# When an airbag is inflated the nitrogen gas has a pressure of 1.30 atmospheres, a temperature of 301 K, and a volume of 40.0 liters. What is the volume of the nitrogen at STP?

Jun 27, 2016

$\text{47.8 L}$

#### Explanation:

You can actually use two approaches to solve this problem.

You can use the ideal gas law to find the number of moles of gas present in that sample, then use the known molar volume of a gas at STP conditions to find the new volume.

As a way to double-check the result you get by using this approach, you can use the combined gas law equation to find the new volume of the gas without finding out how many moles it contains.

So, the ideal gas law equation looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Here you have

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Rearrange the equation to solve for $n$

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

Plug in your values to find

n = (1.30 color(red)(cancel(color(black)("atm"))) * 40.0 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 301color(red)(cancel(color(black)("K")))) = "2.104 moles N"_2

Now, STP conditions are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. Under these conditions for pressure and temperature, one mole of any ideal gas occupies $\text{22.7 L} \to$ this is the aforementioned molar volume of a gas at STP.

In your case, $2.104$ moles of nitrogen gas will occupy

2.104 color(red)(cancel(color(black)("moles N"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))

Now, you can double-check the result by suing the combined gas law

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${P}_{1}$, ${V}_{1}$, ${T}_{1}$ - the pressure, volume, and temperature of the gas at an initial state
${P}_{2}$, ${V}_{2}$, ${T}_{2}$ - the pressure, volume, and temperature of the gas at a final state

At mentioned before, STP conditions are defined as

${P}_{2} = \text{100 kPa" = 101/101.325color(white)(a)"atm}$

${T}_{2} = {0}^{\circ} \text{C" + 273.15 = "273.15 K}$

Rearrange the combined gas law equation to solve for ${V}_{2}$

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2 \implies {V}_{2} = {P}_{1} / {P}_{2} \cdot {T}_{2} / {T}_{1} \cdot {V}_{1}$

Plug in your values to find

V_2 = (1.30 color(red)(cancel(color(black)("atm"))))/(100/101.325color(red)(cancel(color(black)("atm")))) * (273.15color(red)(cancel(color(black)("K"))))/(301color(red)(cancel(color(black)("K")))) * "40.0 L"

${V}_{2} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{47.8 L}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

SIDE NOTE STP conditions are often in accordance to their old definition of a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions, one mole of any ideal gas occupies $\text{22.4 L}$. If this is the STP definition given to you, simply redo the calculations using $\text{22.4 L}$ instead of $\text{22.7 L}$.