Chapter 13, Problem 34QAP

Interpretation Introduction

Interpretation:

Final temperature of gas should be determined.

Concept Introduction:

Charles’s law: It is also known as temperature volume relationship. It states that volume of given mass of gas is directly proportional to its temperature.

Answer to Problem 34QAP

Final temperature of gas is $-\text{237}\text{.75}{}^{\text{0}}\text{C}$.

Explanation of Solution

Relation between volume and temperature is given by Charles’s law.

Charles’s law states that volume of given mass of gas is directly proportional to its temperature. As temperature increases volume also increases.

Mathematical expression is:

$\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\text{=k at constant pressure}$

Initial volume of gas = $\text{2}{\text{.01×10}}^{\text{2}}\text{L}$

Initial temperature of gas ${\text{(T}}_{\text{1}}\text{)}$ = ${\text{1150}}^{\text{0}}\text{C}$ = $273.15+1150=1423.15$ K

Final volume of gas = $\text{5}\text{L}$

Substituting the values in Charles’s equation,

$\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\text{=k at constant pressure}$

$\frac{\text{2}{\text{.01×10}}^{\text{2}}}{\text{1423}\text{.15}}\text{=}\frac{\text{5}}{{\text{T}}_{\text{2}}}$

${\text{T}}_{\text{2}}\text{=}\text{35}\text{.4}\text{K}$

Can be converted into Celsius

$\text{Celcius+273}\text{.15=Kelvin}$

Kelvin = $\text{35-273}\text{.15}\text{=}-\text{237}\text{.75}{}^{\text{0}}\text{C}$.

Interpretation Introduction

Interpretation:

Final volume of gas should be determined.

Concept Introduction:

Charles’s law: It is also known as temperature volume relationship. It states that volume of given mass of gas is directly proportional to its temperature.

Answer to Problem 34QAP

Final volume of gas is 0 mL.

Explanation of Solution

Relation between volume and temperature is given by Charles’s law.

Charles’s law states that volume of given mass of gas is directly proportional to its temperature. As temperature increases volume also increases.

Mathematical expression is:

$\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\text{=k at constant pressure}$

Initial volume of gas = $\text{44}\text{.2}\text{mL}$

Initial temperature of gas ${\text{(T}}_{\text{1}}\text{)}$ = $\text{298}\text{K}$

Final volume of gas = ?

Final Temperature of gas ${\text{(T}}_{\text{2}}\text{)}\text{=}\text{0}\text{K}$

Substituting the values in Charles’s equation,

$\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\text{=k at constant pressure}$

$\begin{array}{l}\frac{\text{44}\text{.2}}{\text{298}}\text{=}\frac{{\text{V}}_{\text{2}}}{\text{0}}\\ {\text{V}}_{\text{2}}\text{=}\text{0}\text{mL}\end{array}$.

Interpretation Introduction

Interpretation:

Final volume of gas should be determined.

Concept Introduction:

Charles’s law: It is also known as temperature volume relationship. It states that volume of given mass of gas is directly proportional to its temperature.

Answer to Problem 34QAP

Final volume of gas is $\text{40}\text{.51}\text{mL}$.

Explanation of Solution

Relation between volume and temperature is given by Charles’s law.

Charles’s law states that volume of given mass of gas is directly proportional to its temperature. As temperature increases volume also increases.

Mathematical expression is:

$\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\text{=k at constant pressure}$

Initial volume of gas = $\text{44}\text{.2}\text{mL}$

Initial temperature of gas ${\text{(T}}_{\text{1}}\text{)}$ = $\text{298}\text{K}$

Final volume of gas = ?

Final Temperature of gas ${\text{(T}}_{\text{2}}{\text{)= 0}}^{\text{0}}\text{C}$ = $\text{0+273}\text{.15}\text{=}\text{273}\text{.15}\text{K}$

Substituting the values in Charles’s equation,

$\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\text{=k at constant pressure}$

$\frac{\text{44}\text{.2}}{\text{298}}\text{=}\frac{{\text{V}}_{\text{2}}}{\text{273}\text{.15}}$

${\text{V}}_{\text{2}}\text{=40}\text{.51}\text{mL}$.