# Question 08265

Dec 24, 2017

$\text{0.3 moles}$

#### Explanation:

The interesting thing to note here is that the volume and the temperature of the gaseous mixture remain constant, so you can use the fact that the pressure and the number of moles of gas present in the mixture have a direct relationship described by

P_1/n_1 = P_2/n_2 " " " "color(darkorange)("(*)")

Here

• ${P}_{1}$ and ${n}_{1}$ represent the pressure and the number of moles of gas present in the mixture at an initial state
• ${P}_{2}$ and ${n}_{2}$ represent the pressure and the number of moles of gas present in the mixture at a final state

Now, use the molar masses of carbon monoxide and sulfur dioxide to calculate the number of moles of each present in the mixture.

2.8 color(red)(cancel(color(black)("g"))) * "1 mole CO"/(28.01color(red)(cancel(color(black)("g")))) ~~ "0.10 moles CO"

3.2 color(red)(cancel(color(black)("g"))) * "1 mole SO"_2/(64.07color(red)(cancel(color(black)("g")))) ~~ "0.050 moles SO"_2

Now, use the ideal gas law to find the initial pressure of the gaseous mixture.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{P}_{1} V = {n}_{1} R T}}}$

Here

• ${P}_{1}$ is the initial pressure of the gas
• $V$ is the volume it occupies
• ${n}_{1}$ is the number of moles of gas present in the mixture
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Rearrange the equation to solve for ${P}_{1}$

${P}_{1} = \frac{{n}_{1} \cdot R T}{V}$

and plug in your values to get

${P}_{1} = \left(\left(0.10 + 0.050\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821 ("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(7color(red)(cancel(color(black)("L}}}}\right)$

${P}_{1} = \text{0.525 atm}$

So, you know that the pressure of the mixture must increase from $\text{0.525 atm}$ to $\text{1.5 atm}$, so you use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to say that the number of moles of gas present in the mixture must increase to

${n}_{2} = {P}_{2} / {P}_{1} \cdot {n}_{1}$

n_2 = (1.5 color(red)(cancel(color(black)("atm"))))/(0.525color(red)(cancel(color(black)("atm")))) * (0.10 + 0.050)color(white)(.)"moles"#

${n}_{2} = \text{0.429 moles}$

This means that you must add

$\text{0.429 moles " - " "(0.10 + 0.050)color(white)(.)"moles" = "0.3 moles}$

of oxygen gas to your flask to get the pressure of the mixture to increase to $\text{1.5 atm}$. The answer is rounded to one significant figure.