapter 7, Problem 7.78P

Interpretation Introduction

(a)

Interpretation:

The nitrox gas tank consists higher than normal level of oxygen and lower than normal level nitrogen and this nitrox gas tank is used for the purpose of the scuba diving. The percentage of N₂ and O₂ gas in the nitrox tank is to be determined.

Concept Introduction:

According to Dalton's law −

  $P_{Total}=P_1+P_2+P_3+\mathrm{..........}$

Here, $P_{Total}$ is total pressure and $P_1+P_2+P_3+\mathrm{..........}$ is sum of partial pressure of all the gases present in the system.

Raoult's law is defined as follows:

  $P_A=X_A{P}_{Total}$

Here, $P_A$ is partial pressure of A, $X_A$ is mole fraction of A and $P_{Total}$ is total pressure.

Answer to Problem 7.78P

The percentage of N₂ gas in nitrox tank = 65%

The percentage of O₂ gas in nitrox tank = 35%

Explanation of Solution

The mole fraction is defined as the ratio of the mole of individual gas with total number of the moles present in the mixture. This expression can be represented as −

  $x_A+x_B=1$

  $x_A=1-x_B$

According Dalton's law the total pressure is the sum of the partial pressure of each gas. This can be represented as −

  $P_{Total}=P_1+P_2+P_3+\mathrm{..........}$

Where,

  $P_{Total}=\text{total pressure}$

  $P_1=\text{partial pressure of gas 1}$

  $P_2=\text{partial pressure of gas 2}$

According to Raoult's law the partial pressure of individual gas is calculated by the multiplication of the total pressure with that of the partial pressure of that gas. in this the partial pressure of individual gas is directly proportional to the mole fraction of that gas.

  $P_A=X_A{P}_{Total}$

The percentage of N₂ gas in the nitrox gas can be calculated as −

  $\%{\text{N}}_{\text{2}}=\frac{{\text{mol of N}}_{\text{2}}}{\text{Total mol of gas}}\times 100\%$

  $\%N_2=\frac{13}{20}\times 100\%$

  $\%{\text{N}}_{\text{2}}=65\%$

The percentage of O₂ gas in the nitrox gas can be calculated as −

  $\%{\text{O}}_{\text{2}}=\frac{{\text{mol of O}}_{\text{2}}}{\text{Total mol of gas}}\text{×100\%}$

  $\%O_2=\frac{7}{20}\times 100\%$

  $\%{\text{O}}_{\text{2}}=35\%$

Interpretation Introduction

(b)

Interpretation:

The partial pressure of each gas in psi in the tank is to be determined.

Concept Introduction:

According to Dalton's law −

  $P_{Total}=P_1+P_2+P_3+\mathrm{..........}$

Here, $P_{Total}$ is total pressure and $P_1+P_2+P_3+\mathrm{..........}$ is sum of partial pressure of all the gases present in the system.

Raoult's law is defined as follows:

  $P_A=X_A{P}_{Total}$

Here, $P_A$ is partial pressure of A, $X_A$ is mole fraction of A and $P_{Total}$ is total pressure.

Answer to Problem 7.78P

The partial pressure of nitrogen gas in nitrox tank = 1120 psi

The partial pressure of oxygen gas in nitrox tank = 2080 psi

Explanation of Solution

The mole fraction of the N₂ gas can be calculated as −

  $X_{N_2}=\frac{{\text{mol of N}}_{\text{2}}}{\text{Total mol of gas}}$ ]

Putting the values,

  $X_{N_2}=\frac{13}{20}=0.65$

The mole fraction of the O₂ gas can be calculated as −

  $X_{{\text{O}}_{\text{2}}}=\frac{{\text{mol of O}}_{\text{2}}}{\text{Total mol of gas}}$

  $X_{O_2}=\frac{7}{20}=0.35$

The mole fraction of N₂ gas = 0.65

The mole fraction of O₂ gas = 0.35

It is given that,

Total pressure = 3200psi

The partial pressure of nitrogen gas can be calculated as −

  $P_{N_2}=X_{N_2}{P}_{Total}$

  $P_{N_2}=0.65\times 3200psi$

  $P_{N_2}=2080psi$

The partial pressure of oxygen gas can be calculated as −

  $P_{O_2}=X_{O_2}{P}_{Total}$

  $P_{{\text{O}}_{\text{2}}}=0.35\times 320\text{0 psi}$

  $P_{{\text{O}}_{\text{2}}}=112\text{0 psi}$

The partial pressure of nitrogen gas in nitrox tank = 1120 psi

The partial pressure of oxygen gas in nitrox tank = 2080 psi