Chapter 6, Problem 6.53AP

(a)

Interpretation Introduction

Interpretation:

The final pressure of the gas has to be given.

Concept Introduction:

Gay Lussac’s Law:

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

  $\frac{\text{P}}{\text{T}}\text{=K}$

Where P is the pressure of the gas.

T is the temperature of the gas in Kelvin.

K is the constant.

The temperature or the pressure of the gas can be calculated using the relation,

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

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

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

Answer to Problem 6.53AP

The final pressure of the gas is $\text{4}\text{.34}\text{atm}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $3.25\text{atm}\text{.}$

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

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is $\text{398}\text{K}\text{.}$

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

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

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

  ${\text{P}}_{\text{2}}\text{=}\frac{\left(3.25\text{atm}\right)\text{×}\left(\text{398K}\right)}{\left(\text{298K}\right)}$

  ${\text{P}}_{\text{2}}\text{=4}\text{.34}\text{atm}$

The final pressure of the gas is $\text{4}\text{.34}\text{atm}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The final pressure of the gas has to be given

Concept Introduction:

Refer to part (a).

Answer to Problem 6.53AP

The final pressure of the gas is $\text{348}\text{.5}\text{mm}\text{Hg}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $\text{550}\text{mm}\text{Hg}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is $\text{273}\text{K}\text{.}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is ${\text{-100}}^{\text{o}}\text{C}\text{.}$

Conversion of temperature from Celsius to Kelvin.

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

  ${\text{-100}}^{\text{o}}\text{C=-100+273}\text{K=173K}$

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

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

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

  ${\text{P}}_{\text{2}}\text{=}\frac{\left(\text{550}\text{mm}\text{Hg}\right)\text{×}\left(\text{173K}\right)}{\left(\text{273K}\right)}$

  ${\text{P}}_{\text{2}}\text{=348}\text{.5}\text{mm}\text{Hg}$

The final pressure of the gas is $\text{348}\text{.5}\text{mm}\text{Hg}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The final temperature of the gas has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.53AP

The final temperature of the gas is $\text{1314}\text{K}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $\text{0}\text{.50}\text{atm}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is ${\text{250}}^{\text{o}}\text{C}\text{.}$

The pressure of the gas at final state $\left({\text{P}}_{\text{2}}\right)$ is $\text{955}\text{mm}\text{Hg}\text{.}$

Conversion of temperature from Celsius to Kelvin.

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

  ${\text{250}}^{\text{o}}\text{C=250+273}\text{K=523K}$

Conversion of $\text{mm}\text{Hg}$ to $\text{atm}\text{.}$

  $\text{955}\text{mm}\text{Hg×}\frac{\text{1atm}}{\text{760}\text{mm}\text{Hg}}\text{=1}\text{.2565}\text{atm}$

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

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

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

  ${\text{T}}_{\text{2}}\text{=}\frac{\left(\text{1}\text{.2565}\text{atm}\right)\text{×}\left(\text{523K}\right)}{\left(\text{0}\text{.50}\text{atm}\right)}$

  ${\text{T}}_{\text{2}}\text{=1314 K}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is $\text{1314}\text{K}\text{.}$