Chapter 5, Problem 5.29QP

(a)

Interpretation Introduction

Interpretation: The $P-V$ work done by the gas against an external pressure of $1.0\text{atm}$ and the change in internal energy $\left(\Delta E\right)$ for the system are to be calculated.

Concept introduction: Internal energy of a system is defined as the sum of all the kinetic energy and the potential energy of all the compounds of the system. From the first law of thermodynamics, the change in internal energy is the sum of heat absorbed by the system and the work done on the system.

To determine: The $P-V$ work done by ${\text{N}}_{\text{2}}$ gas against an external pressure of $1.0\text{atm}$.

Answer to Problem 5.29QP

Solution

The $P-V$ work during the reaction is $\underset{\_}{126.45J}$.

Explanation of Solution

Explanation

Given

Mass of ${\text{NaN}}_{\text{3}}$ is $2.25\text{g}$.

External pressure is $1.0\text{atm}$.

Density of nitrogen is $1.165\text{g/L}$.

Temperature of the reaction is $2\text{0}{}^{\text{ο}}\text{C}$.

The conversion of Celsius to Kelvin is done as,

$\text{0}{}^{\text{ο}}\text{C}=273\text{K}$

Therefore, the conversion of $2\text{0}{}^{\text{ο}}\text{C}$ to Kelvin is done as,

$\begin{array}{c}\text{20}{}^{\text{ο}}\text{C}=20+273\text{K}\\ =293\text{K}\end{array}$

The given chemical reaction is,

${\text{2NaN}}_3\left(s\right)\stackrel{}{\to }\text{2Na}\left(s\right)+3{\text{N}}_{\text{2}}\left(g\right)$

The number of moles of ${\text{NaN}}_{\text{3}}$ is calculated by using the formula,

$\text{Number}\text{of}\text{moles}\text{of}{\text{NaN}}_{\text{3}}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Molar mass of ${\text{NaN}}_{\text{3}}$ is $65\text{g/mol}$.

Substitute the values of mass and molar mass of ${\text{NaN}}_{\text{3}}$ in the above formula to calculate its number of moles.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{NaN}}_{\text{3}}=\frac{2.25\text{g}}{65\text{g/mol}}\\ =0.03462\text{mol}\end{array}$

As per the balanced chemical reaction, two moles of ${\text{NaN}}_{\text{3}}$ produces three moles of ${\text{N}}_{\text{2}}$. Therefore, number of moles of ${\text{N}}_{\text{2}}$ formed for $0.03462\text{mol}$ of ${\text{NaN}}_{\text{3}}$ is,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{N}}_{\text{2}}=0.03462\text{mol}{\text{NaN}}_{\text{3}}\times \frac{3\text{mol}{\text{N}}_{\text{2}}}{\text{2}\text{mol}{\text{NaN}}_{\text{3}}}\\ =0.03462\times \frac{3}{2}\text{mol}\\ =\frac{0.10386}{2}\\ =0.05193\text{mol}\end{array}$

Therefore, grams of ${\text{N}}_{\text{2}}$ formed is calculated by using the formula,

$\text{Mass}=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{moles}$

Substitute the values of molar mass and number of moles in the above formula to calculate the mass of ${\text{N}}_{\text{2}}$.

$\begin{array}{c}\text{Mass}\text{of}{\text{N}}_2=28\text{g/mol}\times 0.05193\text{mol}\\ =1.45404\text{g}\end{array}$

The volume of ${\text{N}}_{\text{2}}$ generated is calculated by using the formula,

$\text{Volume}=\frac{\text{Mass}}{\text{Density}}$

Substitute the value of mass and density of ${\text{N}}_{\text{2}}$ in the above formula to calculate the volume of ${\text{N}}_{\text{2}}$.

$\begin{array}{c}\text{Volume}=\frac{1.45404\text{g}}{1.165\text{g/L}}\\ =1.248\text{L}\end{array}$

The $P-V$ work done during the reaction is calculated by using the expression,

$w=P\Delta V$

Where,

• $w$ is the work done.
• $P$ is the pressure applied on the system.
• $\Delta V$ is the change in volume of the system.

The change in volume, $\Delta V=V_{\text{f}}-V_{\text{in}}$.

Where,

• $V_{\text{f}}$ is the final volume.
• $V_{\text{in}}$ is the initial volume.

Here, initial volume is zero.

Substitute the values of $P$ and $\Delta V$ in the above expression to calculate the work done.

$\begin{array}{c}w=1.0\text{atm}\times \left(V_{\text{f}}-V_{\text{in}}\right)\\ =1.0\text{atm}\times \left(1.248\text{L}-0\right)\\ =1\times 1.248\text{atm}\text{.L}\\ =1.248\text{atm}\text{.L}\end{array}$

The conversion of $\text{atm}\text{.L}$ to joule is done as,

$1\text{atm}\text{.L}=101.33\text{J}$

Therefore, the conversion of $1.248\text{atm}\text{.L}$ to joule is done as,

$\begin{array}{c}1.248\text{atm}\text{.L}=1.248\times 101.33\text{J}\\ =\underset{\_}{126.45J}\end{array}$

Hence, the $P-V$ work during the reaction is $\underset{\_}{126.45J}$.

(b)

Interpretation Introduction

To determine: The change in internal energy $\left(\Delta E\right)$ for the system.

Answer to Problem 5.29QP

Solution

The change in internal energy, $\Delta E$ is $\underset{\_}{-2.466kJ}$.

Explanation of Solution

Explanation

Given

The heat released is $-2.34\text{kJ}$.

The work done is $126.45\text{J}$.

The conversion of J to $\text{kJ}$ is done as,

$1\text{J}={10}^{-3}\text{kJ}$

Therefore, the conversion of $126.45\text{J}$ to $\text{kJ}$ is done as,

$\begin{array}{c}126.45\text{J}=126.45\times {10}^{-3}\text{kJ}\\ =0.126\text{kJ}\end{array}$

The change in internal energy is calculated by using the first law of thermodynamics which is represented by the equation,

$\Delta E=q-w$

Where,

• $\Delta E$ is the change in internal energy.
• $q$ is the heat released by the system.
• $w$ is the work done.

Here, heat is released so $q$ is negative.

Substitute the values of $q$ and $w$ in the above equation to calculate the change in internal energy.

$\begin{array}{c}\Delta E=-2.34\text{kJ}-0.126\text{kJ}\\ =\underset{\_}{-2.466kJ}\end{array}$

Hence, the change in internal energy, $\Delta E$ is $\underset{\_}{-2.466kJ}$.

Conclusion

1. a. The $P-V$ work during the reaction is $\underset{\_}{126.45J}$.
2. b. The change in internal energy, $\Delta E$ is $\underset{\_}{-2.466kJ}$