Chapter 5, Problem 5.43PAE

39 A sample containing only NO₂ and SO₂, has a total pressure of 120. torr. Measurements show that the partial pressure of NO₂ is 43 torr. 1f the vessel has a volume of 800.0 mL and the temperature is 22.0°C, how many moles of each gas are present?

Interpretation Introduction

Interpretation:

The moles number of each gas (NO₂ and SO₂) present in the sample has to be calculated

Concept Introduction:

• An ideal gas has is a hypothetical gas that is formed by a few molecules so that, its density is low, and the intermolecular forces are null. The model of gas facilitates mathematical calculations to study gases.
• The ideal gas law can be expressed in the following way,

  $\text{P x V= n x R x T}$

  Where P is the gas pressure (atm), V is the gas volume (L), T is the gas absolute temperature (K), n is the moles number of the gas and R is the ideal gas constant ( $\text{R=0}\text{.082 }\frac{\text{atm L}}{\text{K mol}}$ )

• $\text{Absolute temperature (K) = T (°C )+ 273}$. where, T (°C) is the temperature in Celsius degrees (°C).

• $\text{1 atm = 760 torr}$
• The ideal gas mixture is a mixture of two or more ideal gases that has the following characteristics,

  Each component of the mixture behaves as ideal gas that occupies all the mixture volume at the mixture temperature.

  Each component exercises a partial pressure. The sum of the partial pressure of the components is equal to the total pressure of the mixture.

• The mole fraction is the ratio between the moles number of the component in the mixture and the total number of moles of gas in the mixture. It can be expressed in the following way,

  ${\text{y}}_{\text{i}}\text{=}\frac{{\text{n}}_{\text{i}}}{{\text{n}}_{\text{total}}}$

  Where, y_i is the molar fraction of component i, n_i is the moles number of component i and n _total is the total moles number of the mixture.

• The partial pressure is the pressure that a gas of the mixture exercises if it occupies all the volume of the mixture.
• The partial pressure of each component in the mixture is directly proportional to its molar fraction. It can be expressed in the following way,

  ${\text{P}}_{\text{i}}={\text{y}}_{\text{i}}{\text{ x P}}_{\text{total}}$

  Where, P_i is the partial pressure of component i, P_total is the total pressure of the mixture and y_i is the molar fraction of component i.

Answer to Problem 5.43PAE

Solution:

$\begin{array}{l}{\text{n}}_{{\text{NO}}_{\text{2}}}\text{= 1}{\text{.865 x 10}}^{\text{-3}}\text{ mol}\\ {\text{n}}_{{\text{SO}}_{\text{2}}}\text{= 3}{\text{.396 x 10}}^{\text{-3}}\text{ mol}\end{array}$

Explanation of Solution

Given Information:

$\begin{array}{l}{\text{P}}_{\text{total}}=120\text{ torr}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}=43\text{ torr}\\ {\text{V}}_{\text{vessel}}=800.0\text{ mL}\\ {\text{T}}_{\text{sample}}=\text{22}\text{.0 °C}\end{array}$

1-Convert the T in K

${\text{T}}_{\text{sample}}=\text{22}\text{.0 °C + 273 =295}\text{.0 °C}$

2-Convert the pressure in atm

$\begin{array}{c}{\text{P}}_{\text{total}}=120\text{ torr x }\frac{1\text{ atm}}{760\text{ torr}}=0.158\text{ atm}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}=43\text{ torr x }\frac{1\text{ atm}}{760\text{ torr}}=0.056\text{ atm}\end{array}$

3-Convert the volume in L

${\text{V}}_{\text{vessel}}=800.0\text{ mL x }\frac{1\text{ L}}{1000\text{ mL}}=0.800\text{ L}$

4-Calculate the partial pressure of SO₂

${\text{P}}_{{\text{SO}}_{\text{2}}}={\text{P}}_{\text{total}}{\text{ - P}}_{{\text{NO}}_{\text{2}}}\text{ =}0.158\text{ atm - }0.056\text{ atm = 0}\text{.102 atm}$

5-Calculate the total moles number of the sample

6-Calculate the moles number of each component

$\begin{array}{l}0.056\text{ atm = }\frac{{\text{n}}_{{\text{NO}}_{\text{2}}}}{\text{5}{\text{.261 x 10}}^{\text{-3}}\text{ mol}}\text{ x 0}\text{.158 atm}\\ \frac{0.056\text{ atm }}{\text{0}\text{.158 atm}}\text{= }\frac{{\text{n}}_{{\text{NO}}_{\text{2}}}}{\text{5}{\text{.261 x 10}}^{\text{-3}}\text{ mol}}\\ {\text{n}}_{{\text{NO}}_{\text{2}}}=\frac{0.056\text{ atm }}{\text{0}\text{.158 atm}}\text{x 5}{\text{.261 x 10}}^{\text{-3}}\text{ mol= 1}{\text{.865 x 10}}^{\text{-3}}\text{ mol}\\ 0.102\text{ atm = }\frac{{\text{n}}_{{\text{SO}}_{\text{2}}}}{\text{5}{\text{.261 x 10}}^{\text{-3}}\text{ mol}}\text{ x 0}\text{.158 atm}\\ \frac{0.102\text{ atm }}{\text{0}\text{.158 atm}}\text{= }\frac{{\text{n}}_{{\text{SO}}_{\text{2}}}}{\text{5}{\text{.261 x 10}}^{\text{-3}}\text{ mol}}\\ {\text{n}}_{{\text{SO}}_{\text{2}}}=\frac{0.102\text{ atm }}{\text{0}\text{.158 atm}}\text{x 5}{\text{.261 x 10}}^{\text{-3}}\text{ mol= 3}{\text{.396 x 10}}^{\text{-3}}\text{ mol}\end{array}$

Conclusion

The sample is formed by $\text{1}{\text{.865 x 10}}^{\text{-3}}\text{ mol}$ of NO₂ and $\text{ 3}{\text{.396 x 10}}^{\text{-3}}\text{ mol}$ of SO₂.