Chapter 6, Problem 6.57AP

(a)

Interpretation Introduction

Interpretation:

The pressure of the gas has to be given.

Concept Introduction:

Combined Gas Law:

Combined gas law relates the pressure, temperature and the volume of the gases.

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

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

${\text{P}}_{\text{2}}{\text{,V}}_{\text{2}}\text{\&}{\text{T}}_{\text{2}}$ are the pressure, temperature and the volume of the gas at the final state respectively

Answer to Problem 6.57AP

The pressure of the gas at final state $\left({\text{P}}_{\text{2}}\right)$ is $\text{1}\text{.403}\text{atm}\text{.}$

Explanation of Solution

Given,

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $\text{4}\text{.0}\text{L}\text{.}$

The volume of the gas at final state $\left({\text{V}}_{\text{2}}\right)$ is $\text{3}\text{.0}\text{L}\text{.}$

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $\text{0}\text{.90}\text{atm}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is $\text{265}\text{K}\text{.}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is $310\text{K}\text{.}$

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

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

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

  ${\text{P}}_{\text{2}}\text{=}\frac{\left(\text{0}\text{.90}\text{atm}\right)\left(\text{4}\text{.0}\text{L}\right)\left(\text{310K}\right)}{\left(\text{265K}\right)\left(\text{3}\text{.0L}\right)}$

  ${\text{P}}_{\text{2}}\text{=1}\text{.403 atm}$

The pressure of the gas at final state $\left({\text{P}}_{\text{2}}\right)$ is $\text{1}\text{.403}\text{atm}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The volume of the gas has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.57AP

The volume of the gas is $\text{113}\text{.53L}\text{.}$

Explanation of Solution

Given,

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $75\text{L}\text{.}$

The pressure of the gas at final state $\left({\text{P}}_{\text{2}}\right)$ is $\text{700}\text{mm}\text{Hg}\text{.}$

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $1.2\text{atm}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is ${\text{5}}^{\text{o}}\text{C}\text{.}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is ${\text{50}}^{\text{o}}\text{C}\text{.}$

Conversion of pressure from millimeter mercury to atmosphere.

  $\text{700}\text{}\text{mm}\text{Hg×}\frac{\text{1atm}}{\text{760}\text{mm}\text{Hg}}\text{=0}\text{.9210}\text{atm}$

Conversion of temperature from Celsius to Kelvin.

  ${\text{1}}^{\text{o}}\text{C=273}\text{K}$

  $\text{5}{\text{.0}}^{\text{o}}\text{C=5+273}\text{K=278}\text{K}$

  ${\text{50}}^{\text{o}}\text{C=50+273}\text{K=323}\text{K}$

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

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

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

  ${\text{V}}_{\text{2}}\text{=}\frac{\left(\text{1}\text{.2}\text{atm}\right)\left(\text{75}\text{L}\right)\left(\text{323K}\right)}{\left(\text{278K}\right)\left(\text{0}\text{.9210}\text{atm}\right)}$

  ${\text{V}}_{\text{2}}\text{=113}\text{.53 L}$

The volume of the gas is $\text{113}\text{.53L}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The temperature of the gas has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.57AP

The temperature of the gas is $\text{739}\text{.04K}\text{.}$

Explanation of Solution

Given,

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $\text{125ml}\text{.}$

The pressure of the gas at final state $\left({\text{P}}_{\text{2}}\right)$ is $\text{100}\text{mm}\text{Hg}\text{.}$

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $\text{200}\text{mm}\text{Hg}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is $\text{298K}\text{.}$

The volume of the gas at final state $\left({\text{V}}_{\text{2}}\right)$ is $\text{0}\text{.62}\text{L}\text{.}$

Conversion of volume from milliliter to liter.

  $\begin{array}{l}\text{1L=1000ml}\\ \text{125}\text{ml=0}\text{.125L}\end{array}$

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

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

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

  ${\text{T}}_{\text{2}}\text{=}\frac{\left(\text{100}\text{mm}\text{Hg}\right)\left(\text{0}\text{.62}\text{L}\right)\left(\text{298K}\right)}{\left(\text{200}\text{mm}\text{Hg}\right)\left(\text{0}\text{.125}\text{L}\right)}$

  ${\text{T}}_{\text{2}}\text{=739}\text{.04}\text{K}$

The temperature of the gas is $\text{739}\text{.04K}\text{.}$