# Question 62744

Jul 24, 2017

The sample contains 0.80 mol of ${\text{N}}_{2}$.

#### Explanation:

We can use the Ideal Gas Law to solve this problem:.

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

We can rearrange this equation to get

$n = \frac{p V}{R T}$

In this problem,

$p = \text{5.0 atm}$
$V = {\text{4.0 dm}}^{3}$
$R = \text{0.082 06 atm·dm"^3"K"^"-1""mol"^"-1}$
$T = \text{(32 + 273.15) K" = "305.15 K}$

n = (5.0 color(red)(cancel(color(black)("atm"))) × 4.0 color(red)(cancel(color(black)("dm"^3))))/("0.082 06" color(red)(cancel(color(black)("atm·dm"^3"K"^"-1")))"mol"^"-1" × 305.15 color(red)(cancel(color(black)("K")))) = "0.80 mol"

Jul 25, 2017

There are 0.80 mol of nitrogen.

#### Explanation:

"But sir, what if the pressure unit is in pascals and the volume is in cubic metres instead of cubic decimetres?"

If

 p = 5.1 × 10^5color(white)(l) "Pa"
$V = {\text{0.0040 m}}^{3}$
$R = \text{8.314 Pa·m"^3"K"^"-1""mol"^"-1}$
$T = \text{305.15 K}$

Then

n = (5.1 × 10^5 color(red)(cancel(color(black)("Pa"))) × 0.0040 color(red)(cancel(color(black)("m"^3))))/( 8.314 color(red)(cancel(color(black)("Pa·m"^3"K"^"-1""mol"^"-1"))) × 305.15 color(red)(cancel(color(black)("K")))) = "0.80 mol"#