Chapter U3.6, Problem 7E

(a)

Interpretation Introduction

Interpretation : The proportionality constant, $\text{k=}\frac{\text{V}}{\text{T}}$ , at the beginning of the dayis to be calculated.

Concept Introduction : According to Charles’s law, if pressure and the number of particles of a gas remain same, then volume is proportional to the Kelvin temperature. $\text{V=kT}$

Answer to Problem 7E

The proportionality constant is, $\text{k=83}\text{.3L/K}$ at the beginning of the day.

Explanation of Solution

As the temperature decreases, the volume of gas also decreases. The volume is directly proportional to the temperature.

Given:

  $\begin{array}{l}{\text{Initialvolume=V}}_{\text{1}}\text{=25000L}\\ {\text{Initial temperature=T}}_{\text{1}}{\text{=27}}^{\text{0}}\text{C}\end{array}$

First change the Celsius temperature to Kelvin

So

  $\begin{array}{l}{\text{T}}_1{\text{=27}}^{\text{0}}\text{C}\\ \text{=27+273}\\ \text{=300K}\end{array}$

Using the formula

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

Find the value of k;

  $\text{k=}\frac{\text{V}}{\text{T}}\text{=}\frac{\text{25000L}}{\text{300K}}\text{=83}\text{.3L/K}$

(b)

Interpretation Introduction

Interpretation : The volume of the balloon when the temperature has dropped to 22⁰Chas to be calculated.

Concept Introduction : Volume is directly proportional to temperature. As the temperature decreases, the volume of gas also decreases.

Answer to Problem 7E

The volume of the balloon when the temperature has dropped to 22⁰C $\text{=21,476L}$

Explanation of Solution

As the temperature decreases, the volume of gas also decreases because volume is directly proportional to temperature.

Given:

The new temperature ${\text{T}}_2{\text{=22}}^{\text{0}}\text{C}$

First, change the Celsius temperature to Kelvin

  $\begin{array}{l}{\text{T}}_2{\text{=22}}^{\text{0}}\text{C}\\ \text{=22+273}\\ \text{=295K}\end{array}$

Let final volume be ${\text{V}}_{\text{2}}$

Now solving for the volume at 295K;

  $\begin{array}{l}{\text{V}}_2{\text{=kT}}_2\\ \text{=72}\text{.8L/K}\times \text{295}\\ \text{=21,476L}\end{array}$

(c)

Interpretation Introduction

Interpretation : The proportionality constant, $\text{k=}\frac{\text{V}}{\text{T}}$ at the end of the day when the temperature is 15 ^oC has to be calculated.

Concept Introduction : According to Charles’s law, if pressure and the number of particles of a gas remain same, then volume is proportional to the Kelvin temperature.

  $\text{V=kT}$

Answer to Problem 7E

The proportionality constant, $\text{k=}\frac{\text{V}}{\text{T}}$ , at the end of the day when the temperature is 15⁰Cissame as that at the beginning of the day, i.e., $\text{k=83}\text{.3L/K}$ .

Explanation of Solution

The temperatureat the end of the day is 15⁰C.

The proportionality constant is a constant value. It remains same at all times of the day. The value of $\text{k=}\frac{\text{V}}{\text{T}}$ the same at the beginning of the day and at the end of the day.

So, $\text{k=83}\text{.3L/K}$