What is the density of nitrogen gas at 90.5 kPa and 43.0 °C?

Oct 16, 2016

0.965 g/L

Explanation:

Known:
P= 90.5 kPa = 90500 Pa
T= ${43}^{o}$ C = 316.15 K

We can safely assume that when working with gases, we will be using the ideal gas law. In this case, we combine the ideal gas formula with the mole formula and the density formula:

$P \cdot V = n \cdot R \cdot T$

$n = \frac{m}{M}$

$d = \frac{m}{V}$
$m = d \cdot V$

$n = \frac{d \cdot V}{M}$

$P \cdot V = \frac{d \cdot V}{M} \cdot R \cdot T$

$P \cdot V \cdot M = d \cdot V \cdot R \cdot T$

$\frac{P \cdot V \cdot M}{V \cdot R \cdot T} = d$

$\frac{P \cdot M}{R \cdot T} = d$

Calculate molar mass of ${N}_{2}$:

$M = 14.01 \cdot 2$
$= 28.02$

Sub the values we have in to get density:

$\frac{P \cdot M}{R \cdot T} = d$

(90500 Pa*28.02g·mol^"-1")/(8.314 Pa·m^3K^"-1"mol^"-1"*316.15 K) = d

$964.75 \frac{g}{m} ^ 3 = d$

Note: the value of R, the gas constant, changes depending on the units you use for other measurements. This is why the answer was in $\frac{g}{m} ^ 3$ and has to be converted to $\frac{g}{L}$. It is VERY IMPORTANT to choose the right R value or to adjust other units to accommodate the value of R you picked.

$\frac{964.75 \frac{g}{m} ^ 3}{1000 \frac{L}{m} ^ 3} = d$

$0.965 \frac{g}{L} = d$