Chapter 11, Problem 11.93QP

Interpretation Introduction

Interpretation:

The total volume of gases and the partial pressure of the given gases have to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T.  An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$\text{PV = nRT}$

The relationship between partial pressure and ${\text{P}}_{\text{total}}$ is

${\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}$

Answer to Problem 11.93QP

The total volume of products is $\text{1}\text{.7×}{\text{10}}^{\text{2}}\text{L}$.

The partial pressure of given reaction

$\begin{array}{l}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}{\text{χ}}_{{\text{CO}}_{\text{2}}}{\text{P}}_{\text{T}}\\ {\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}\text{(0}\text{.41)(1}\text{.2}\text{atm)}\text{=}\text{0}\text{.49}\text{atm}\\ {\text{P}}_{\text{H2O}}\text{=}\text{(0}\text{.34)(1}\text{.2}\text{atm)}\text{=}\text{0}\text{.41}\text{atm}\\ {\text{P}}_{{\text{N}}_{\text{2}}}\text{=}\text{(0}\text{.21)(1}\text{.2}\text{atm)}\text{=}\text{0}\text{.25}\text{atm}\\ {\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\text{(0}\text{.034)(1}\text{.2}\text{atm)}\text{=}\text{0}\text{.041}\text{atm}\\ \end{array}$

Explanation of Solution

To calculate the moles Nitro-glycerine

You can map out the following strategy to solve for the total volume of gas.

$\begin{array}{l}\text{Grams nitro-glycerine}\to \text{moles nitro-glycerine}\to \text{moles products}\to \text{volume of products}\hfill \\ \hfill \\ \text{?mol products = 2}{\text{.6 10}}^{\text{2}}\text{ g nitroglycerin}\text{×}\frac{\text{1mol nitroglycerin}}{\text{227}\text{.09g nitroglycerin}}\text{×}\frac{\text{29mol product}}{\text{4mol nitroglycerin}}\hfill \\ \hfill \\ \text{=8}\text{.3mol}\hfill \end{array}$

The moles Nitro-glycerine is calculated by plugging in the values of the given grams of Nitro-glycerine and molecular weight of Nitro-glycerine.  The moles Nitro-glycerine was found to be $\text{8}\text{.3mol}$.

To calculate the volume of products

$\begin{array}{l}\text{PV = nRT}\\ {\text{ V}}_{\text{product}}\text{ =}\frac{{\text{n}}_{\text{product}}\text{RT}}{\text{P}}\\ {\text{ V}}_{\text{product}}\text{=}\frac{\text{(8}\text{.3}\text{mol)}\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(298}\text{K)}}{\text{(1}\text{.2}\text{atm)}}\text{=}\text{1}\text{.7×}{\text{10}}^{\text{2}}\text{L}\end{array}$

To calculate the mole fraction of each gaseous product

$\begin{array}{l}{\text{χ}}_{\text{component}}\text{=}\frac{\text{moles}\text{component}}{\text{total}\text{moles}\text{of}\text{all}\text{components}}\\ \\ {\text{χ}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{12}\text{mol}{\text{CO}}_{\text{2}}}{\text{29}\text{mol}\text{product}}\text{=}\text{0}\text{.41}\\ {\text{χ}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\frac{\text{12}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{29}\text{mol}\text{product}}\text{=}\text{0}\text{.34}\\ {\text{χ}}_{{\text{N}}_{\text{2}}}\text{=}\frac{\text{12}\text{mol}{\text{N}}_{\text{2}}}{\text{29}\text{mol}\text{product}}\text{=}\text{0}\text{.21}\\ {\text{χ}}_{{\text{O}}_{\text{2}}}\text{=}\frac{\text{12}\text{mol}{\text{O}}_{\text{2}}}{\text{29}\text{mol}\text{product}}\text{=}\text{0}\text{.034}\end{array}$

To calculate the partial pressure using above equation

$\begin{array}{l}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}{\text{χ}}_{{\text{CO}}_{\text{2}}}{\text{P}}_{\text{T}}\\ {\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}\text{(0}\text{.41)(1}\text{.2}\text{atm)}\text{=}\text{0}\text{.49}\text{atm}\\ {\text{P}}_{\text{H2O}}\text{=}\text{(0}\text{.34)(1}\text{.2}\text{atm)}\text{=}\text{0}\text{.41}\text{atm}\\ {\text{P}}_{{\text{N}}_{\text{2}}}\text{=}\text{(0}\text{.21)(1}\text{.2}\text{atm)}\text{=}\text{0}\text{.25}\text{atm}\\ {\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\text{(0}\text{.034)(1}\text{.2}\text{atm)}\text{=}\text{0}\text{.041}\text{atm}\\ \end{array}$

Conclusion

The total volume of gases and the partial pressure of the given gases were calculated.