Chapter 14, Problem 96A

(a)

Interpretation Introduction

Interpretation: The graph using pressure as the dependent variable is to be calculated.

Concept Introduction: Pressure is the amount of force exerted per unit area.

Answer to Problem 96A

The graph of pressure as a dependent variable is as follows:

  

Explanation of Solution

The given data table is as follows:

Temperature $\left(\text{C}\right)$	Pressure (mm Hg)
$10$	$726$
$20$	$750$
$40$	$800$
$70$	$880$
$100$	$960$

The graph of pressure as a dependent variable is as follows:

  

(b)

Interpretation Introduction

Interpretation: The pressure at $0\text{C}$ is to be determined.

Concept Introduction: Pressure is the amount of force exerted per unit area.

Answer to Problem 96A

The pressure at $0\text{C}$ is $\text{698 mm Hg}$ .

Explanation of Solution

The graph of pressure as a dependent variable is as follows:

  

From the graph, the pressure at $0\text{C}$ is $\text{698 mm Hg}$ .

(c)

Interpretation Introduction

Interpretation: The relationship between variables directly or inversely proportional is to be explained.

Concept Introduction: Pressure is the amount of force exerted per unit area.

Answer to Problem 96A

The pressure and temperature are directly proportional to each other.

Explanation of Solution

According to Gay-Lussac's law, the pressure of an enclosed gas will increase with the increase in temperature of the gas, provided that the volume remains the same. Hence, the pressure and temperature are directly proportional to each other.

(d)

Interpretation Introduction

Interpretation: The change in the pressure of the gas with each degree of Celcius change in the temperature is to be explained.

Concept Introduction: According to Gay-Lussac's law, the pressure of a gas will increase with the increase in temperature of the gas, provided that the volume remains the same.

Answer to Problem 96A

When there is an increase in the temperature of a system, there is also an increase in the pressure and vice versa.

Explanation of Solution

The pressure is directly proportional to the temperature of a given volume of gas. When a system's temperature increases, its pressure increases, and vice versa. Gay-Lussac's law describes the relationship between a gas's pressure and its temperature.

(e)

Interpretation Introduction

Interpretation: The equation relating the pressure and temperature of a gas is to be written.

Concept Introduction: According to Gay-Lussac's law, the pressure of a gas will increase with the increase in temperature of the gas, provided that the volume remains the same.

Answer to Problem 96A

The equation relates the pressure and temperature of a gas is as follows:

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Explanation of Solution

According to Gay-Lussac's law, pressure and temperature are directly proportional to each other.

The equation relates the pressure and temperature of a gas is as follows:

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Where,

• $T_1$ is the initial temperature.
• $T_2$ is the final temperature.
• $P_1$ is the initial pressure.
• $P_2$ is the final pressure.

(f)

Interpretation Introduction

Interpretation: The gas illustrated by the given data is to be explained.

Concept Introduction: The pressure of a gas will increase with the increase in temperature of the gas, provided that the volume remains the same.

Answer to Problem 96A

The given data illustrate Gay-Lussac's law.

Explanation of Solution

According to Gay-Lussac's law, pressure and temperature are directly proportional when volume remains constant.

The equation relates the pressure and temperature of a gas is as follows:

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Where,

• $T_1$ is the initial temperature.
• $T_2$ is the final temperature.
• $P_1$ is the initial pressure.
• $P_2$ is the final pressure.

The initial pressure is $726\text{ mm Hg}$ .

The final pressure is $750\text{ mm Hg}$

The initial temperature is $10\text{C}$ .

The final temperature is $20\text{C}$ .

The graph is as follows:

  

The conversion of degree Celcius into kelvin is done as follows:

  $T\left(\text{K}\right)=T\left({}^{\circ }\text{C}\right)+273.15$

Convert $10\text{C}$ into K as follows:

  $\begin{array}{c}T\left(\text{K}\right)=T\left({}^{\circ }\text{C}\right)+273.15\\ \text{=1}0+273.15\\ =283.15\text{ K}\end{array}$

Convert ${20}^{\circ }\text{C}$ into K as follows:

  $\begin{array}{c}T\left(\text{K}\right)=T\left({}^{\circ }\text{C}\right)+273.15\\ \text{=}20+273.15\\ =293.15\text{ K}\end{array}$

Substitute the values of $T_1$ , $T_2$ , $P_1$ and $P_2$ in the above equation as follows:

  $\begin{array}{c}\frac{P_1}{T_1}=\frac{P_2}{T_2}\\ \frac{726\text{ mm Hg}}{283.15\text{ K}}=\frac{750\text{ mm Hg}}{293.15\text{ K}} \\ 2.56{\text{ mm Hg K}}^{-1}=2.56{\text{ mm Hg K}}^{-1}\end{array}$

The value of LHS is equal to the value of RHS. Hence, it follows Gay-Lussac's law.