Question

a sealable bag holds 3.79 l of gas. how many moles of h₂ gas are needed to fill the bag at 1.0 atm and 293 k? ? mol h₂ hint: use pv = nrt.

Explanation:

Step1: List known values

$P=1.0\ \text{atm}$, $V=3.79\ \text{L}$, $T=293\ \text{K}$, $R=0.0821\ \frac{\text{L·atm}}{\text{mol·K}}$

Step2: Rearrange ideal gas law for n

$n=\frac{PV}{RT}$

Step3: Calculate moles of H₂

$n=\frac{(1.0)(3.79)}{(0.0821)(293)}=\frac{3.79}{24.0553}\approx0.158$

Answer:

0.16 (rounded to two decimal places) or 0.158 (unrounded)