Chapter 5, Problem 117AE

Interpretation Introduction

Interpretation:Pressure of tank after experiment at $25°\text{C}$ and $125°\text{C}$ should be determined.

Concept introduction: State of any gas is determined by some of its parameters. These parameters are pressure, amount of gas, temperature and volume. Ideal gas law helps to govern state of gas with help of relationships between these gas parameters.

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Answer to Problem 117AE

Pressure of tank after experiment at $25°\text{C}$ and $125°\text{C}$ is $\text{2}\text{.00 atm}$ and $4.00\text{ atm}$ respectively.

Explanation of Solution

Reaction for formation of water from hydrogen and oxygen is as follows:

  $2{\text{H}}_{\text{2}}+{\text{O}}_{\text{2}}\to 2{\text{H}}_{\text{2}}\text{O}$

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 gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for ${\text{H}}_2$ .

  $n=\frac{PV}{RT}$

Value of $P$ is $2.00\text{ atm}$ .

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 $20.0\text{ L}$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(2.00\text{ atm}\right)\left(20.0\text{ L}\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}\\ =1.64\text{ mol}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $n$ for ${\text{O}}_2$ .

  $n=\frac{PV}{RT}$

Value of $P$ is $3.00\text{ atm}$ .

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 $20.0\text{ L}$ .

Substitute the value in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(3.00\text{ atm}\right)\left(20.0\text{ L}\right)}{\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}\\ =2.45\text{ mol}\end{array}$

Since $2{\text{ mol H}}_2$ is required to react with ${\text{1 mol O}}_2$ thus $1.64{\text{ mol H}}_2$ is limiting reagentin reaction. Thus amount of product ${\text{H}}_2\text{O}$ is formed according to limiting reagent.

Moles of ${\text{H}}_2\text{O}$ that would produce from $1.64{\text{ mol H}}_2$ can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\left(1.64{\text{ mol H}}_2\right)\left(\frac{2{\text{ mol H}}_2\text{O}}{2{\text{ mol H}}_2}\right)\\ =1.64{\text{ mol H}}_2\text{O}\end{array}$

Moles of ${\text{O}}_2$ that would consumed in reactioncan be calculated as follows:

  $\begin{array}{c}\text{Moles}=\left(1.64{\text{ mol H}}_2\right)\left(\frac{1{\text{ mol O}}_2}{2{\text{ mol H}}_2}\right)\\ =0.82{\text{ mol O}}_2\end{array}$

Moles of ${\text{O}}_2$ remain in tank after completion of reaction can be calculated as follows:

  $\begin{array}{c}\text{Moles}=2.45\text{ mol}-0.82\text{ mol}\\ =1.63\text{ mol}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $P$ for $1.63{\text{ mol O}}_2$ .

  $P=\frac{nRT}{V}$

Value of $n$ is $1.63\text{ mol}$ .

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 $20.0\text{ L}$ .

Substitute the value in above equation.

  $\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{\left(1.63\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}{20.0\text{ L}}\\ =2.00\text{ atm}\end{array}$

Hence pressure of tank after experiment at $25°\text{C}$ is $2.00\text{ atm}$ .

Conversion of $125°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =125°\text{C}+273\\ =398\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $P$ at $\text{125 }°\text{C}$ for $1.63{\text{ mol O}}_2$ .

  $P=\frac{nRT}{V}$

Value of $n$ is $1.63\text{ mol}$ .

Value of $T$ is $398\text{K}$ .

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

Value of $V$ is $20.0\text{ L}$ .

Substitute the value in above equation.

  $\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{\left(1.63\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(398\text{K}\right)}{20.0\text{ L}}\\ =2.66\text{ atm}\end{array}$

Pressure of ${\text{H}}_2\text{O}\left(g\right)$ at $373\text{ K}$ is $1.00\text{ atm}$ thus pressure of ${\text{H}}_2\text{O}\left(g\right)$ at $398\text{ K}$ can be calculated as follows:

  $\begin{array}{c}\text{Pressure}=\left(1\text{ atm}\right)\left(\frac{398\text{ K}}{373\text{ K}}\right)\\ =1.06\text{ atm}\end{array}$

Total pressure of tank at $\text{125 }°\text{C}$ can be calculated as follows:

  $\begin{array}{c}\text{Total pressure}=1.06+2.68\text{ atm}\\ =4.00\text{ atm}\end{array}$

Hence pressure of tank $125°\text{C}$ is $4.00\text{ atm}$ .

Conclusion

Chemical formula for manganese chlorideis ${\text{MnCl}}_4$ .