Chapter 5, Problem 5.30QP

(a)

Interpretation Introduction

Interpretation: The $P-V$ work done by the gases against an external pressure of $1.00\text{atm}$ 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 the gases against an external pressure of $1.00\text{atm}$.

Answer to Problem 5.30QP

Solution

The $P-V$ work during the reaction is $\underset{\_}{-48.23J}$.

Explanation of Solution

Explanation

Given

Mass of ${\text{KNO}}_{\text{3}}$ is $1.00\text{g}$.

External pressure is $1.00\text{atm}$.

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

Density of ${\text{CO}}_{\text{2}}$ is $1.830\text{g/L}$.

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,

${\text{2KNO}}_{\text{3}}\left(s\right)+\frac{1}{8}{\text{S}}_{\text{8}}\left(s\right)+\text{3C}\left(s\right)\stackrel{}{\to }{\text{K}}_{\text{2}}\text{S}\left(s\right)+{\text{N}}_{\text{2}}\left(g\right)+{\text{3CO}}_{\text{2}}\left(g\right)$

The number of moles potassium nitrate is calculated by using the formula,

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

Molar mass of ${\text{KNO}}_{\text{3}}$ is $101.1\text{g/mol}$.

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

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{KNO}}_{\text{3}}=\frac{1.00\text{g}}{101.1\text{g/mol}}\\ =0.00989\text{mol}\end{array}$

As per the balanced chemical reaction, two moles of ${\text{KNO}}_{\text{3}}$ produces one mole of nitrogen and three moles of carbon dioxide that is four moles of gases. Therefore, number of moles of gases formed is,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{gases}=0.00989\text{mol}{\text{KNO}}_{\text{3}}\times \frac{\text{4}\text{mol}\text{gases}}{\text{2}\text{mol}{\text{KNO}}_{\text{3}}}\\ =0.00989\times \frac{4}{2}\text{mol}\\ =\frac{0.03956}{2}\\ =0.01978\text{mol}\end{array}$

The volume occupied by the gas is calculated by using ideal gas equation,

$\begin{array}{c}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

Where,

• $P$ is the pressure of the container.
• $V$ is the volume of the container.
• $n$ is the number of moles.
• $R$ is the universal gas constant $\left(0.0821\text{L}\text{atm}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$.
• $T$ is the temperature.

Substitute the values of $P$, $n$, $R$ and $T$ in the ideal gas equation to calculate the volume.

$\begin{array}{c}V=\frac{nRT}{P}\\ =\frac{0.01978\text{mol}\times \text{0}\text{.0821}\text{L}\text{.atm/mol}\text{.K}\times \text{293}\text{K}}{1.00\text{atm}}\\ =\frac{1.623\times {10}^{-3}\times 293\text{L}\text{.atm}}{1.00\text{atm}}\\ =\frac{0.476\text{L}\text{.atm}}{1.00\text{atm}}\end{array}$

Simplify the above equation,

$V=0.476\text{L}$

The volume of a gas is increased by $0.476\text{L}$ that is $\Delta V=0.476\text{L}$.

The conversion of L to joule is done as,

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

Therefore, the conversion of $0.476\text{L}$ to joule is done as,

$\begin{array}{c}0.476\text{L}=0.476\times 101.33\text{J}\\ =48.23\text{J}\end{array}$

So 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.

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

$\begin{array}{c}w=-1.00\text{atm}\times \text{48}\text{.23}\text{J/atm}\\ =\underset{\_}{-48.23J}\end{array}$

Hence, the $P-V$ work during the reaction is $\underset{\_}{-48.23J}$.

(b)

Interpretation Introduction

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

Answer to Problem 5.30QP

Solution

The change in internal energy, $\Delta E$ is $\underset{\_}{-21.64kJ}$.

Explanation of Solution

Explanation

Given

The heat released is $-21.6\text{kJ}$.

The work done is $-48.23\text{J}$.

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

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

Therefore, the conversion of $48.23\text{J}$ to $\text{kJ}$ is done as,

$\begin{array}{c}48.23\text{J}=48.23\times {10}^{-3}\text{kJ}\\ =0.04823\text{kJ}\end{array}$

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 absorbed by the system.
• $w$ is the work done on the system.

Here, heat is released so $q$ is negative and work is done by the system on the surroundings so $w$ is also negative.

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

$\begin{array}{c}\Delta E=-21.6\text{kJ}+\left(-0.04823\text{kJ}\right)\\ =-21.6\text{kJ}-0.04823\text{kJ}\\ =\underset{\_}{-21.64kJ}\end{array}$

Hence, the change in internal energy, $\Delta E$ is $\underset{\_}{-21.64kJ}$.

Conclusion

1. a. The $P-V$ work during the reaction is $\underset{\_}{-48.23J}$.
2. b. The change in internal energy, $\Delta E$ is $\underset{\_}{-21.64kJ}$