Chapter 12, Problem 12.62QE

(a)

Interpretation Introduction

Interpretation:

Solubility of ozone will be greater or lesser than $0.0052m$ at $\text{0}\text{°C}$ and $\text{1}\text{.0}\text{atm}$ has to be stated.

Concept Introduction:

Henry’s law states that the solubility of the gas at a given temperature is directly proportional to the partial pressure of the gas.  This can be represented in equation form as shown below;

    $\begin{array}{l}C\propto P\\ C=kP\end{array}$

Where,

    $C$ is the concentration of gas.

    $k$ is the proportionality constant.

    $P$ is the partial pressure.

Solubility of gas is directly proportional to the partial pressure of the gas.  Therefore, as the pressure increases, the solubility of the gas also increases.

Temperature also plays an important role in solubility of gas.  If the enthalpy of the solution is positive, then the solubility will increase with the increase in temperature.  If the enthalpy of the solution is negative, then the solubility will decrease with increase in temperature.

(b)

Interpretation Introduction

Interpretation:

Solubility of ozone will be greater or lesser than $0.0052m$ at $\text{40}\text{°C}$ and $\text{1}\text{.00}\text{atm}$ has to be stated.

Concept Introduction:

Refer part (a)

(c)

Interpretation Introduction

Interpretation:

Solubility of ozone will be greater or lesser than $0.0052m$ at $\text{25}\text{°C}$ and $\text{2}\text{.0}\text{atm}$ has to be stated.

Concept Introduction:

Refer part (a)

(d)

Interpretation Introduction

Interpretation:

Solubility of ozone will be greater or lesser than $0.0052m$ at $\text{50}\text{°C}$ and $\text{0}\text{.50}\text{atm}$ has to be stated.

Concept Introduction:

Refer part (a)