# Question 4cfd2

Feb 9, 2017
1. The density of argon is 1.57 g/L.

#### Explanation:

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

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

Since $n = \frac{m}{M}$, we can substitute this to get

$P V = \left(\frac{m}{M}\right) R T$

We can rearrange this to

$P M = \frac{m}{V} R T$

But $\text{density"= "mass"/"volume}$ or color(brown)(bar(ul(|color(white)(a/a)ρ = m/Vcolor(white)(a/a)|)))" "

PM = ρRT

and

color(brown)(bar(ul(|color(white)(a/a)ρ = (PM)/(RT)color(white)(a/a)|)))" "

P = 748.0 color(red)(cancel(color(black)("torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("torr")))) = "0.9842 atm"
$M = \text{39.95 g/mol}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{20.65 °C" = "293.80 K}$

ρ = (0.9482color(red)(cancel(color(black)(·"atm"))) × 39.95color(white)(l) "g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.082 06" "L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 293.80color(red)(cancel(color(black)("K")))) = "1.57 g/L"

Feb 9, 2017

2. The value of $\text{R }$is $\text{8.314 Pa·m"^3"K"^"-1""mol"^"-1}$

STP is defined as a pressure of exactly 1 bar (${10}^{5} \textcolor{w h i t e}{l} \text{Pa}$) and 0 °C ($\text{273.15 K}$).

At STP the volume of 1 mol of an ideal gas is 22.71 L (${\text{0.022 71 m}}^{3}$).

The Ideal Gas Law is

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

We can rearrange this to get

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

R = (10^5 color(white)(l)"Pa" × "0.022 71 m"^3)/("1 mol × 273.15 K") = "8.314 Pa·m"^3"K"^"-1""mol"^"-1"#

Feb 9, 2017

3. You will need 15.0 L of ${\text{N}}_{2}$

#### Explanation:

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

The balanced equation for the reaction is

$\textcolor{w h i t e}{l l} {\text{N"_2 + "3H"_2 → "2NH}}_{3}$
$\textcolor{w h i t e}{l l} \text{1 L"color(white)(ml)"3 L}$

According to Gay-Lussac, ${\text{1 L"color(white)(l) "of N}}_{2}$ reacts with ${\text{3 L of H}}_{2}$.

${\text{Volume of N"_2 = 45.0 color(red)(cancel(color(black)("L" color(white)(l)"H"_2))) × ("1 L"color(white)(l) "N"_2)/(3 color(red)(cancel(color(black)("L"color(white)(l) "H"_2)))) = "15.0 L"color(white)(l)"N}}_{2}$