Chapter 10, Problem 11E

Interpretation Introduction

(a)

Interpretation:

The change in the pressure $\left(P\right)$ when volume $\left(V\right)$ of a gas is increased from $\text{25}\text{.0 mL}$ to $\text{50}\text{.0 mL}$ 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 the inverse relation. According to Boyle’s law, relation between pressure and volume is written below.

Answer to Problem 11E

The pressure decreases with increase in volume from $\text{25}\text{.0 mL}$ to $\text{50}\text{.0 mL}$ by $1/\text{2}{P}_1$.

Explanation of Solution

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 keep the $PV$ constant Boyle’s law for the gaseous molecule is stated below.

$\begin{array}{rll}P& \propto & \frac{1}{V},T\text{ is constant}\\ P& =& \frac{k}{V}\\ \text{P}V& =& k\\ P_1{V}_1& =& P_2{V}_2\end{array}$ …(1)

Where,

• $k$ is a proportionality constant.

• $P_1$ is the initial pressure of the gas.

• $P_2$ is the final pressure of the gas.

• $V_1$ is the initial volume of the container or gas.

• $V_2$ is the final volume of the container or gas.

The given values of the initial volume $\left(V_1\right)$ and final volume $\left(V_2\right)$ are $2\text{5}\text{.0 mL}$ and $\text{50}\text{.0 mL}$ respectively.

Substitute the values of $V_1$ and $V_2$ in equation (1).

$\begin{array}{rll}{P}_1{V}_1& =& P_2{V}_2\\ \frac{P_2}{P_1}& =& \frac{\text{25}\text{.0 mL}}{\text{50}\text{.0 mL}}\\ P_2& =& \frac{\text{25}\text{.0 }}{\text{50}\text{.0}}\times P_1\\ & =& \frac{1}{2}{P}_1\end{array}$

Therefore, decrease in pressure is $\frac{1}{2}{P}_1$.

Conclusion

The pressure decreases by the factor of $1/\text{2}{P}_1$ with increase in volume from $\text{25}\text{.0 mL}$ to $\text{50}\text{.0 mL}$.

Interpretation Introduction

(b)

Interpretation:

Change in the pressure $\left(P\right)$ when temperature $\left(T\right)$ of a gas is increased from $25{}^{\text{o}}\text{C}$ to $\text{50}{}^{\text{o}}\text{C}$ is to be stated.

Concept introduction:

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

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

Answer to Problem 11E

The pressure $\left(P\right)$ is increased when temperature $\left(T\right)$ of a gas is increased from $25{}^{\text{o}}\text{C}$ to $\text{50}{}^{\text{o}}\text{C}$ by the factor of $1.08\times P_1$.

Explanation of Solution

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

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$ ...(2)

Where,

• $k$ is a proportionality constant

• $P_1$ is the initial pressure of the gas.

• $P_2$ is the final pressure of the gas.

• $T_1$ is the initial temperature of the gas.

• $T_2$ is the final temperature of the gas.

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

Conversion of initial temperature $\left(T_1\right)$ into Kelvin is shown below.

$\begin{array}{rll}{T}_{\text{1}}& =& 25{}^{\text{ο}}\text{C}\\ T_1& =& \left(25+\text{273}\right)\text{K}\\ T_{\text{1}}& =& \text{298}\text{K}\end{array}$

Conversion of final temperature $\left(T_2\right)$ into Kelvin is shown below.

$\begin{array}{rll}{T}_2& =& 50.0{}^{\text{ο}}\text{C}\\ T_2& =& \left(50.0+\text{273}\right)\text{K}\\ T_2& =& 323\text{K}\end{array}$

Substitute the value of $T_1$ and $T_2$ in equation (2) as shown below.

$\begin{array}{rll}\frac{P_1}{T_1}& =& \frac{P_2}{T_2}\\ \frac{P_2}{P_1}& =& \frac{\text{323 K}}{\text{298 K}}\\ P_2& =& 1.08\times P_1\end{array}$

Therefore, the pressure will increase in the container.

Conclusion

Increase in pressure $\left(P\right)$ when temperature $\left(T\right)$ of a gas is increased from $\text{25}{}^{\text{ο}}\text{C}$ to $\text{50}{}^{\text{ο}}\text{C}$ is $1.08\times P_1$.

Interpretation Introduction

(c)

Interpretation:

The change in the pressure $\left(P\right)$ when moles of a gas $\left(n\right)$ are increased from $\text{0}\text{.250 mol}$ to $\text{0}\text{.500 mol}$ 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.

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

Answer to Problem 11E

Increase in pressure $\left(P\right)$ when moles $\left(n\right)$ of a gas are increased from $\text{0}\text{.250 mol}$ to $\text{0}\text{.500 mol}$ is $\text{2}{P}_{\text{1}}$.

Explanation of Solution

The relation given by all the variables of a gas is stated as below.

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

According to the relation given above pressure is directly proportional to the number of moles. Pressure will be decreased with the decrease in number of moles to maintain the gas law when all other variables are kept constant.

Gas law is written below.

$\begin{array}{rll}PV& =& nRT\\ \frac{P_1{V}_1}{n_1{T}_{1}R}& =& \frac{P_2{V}_2}{n_2{T}_{2}R}\end{array}$ …(3)

Where,

• $P_1$ is the initial pressure exerted by gas molecules in $\left(\text{Pa}\right)$.

• $V_1$ is the initial volume of the container or gas.

• $V_2$ is the final volume of the container or gas.

• $T_1$ is the initial temperature of the gas.

• $T_2$ is the final temperature of the gas.

• $n$ is the number of moles of gas.

• $R$ is the gas constant which has a constant value of 8.314 J/mol × K.

The number of moles, $n_1=0.250\text{mol}$.

The number of moles, $n_2=0.500\text{mol}$.

Substitute the given values of $n_1$ and $n_2$ in equation (3).

$\begin{array}{rll}\frac{P_1}{n_1}& =& \frac{P_2}{n_2}\\ \frac{P_2}{P_1}& =& \frac{0.500\text{mol}}{0.250\text{mol}}\\ \frac{P_2}{P_1}& =& \frac{2}{1}\\ P_2& =& 2P_1\end{array}$

Therefore, the pressure will increase in the container.

Conclusion

The pressure will increase in the container.