Chapter 5, Problem 92E

a)

Interpretation Introduction

Interpretation: Pressure exerted by $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container at $25.0°\text{C}$ through ideal gas equation is to be calculated.

Concept introduction:Ideal gas law is used to represent hypothetical gas to relate its volume, pressure, and temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.

Answer to Problem 92E

Pressure of $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container is $1.223\text{ atm}$ .

Explanation of Solution

Conversion of $25°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =25°\text{C}+273\\ =298\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.

Rearrange above equation to calculate value of $P$ .

  $P=\frac{nRT}{V}$

Value of $n$ is $0.5000\text{ mol}$ .

Value of $T$ is $298\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $10.000\text{ L}$ .

Substitute the value in above equation.

  $\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{\left(0.5000\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}{\left(10.000\text{ L}\right)}\\ =1.223\text{ atm}\end{array}$

Hence pressure of $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container is $1.223\text{ atm}$ .

b)

Interpretation Introduction

Interpretation: Pressure exerted by $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container at $25.0°\text{C}$ through van der Waals equation is to be calculated.

Concept introduction:Real gases with atoms or molecules have intermolecular attractions as well as repulsions. These gases do not follow ideal gas equation. Such gases obey van der Waals equation. Van der Waals equation is as follows:

  $\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.
• $a$ and $b$ are van der Waals parameters.

Answer to Problem 92E

Pressure of $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container is $1.221\text{ atm}$ .

Explanation of Solution

Conversion of $25°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =25°\text{C}+273\\ =298\text{K}\end{array}$

Van der Waals equation is as follows:

  $\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.
• $a$ and $b$ are van der Waals parameters.

Rearrange above equation to calculate value of $P$ .

  $P=\left(\frac{nRT}{\left(V-nb\right)}\right)-\left(a\frac{n^2}{V^2}\right)$

Value of $n$ is $0.5000\text{ mol}$ .

Value of $T$ is $298\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $V$ is $10.000\text{ L}$ .

Value of $a$ for ${\text{N}}_2$ gas is $1.39\text{ atm}\cdot {\text{L}}^2\cdot {\text{mol}}^{-2}$

Value of $b$ for ${\text{N}}_2$ gas is $0.0139\text{ L}\cdot {\text{mol}}^{-1}$

Substitute the value in above equation.

  $\begin{array}{c}P=\left(\frac{nRT}{\left(V-nb\right)}\right)-\left(a\frac{n^2}{V^2}\right)\\ =\left(\begin{array}{l}\left(\frac{\left(0.5000\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}{\left(\left(10.000\text{ L}\right)-\left(0.5000\text{ mol}\right)\left(0.0139\text{ L}\cdot {\text{mol}}^{-1}\right)\right)}\right)-\\ \left(\left(1.39\text{ atm}\cdot {\text{L}}^2\cdot {\text{mol}}^{-2}\right)\frac{{\left(0.5000\text{ mol}\right)}^2}{{\left(10.000\text{ L}\right)}^2}\right)\end{array}\right)\\ =1.221\text{ atm}\end{array}$

Hence pressure of $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container is $1.221\text{ atm}$ .

c)

Interpretation Introduction

Interpretation: Result of pressure exerted by $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container calculated by ideal gas and van der Waals equation should be compared.

Concept introduction:Ideal gas law is used to represent hypothetical gas to relate its volume, pressure, and temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.

Explanation of Solution

In ideal conditions, pressure of $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container is $1.223\text{ atm}$ but real gas exert less pressure than this it is $1.221\text{ atm}$ . This is due to intermolecular forces between gas molecules.

d)

Interpretation Introduction

Interpretation: Result of pressure exerted by $0.5000{\text{ mol N}}_2$ in $10.000\text{ L}$ container calculated by ideal gas and van der Waals and pressure exerted by $0.5000{\text{ mol N}}_2$ in $1.0000\text{ L}$ container calculated by ideal gas and van der Waals equation should be compared.

Concept introduction:Ideal gas law is used to represent hypothetical gas to relate its volume, pressure, and temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of the gas.
• $V$ is volume of gas.
• $n$ denotes moles of gas.
• $R$ is gas constant.
• $T$ is temperature of gas.

Explanation of Solution

In ideal conditions, pressure of $0.5000{\text{ mol N}}_2$ in $1.0000\text{ L}$ container is $12.23\text{ atm}$ and real gas exerts less pressure than this it is $12.13\text{ atm}$ . This difference in pressure of ideal gas and real gas becomes negligible as the volume of container increases. This is because interaction between molecules becomes less as volume of container increases.