Chapter 6, Problem 6.50AP

(a)

Interpretation Introduction

Interpretation:

The volume of the gas has to be given.

Concept Introduction:

Charles Law:

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

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

Where V is the volume of the gas

T is the temperature of the gas in Kelvin.

K is the constant.

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

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

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

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

Answer to Problem 6.50AP

The volume of the gas is $\text{0}\text{.0214}\text{L}\text{.}$

Explanation of Solution

Given,

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $\text{10}\text{ml}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is $\text{210}\text{K}\text{.}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is $\text{450}\text{K}\text{.}$

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

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

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

  ${\text{V}}_{\text{2}}\text{=}\frac{\left(\text{10}\text{ml}\right)\text{×}\left(\text{450}\text{K}\right)}{\text{210}\text{K}}$

  ${\text{V}}_{\text{2}}\text{=21}\text{.43}\text{ml}$

Conversion of volume in ml to liter

  $\text{1L=1000ml}$

  $\text{21}\text{.43ml=0}\text{.0214}\text{L}$

The volume of the gas is $\text{0}\text{.0214}\text{L}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The volume of the gas has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.50AP

The volume of the gas is $\text{0}\text{.1166}\text{L}\text{.}$

Explanation of Solution

Given,

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $\text{255ml}\text{.}$

The temperature of the gas at initial state $\left({\text{T}}_{\text{1}}\right)$ is ${\text{55}}^{\circ }\text{C}\text{.}$

The temperature of the gas at final state $\left({\text{T}}_{\text{2}}\right)$ is $\text{150}\text{K}\text{.}$

Conversion of temperature in Celsius to Kelvin

  $0^{\text{o}}\text{C=273K}$

  ${\text{55}}^{\text{o}}\text{C=55+273K=328}\text{K}$

Conversion of volume in ml to liter

  $\text{1L=1000ml}$

  $\text{255ml=0}\text{.255L}$

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

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

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

  ${\text{V}}_{\text{2}}\text{=}\frac{\left(\text{0}\text{.255L}\right)\text{×}\left(\text{150K}\right)}{\text{328K}}$

  ${\text{V}}_{\text{2}}\text{=0}\text{.1166}\text{L}$

The volume of the gas is $\text{0}\text{.1166}\text{L}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The temperature of the gas has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.50AP

The temperature of the gas is $\text{492}\text{K}\text{.}$

Explanation of Solution

Given,

The volume of the gas at initial state $\left({\text{V}}_{\text{1}}\right)$ is $13\text{L}\text{.}$

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

The volume of the gas at final state $\left({\text{V}}_{\text{2}}\right)$ is $\text{52}\text{L}\text{.}$

Conversion of temperature in Celsius to Kelvin

  $0^{\text{o}}\text{C=273K}$

  ${\text{-150}}^{\text{o}}\text{C}\text{=-150+273K=123K}$

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

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

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

  ${\text{T}}_{\text{2}}\text{=}\frac{\left(52\text{L}\right)\text{×}\left(\text{123K}\right)}{\text{13}\text{L}}$

  ${\text{T}}_{\text{2}}\text{=492}\text{K}$

The temperature of the gas is $\text{492}\text{K}\text{.}$