Question 5ea04

Apr 2, 2017

$\text{2.9 L}$

Explanation:

Your tool of choice here will be the ideal gas law equation, which looks like this

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

Here

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

Notice that the equation uses moles of gas, but that you are given the mass of the sample. The first thing to do here will be to convert the sample from grams to moles by using the molar mass of oxygen gas.

3.8 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.119 moles O"_2

Now, room temperature is usually taken to mean a temperature of ${20}^{\circ} \text{C}$, which is equivalent to

T["K"] = 20^@"C" + 273.15 = "293.15 K"#

No mention of the pressure at which the gas is being kept was made, but you can assume that you're dealing with normal pressure, i.e. with the pressure at sea level.

$P = \text{1.0 atm}$

Rearrange the ideal gas law equation to solve for $V$

$P V n R T \implies V = \frac{n R T}{P}$

Plug in your values to find

$V = \left(0.119 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 293.15color(red)(cancel(color(black)("K"))))/(1.0color(red)(cancel(color(black)("atm}}}}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{V = \text{2.9 L}}}}$

The answer is rounded to two sig figs.