# Question 6f85b

May 31, 2016

$m = 1.65 g$ of Neon in each tube.

#### Explanation:

We can modify the ideal gas law: $P V = n R T$, by using the expression of number of mole as:

$n = \frac{m}{M M}$ where, $m$ is the given mass and $M M = 20.2 \frac{g}{m o l}$ is the molar mass of neon.

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

We will need first to convert $157 k P a$ to $a t m$:

?atm=157cancel(kPa)xx(1atm)/(101.325cancel(kPa))=1.55atm#

$\implies m = \frac{1.55 \cancel{a t m} \times 1.3 \cancel{L} \times 20.2 \frac{g}{\cancel{m o l}}}{0.0821 \frac{\cancel{L} \cdot \cancel{a t m}}{\cancel{K} \cdot \cancel{m o l}} \times 301 \cancel{K}} = 1.65 g$