Chapter 13, Problem 57A

(a)

Interpretation Introduction

Interpretation:

Whether the total pressure in the apparatus is less than 5atm or greater than 5 atm or equal to 5 atm when valve is opened needs to be determined.

Concept Introduction :

Boyle’s law states that the pressure inversely varies with the volume of the gas at a constant temperature.

Answer to Problem 57A

The pressure of the apparatus after open the valve is less than $\text{5}\text{atm}$ .

Explanation of Solution

Boyle’s Law states that, as pressure decreases then the volume is increased or vice versa. So, when the valve is opened, the volume is a change with due respect pressure is also changed. When the valve opens, the volume changes from the high pressure region to low pressure region. As reaching the state where the volume of both bulbs becomes equal then pressure is also equal on both bulbs.

(b)

Interpretation Introduction

Interpretation:

When the valve is opened, the final pressure is to be determined.

Concept Introduction :

The pressure varies inversely with the volume of the gas at a constant temperature on the basis of Boyle’s law.

So, the final pressure inside the apparatus is

  ${\text{P}}_{\text{3}}{\text{V}}_{\text{3}}\text{=}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{+}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}$

Answer to Problem 57A

The total volume of the gas 5.4 L and the total pressure of the gas is 5 atm.

Explanation of Solution

Given information: pressure of left bulb = $\text{2}\text{atm}$

Pressure of right bulb = $\text{3}\text{atm}$

Volume of left bulb = $\text{9}\text{.00L}$

Volume of right bulb = $\text{3}\text{.00L}$

For left bulb :-

  $\begin{array}{l}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\\ {\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}}\\ \text{=}\frac{\text{2×9}}{\text{5}}\\ \text{=}\text{3}\text{.6}\text{L}\end{array}$

For right bulb:-

  $\begin{array}{l}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\\ {\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}}\\ \text{=}\frac{\text{3×3}}{\text{5}}\\ \text{=}\text{1}\text{.8}\text{L}\end{array}$

The total volume is,

  $\begin{array}{l}{\text{V}}_{\text{t}}\text{=}{\text{V}}_{\text{1}}\text{+}{\text{V}}_{\text{2}}\\ \text{=}\text{5}\text{.4}\text{L}\end{array}$

Now, the total pressure is,

  ${\text{P}}_{\text{t}}{\text{V}}_{\text{t}}\text{=}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{+}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}$

  $\begin{array}{l}{\text{P}}_{\text{t}}{\text{V}}_{\text{t}}\text{=}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{+}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\\ {\text{P}}_{\text{t}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{+}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{V}}_{\text{t}}}\\ \text{=}\frac{\left(\text{2×9}\right)\text{+}\left(\text{3×3}\right)}{\text{5}\text{.4}}\\ \text{=}\frac{\text{27}}{\text{5}\text{.4}}\\ \text{=}\text{5}\text{.0}\text{atm}\end{array}$