Chapter 12, Problem 12.60QE

(a)

Interpretation Introduction

Interpretation:

Solubility of neon will be greater or lesser than $9.07\times {10}^{-4}m$ at $\text{0}\text{°C}$ and $\text{3}\text{.0}\text{atm}$ 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.

Answer to Problem 12.60QE

Solubility of neon in water will be greater at $\text{0}\text{°C}$ and $\text{3}\text{.0}\text{atm}$.

Explanation of Solution

Solubility of neon is given as $9.07\times {10}^{-4}m$ at $\text{25}\text{°C}$ and $\text{2}\text{.0}\text{atm}$.  The given condition is $\text{0}\text{°C}$ and $\text{3}\text{.0}\text{atm}$.  The temperature is decreased from $\text{25}\text{°C}$ to $\text{0}\text{°C}$ and pressure is increased from $\text{2}\text{.0}\text{atm}$ to $\text{3}\text{.0}\text{atm}$.  The solubility of the gas increases with decrease in temperature and increase in pressure.  Therefore, the solubility of neon increases and it will be greater than $9.07\times {10}^{-4}m$.

(b)

Interpretation Introduction

Interpretation:

Solubility of neon will be greater or lesser than $9.07\times {10}^{-4}m$ at $\text{25}\text{°C}$ and $\text{1}\text{.00}\text{atm}$ has to be stated.

Concept Introduction:

Refer part (a)

Answer to Problem 12.60QE

Solubility of neon in water will be lesser at $\text{25}\text{°C}$ and $\text{1}\text{.0}\text{atm}$.

Explanation of Solution

Solubility of neon is given as $9.07\times {10}^{-4}m$ at $\text{25}\text{°C}$ and $\text{2}\text{.0}\text{atm}$.  The given condition is $\text{25}\text{°C}$ and $\text{1}\text{.0}\text{atm}$.  The temperature remains the same while the pressure is decreased from $\text{2}\text{.0}\text{atm}$ to $\text{1}\text{.0}\text{atm}$.  The solubility of the gas decreases with increase in temperature and decrease in pressure.  Therefore, the solubility of neon decreases and it will be lesser than $9.07\times {10}^{-4}m$.

(c)

Interpretation Introduction

Interpretation:

Solubility of neon will be greater or lesser than $9.07\times {10}^{-4}m$ at $\text{10}\text{°C}$ and $\text{2}\text{.0}\text{atm}$ has to be stated.

Concept Introduction:

Refer part (a)

Answer to Problem 12.60QE

Solubility of neon in water will be greater at $\text{10}\text{°C}$ and $\text{2}\text{.0}\text{atm}$.

Explanation of Solution

Solubility of neon is given as $9.07\times {10}^{-4}m$ at $\text{25}\text{°C}$ and $\text{2}\text{.0}\text{atm}$.  The given condition is $\text{10}\text{°C}$ and $\text{2}\text{.0}\text{atm}$.  The pressure remains the same while the temperature is decreased from $\text{25}°\text{C}$ to $\text{10}°\text{C}$.  The solubility of the gas increases with decrease in temperature and increase in pressure.  Therefore, the solubility of neon increases and it will be greater than $9.07\times {10}^{-4}m$.

(d)

Interpretation Introduction

Interpretation:

Solubility of neon will be greater or lesser than $9.07\times {10}^{-4}m$ at $\text{50}\text{°C}$ and $\text{2}\text{.0}\text{atm}$ has to be stated.

Concept Introduction:

Refer part (a)

Answer to Problem 12.60QE

Solubility of neon in water will be lesser at $\text{50}\text{°C}$ and $\text{2}\text{.0}\text{atm}$.

Explanation of Solution

Solubility of neon is given as $9.07\times {10}^{-4}m$ at $\text{25}\text{°C}$ and $\text{2}\text{.0}\text{atm}$.  The given condition is $\text{50}\text{°C}$ and $\text{2}\text{.0}\text{atm}$.  The pressure remains the same while the temperature is increased from $\text{25}\text{°C}$ to $\text{50}°\text{C}$.  The solubility of the gas decreases with increase in temperature and decrease in pressure.  Therefore, the solubility of neon decreases and it will be lesser than $9.07\times {10}^{-4}m$.

(e)

Interpretation Introduction

Interpretation:

Solubility of neon will be greater or lesser than $9.07\times {10}^{-4}m$ at $\text{50}\text{°C}$ and $\text{1}\text{.5}\text{atm}$ has to be stated.

Concept Introduction:

Refer part (a)

Answer to Problem 12.60QE

Solubility of neon in water will be lesser at $\text{50}\text{°C}$ and $\text{1}\text{.5}\text{atm}$.

Explanation of Solution

Solubility of neon is given as $9.07\times {10}^{-4}m$ at $\text{25}\text{°C}$ and $\text{1}\text{.0}\text{atm}$.  The given condition is $\text{50}\text{°C}$ and $\text{1}\text{.5}\text{atm}$.  The pressure is decreased from $\text{2}\text{.0}\text{atm}$ to $\text{1}\text{.5}\text{atm}$, while the temperature is increased from $\text{25}\text{°C}$ to $\text{50}°\text{C}$.  The solubility of the gas decreases with increase in temperature and decrease in pressure.  Therefore, the solubility of neon decreases and it will be lesser than $9.07\times {10}^{-4}m$.