Chapter 14.3, Problem 36PP

Interpretation Introduction

Interpretation:

The amount of gas is to be calculated at different pressure that is 110 kPa.

Concept Introduction:

Henry’s Law:

At a given temperature, the solubility of the gas in a liquid is directly proportional to the pressure above the liquid:

  $\frac{S_1}{P_1}=\frac{S_2}{P_2}$

Where S represents solubility while P represents pressure.

After cross multiplying, the equation can be written as;

$S_1{P}_2=\hspace{0.17em}{S}_2{P}_1$

Where

S₁ = Solubility of gas

S₂ = Solubility at the new pressure

P₁ = Initial Pressure

P₂ = New Pressure

Answer to Problem 36PP

3g of gas is dissolved at 110 kPa.

Explanation of Solution

Given information:

Mass of gas = 0.55 g

Volume =1 L

Initial pressure P₁ = 20 kPa

New Pressure P₂ = 110 kPa

At a given temperature, the solubility of the gas in a liquid is directly proportional to the pressure above the liquid:

  $\frac{S_1}{P_1}=\frac{S_2}{P_2}$

Where S represents solubility while P represents pressure.

After cross multiplying, the equation can be written as;

  $S_1{P}_2=\hspace{0.17em}{S}_2{P}_1$

Where,

S₁ = Solubility of gas

S₂ = Solubility at the new pressure

P₁ = Initial Pressure

P₂ = New Pressure

Write the explanation of solubility,

S₁ = Solubility of gas= $\frac{Mass}{Volume}$

  = $\frac{\text{0}\text{.55 g}}{\text{1 L}}$

 = $0.55g/L$

As it can be seen from the equation,

  $\begin{array}{c}{S}_1{P}_2=S_2{P}_1\end{array}$

  $\begin{array}{c}{S}_1{P}_2=S_2{P}_1{}_{ }\end{array}$

   $\begin{array}{c}{S}_2=S_1\frac{P_2}{P_1}\end{array}$

Putting all the values in the equation:

  $\begin{array}{l}{S}_2=\frac{0.55\times 110}{20}\end{array}$

  $\begin{array}{c}{S}_2 =3g/L\end{array}$

So 3g of gas is dissolved at 110 kPa.