Chapter 8, Problem 8.26UKC

(a)

Interpretation Introduction

Interpretation:

The apparatus has to be redrawn if the temperature is increased from $\text{300}\text{K}\text{to}\text{450}\text{K}$ at constant pressure.

Concept introduction:

Charles’s law: States that volume is directly proportional to temperature when the gas is held at constant pressure and number of molecules.

$\begin{array}{l}\text{V}\propto \text{T}\\ \text{V}\text{=}\text{Constant}\text{×}\text{T}\\ \frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\end{array}$

Answer to Problem 8.26UKC

According to Charles’s law, if temperature increases at constant pressure, then volume also increases. Therefore, the apparatus is redrawn as given below:

Explanation of Solution

Given data is that the temperature increases from $\text{300}\text{K}\text{to}\text{450}\text{K}$ at constant pressure.

Charles’s law is the gas law which relates volume and temperature at constant pressure and number of molecules.

From Charles’s law we know that $\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

According to Charles’s law, if temperature increases at constant pressure, then volume also increases. Therefore, the apparatus is redrawn as given below:

(b)

Interpretation Introduction

Interpretation:

The apparatus has to be redrawn if the pressure is increased from $\text{1}\text{atm}\text{to}\text{2}\text{atm}$ at constant temperature.

Concept introduction:

Boyle’s law:

At fixed temperature, the volume of a fixed amount of gas is inversely proportional to the pressure exerted by the gas.

$\begin{array}{l}\text{P}\propto \frac{\text{1}}{\text{V}}\left(\text{n}\text{,}\text{T}\text{will}\text{be}\text{constant}\right)\\ \text{PV}\text{=}\text{constant}\\ {\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\end{array}$

Answer to Problem 8.26UKC

According to Boyle’s law, if pressure increases at constant temperature, then volume decreases. Therefore, the apparatus is redrawn as given below:

Explanation of Solution

Given data is that the pressure is increased from $\text{1}\text{atm}\text{to}\text{2}\text{atm}$ at constant temperature

Boyle’s law is the law which relates pressure and volume at a constant temperature and number of molecules.

From Boyle’s law:

${\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}$

According to Boyle’s law, if pressure increases at constant temperature, then volume decreases. Therefore, the apparatus is redrawn as given below:

(c)

Interpretation Introduction

Interpretation:

The apparatus has to be redrawn if the temperature is decreased from $\text{300}\text{K}\text{to}200\text{K}$ and the pressure is decreased from $3\text{atm}\text{to}\text{2}\text{atm}$.

Concept introduction:

General Gas Law:

Combining Charles’s law and Boyle’s law we get the General gas law or combined gas law.

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

Answer to Problem 8.26UKC

The volume remains unchanged and the apparatus is redrawn as given below:

Explanation of Solution

Given data is that the temperature is decreased from $\text{300}\text{K}\text{to}200\text{K}$ and the pressure is decreased from $3\text{atm}\text{to}\text{2}\text{atm}$.

$\begin{array}{c}{\text{P}}_{\text{1}}\text{=}\text{3}\text{atm}\\ {\text{T}}_{\text{1}}\text{=}\text{300}\text{K}\\ {\text{T}}_{\text{2}}\text{=}\text{200}\text{K}\\ {\text{P}}_{\text{2}}\text{=}\text{2}\text{atm}\end{array}$

Combined gas law is proposed by combining the Boyle’s law and Charle’s law and is given by:

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

$\begin{array}{c}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\\ \frac{\text{3}\text{atm}{\text{V}}_{\text{1}}}{\text{300}\text{K}}\text{=}\frac{\text{2}\text{atm}{\text{V}}_{\text{2}}}{\text{200}\text{K}}\\ \frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\text{=}\text{1}\end{array}$

Therefore, the volume remains unchanged and the apparatus is redrawn as given below: