Chapter 10, Problem 9E

Interpretation Introduction

(a)

Interpretation:

The change in the pressure if the volume of a gas is increased is to be stated.

Concept introduction:

Boyle’s law which relates pressure and temperature states that when temperature is kept constant for a gas then the pressure and temperature follows an inverse relation. According to Boyle’s law relation between pressure and volume is written below.

$\begin{array}{r}\text{P}\propto \frac{\text{1}}{\text{V}}\text{, T is constant}\\ \text{P}& =& \frac{\text{k}}{\text{V}}\\ \text{PV}& =& \text{k}\end{array}$

Answer to Problem 9E

With increase in the volume, the net value of pressure decreases.

Explanation of Solution

Boyle’s law for the gaseous molecule is stated below.

$\begin{array}{r}\text{P}\propto \frac{\text{1}}{\text{V}}\text{, T is constant}\\ \text{P}& =& \frac{\text{k}}{\text{V}}\\ \text{PV}& =& \text{k}\end{array}$

Where,

• $\text{k}$ is a proportionality constant

• $\text{P}$ is the pressure of gas.

• $\text{V}$ is the volume of the container or gas.

According to the equation of Boyle’s law if volume is increased for a particular or constant temperature then the pressure of the gas will be decreased to maintain the $\text{PV}$ constant.

Therefore, pressure decreases with increase in volume.

Conclusion

Pressure decreases with increase in volume of a gas.

Interpretation Introduction

(b)

Interpretation:

The change in pressure when the temperature of a gas is increased is to be stated.

Concept introduction:

Gay Lussac’s law relates the pressure and temperature for a particular gas when rest of the variables is constant. Gay Lussac’s law or given relation between pressure and temperature is written below.

$\begin{array}{c}\text{P}\propto \text{T},\text{ V is constant}\\ \text{P}=\text{kT},\text{V is constant}\\ \frac{\text{P}}{\text{T}}=\text{k}\end{array}$

Answer to Problem 9E

If the temperature $\left(\text{T}\right)$ of a gas is increased then, the pressure of the gas also increases.

Explanation of Solution

Relation given by Gay Lussac’s for the gaseous molecule is stated below.

$\begin{array}{r}\text{P}\propto \text{T},\text{ V is constant}\\ \text{P}& =& \text{kT},\text{V is constant}\\ \frac{\text{P}}{\text{T}}& =& \text{k}\end{array}$

Where,

• $\text{k}$is a proportionality constant

• $\text{P}$ is the pressure of gas.

• $\text{T}$ is the volume of the container or gas.

Ratio of pressure and temperature is constant. Pressure will be increased with the increase in value of temperature to maintain the $\frac{\text{P}}{\text{T}}$ ratio constant.

Therefore, pressure increases with increase in temperature.

Conclusion

With increased temperature of a gas, the pressure of the gas also increases.

Interpretation Introduction

(c)

Interpretation:

The change in the pressure when moles of a gas are increased is to be stated.

Concept introduction:

Avogadro’s law relates volume of a gas with the number of moles or molecules of the same gas. According to this law when temperature and pressure of a gas is kept constant then the equal volumes of gas contains equal number of molecules of that gas. Combined gas law which is the result of Boyle’s law, Charles’s law is, Gay Lussac’s law and Avogadro’s law is written below.

$\begin{array}{l}\text{P}\propto \text{T},\text{ V is constant}\\ \text{P}\propto \frac{1}{\text{V}},\text{T is constant}\\ \text{V}\propto \text{n, }\left(\text{P},\text{T are constant}\right)\\ \text{P}\propto \frac{\text{n}\times \text{T}}{\text{V}}\end{array}$

Answer to Problem 9E

The pressure of a gas increases with increase in number of moles of a gas.

Explanation of Solution

Relation given by all the variables of a gas is stated below.

$\begin{array}{l}\text{P}\propto \text{T},\text{ V is constant}\\ \text{P}\propto \frac{1}{\text{V}},\text{T is constant}\\ \text{V}\propto \text{n, }\left(\text{P},\text{T are constant}\right)\\ \text{P}\propto \frac{\text{n}\times \text{T}}{\text{V}}\end{array}$

Where,

• $\text{T}$is a temperature of the gas.

• $\text{P}$ is the pressure of gas.

• $\text{V}$ is the volume of the container or gas.

• $\text{n}$ is the number of moles of a gas.

Combined gas law is written below.

$\begin{array}{l}\text{P}\propto \frac{\text{n}\times \text{T}}{\text{V}}\\ \text{P}=\frac{\text{n}\times \text{T}\times \text{R}}{\text{V}}\end{array}$

Where,

• $\text{R}$ is the constant of proportionality.

According to the relation given above pressure is directly proportional to the number of moles. Pressure will be increased with the increase in value of number of moles to maintain the gas law.

Therefore, pressure increases with increase in number of moles of a gas.

Conclusion

The increase in the number of moles $\left(\text{n}\right)$ of a gas causes increase in the value of pressure.