Chapter 6, Problem 6.46AP

(a)

Interpretation Introduction

Interpretation:

The volume of the gas has to be given.

Concept Introduction:

Boyle’s Law:

At constant temperature, the pressure and the volume of the gas are inversely related.

  $\text{PV=K}$

Where P is pressure of the gas at constant temperature.

V is volume of the gas at constant temperature.

K is the constant.

The volume or pressure of the gas can be calculated using the relation.

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

Where ${\text{P}}_{\text{1}}\text{\&}{\text{V}}_{\text{1}}$ are pressure and volume of gas at initial state.

${\text{P}}_{\text{2}}\text{\&}{\text{V}}_{\text{2}}$ are pressure and volume of gas at final state.

Answer to Problem 6.46AP

The volume of the gas is $\text{0}\text{.9868}\text{L}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $2.5\text{atm}\text{.}$

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $\text{1}\text{.5}\text{}\text{L}\text{.}$

The pressure of the gas at final state $\left({\text{P}}_{\text{2}}\right)$ is $\text{3}\text{.8}\text{atm}\text{.}$

The volume of the gas at final state $\left({\text{V}}_{\text{2}}\right)$ can be calculated as,

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

  ${\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}}$

  ${\text{V}}_{\text{2}}\text{=}\frac{\left(\text{2}\text{.5}\text{atm}\right)\left(\text{1}\text{.5}\text{L}\right)}{\left(\text{3}\text{.8}\text{atm}\right)}$

  ${\text{V}}_{\text{2}}\text{=0}\text{.9868}\text{L}$

The volume of the gas is $\text{0}\text{.9868}\text{L}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The volume of the gas has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.46AP

The volume of the gas is $\text{0}\text{.7093}\text{L}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $\text{2}\text{.0}\text{atm}\text{.}$

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $\text{350ml}\text{.}$

The pressure of the gas at final state $\left({\text{P}}_{\text{2}}\right)$ is $\text{750mm}\text{Hg}\text{.}$

  $\text{1}\text{L=1000}\text{}\text{ml}$

  $\text{350ml=0}\text{.35L}$

The pressure in $\text{mm}\text{Hg}$ is converted to $\text{atm}$ as,

  $750\text{mm}\text{Hg×}\frac{\text{1}\text{atm}}{\text{760}\text{mm}\text{Hg}}\text{=0}\text{.9868}\text{atm}$

The volume of the gas at final state $\left({\text{V}}_{\text{2}}\right)$ can be calculated as,

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

  ${\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}}$

  ${\text{V}}_{\text{2}}\text{=}\frac{\left(\text{2}\text{.0atm}\right)\left(\text{0}\text{.35L}\right)}{\left(\text{0}\text{.9868}\text{atm}\right)}$

  ${\text{V}}_{\text{2}}\text{=0}\text{.7093}$

The volume of the gas is $\text{0}\text{.7093}\text{L}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The pressure of the gas has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.46AP

The pressure of the gas is $\text{0}\text{.7668}\text{mm}\text{Hg}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $\text{75}\text{mmHg}\text{.}$

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $\text{9}\text{.1}\text{ml}\text{.}$

The volume of the gas at final state $\left({\text{V}}_{\text{2}}\right)$ is $\text{890ml}\text{.}$

The pressure of the gas can be calculated as,

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

  ${\text{P}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}$

  ${\text{P}}_{\text{2}}\text{=}\frac{\left(\text{75}\text{mmHg}\right)\left(\text{9}\text{.1ml}\right)}{\left(\text{890ml}\right)}$

  ${\text{P}}_2\text{=0}\text{.7668}\text{mm}\text{Hg}$

The pressure of the gas is $\text{0}\text{.7668}\text{mm}\text{Hg}\text{.}$