Chapter 6, Problem 6.54AP

(a)

Interpretation Introduction

Interpretation:

The final state 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 6.54AP

The final state pressure of the gas is $1.297\text{atm}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $1.74\text{atm}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is ${\text{120}}^{\circ }\text{C}\text{.}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is ${\text{20}}^{\text{o}}\text{C}\text{.}$

Conversion of temperature from Celsius to Kelvin.

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

  ${\text{120}}^{\text{o}}\text{C=120+273}\text{K=393K}$

  ${\text{20}}^{\text{o}}\text{C=20+273}\text{K=293K}$

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

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

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

  ${\text{P}}_{\text{2}}\text{=}\frac{\left(1.74\text{atm}\right)\text{×}\left(\text{293K}\right)}{\left(\text{393K}\right)}$

  ${\text{P}}_{\text{2}}\text{=1}\text{.297}\text{atm}$

The final state pressure of the gas is $1.297\text{atm}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The final state pressure of the gas has to be given

Concept Introduction:

Refer to part (a).

Answer to Problem 6.54AP

The final state pressure of the gas is $156\text{mm}\text{Hg}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $220\text{mm}\text{Hg}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is ${\text{150}}^{\text{o}}\text{C}\text{.}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is $\text{300}\text{K}\text{.}$

Conversion of temperature from Celsius to Kelvin.

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

  ${\text{150}}^{\text{o}}\text{C=150+273}\text{K=423}\text{K}$

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

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

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

  ${\text{P}}_{\text{2}}\text{=}\frac{\left(220\text{mm}\text{Hg}\right)\text{×}\left(\text{300}\text{K}\right)}{\left(\text{423K}\right)}$

  ${\text{P}}_{\text{2}}\text{=156}\text{mm}\text{Hg}$

The final state pressure of the gas is $156\text{mm}\text{Hg}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The final state temperature of the gas has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.54AP

The final state temperature of the gas is $\text{181}\text{.7}\text{K}\text{.}$

Explanation of Solution

Given,

The pressure of the gas at initial state $\left({\text{P}}_{\text{1}}\right)$ is $\text{0}\text{.75}\text{atm}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is ${198}^{\text{o}}\text{C}\text{.}$

The pressure of the gas at final state $\left({\text{P}}_{\text{2}}\right)$ is $220\text{mm}\text{Hg}\text{.}$

Conversion of temperature from Celsius to Kelvin.

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

  ${198}^{\text{o}}\text{C=198+273}\text{K=471}\text{K}$

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

  $\text{220}\text{mm}\text{Hg×}\frac{\text{1atm}}{\text{760}\text{mm}\text{Hg}}\text{=0}\text{.2894}\text{atm}$

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

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

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

  ${\text{T}}_{\text{2}}\text{=}\frac{\left(\text{0}\text{.2894 atm}\right)\left(\text{471K}\right)}{\left(\text{0}\text{.75}\text{atm}\right)}$

  ${\text{T}}_{\text{2}}\text{=181}\text{.7K}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is $\text{181}\text{.7}\text{K}\text{.}$