Question

3.60 moles of gas are held in a sealed 5.73 l container at 301 k. what is the pressure of the gas in the container? ? atm hint: use pv = nrt.

Explanation:

Step1: List given values and R

n=3.60 mol, V=5.73 L, T=301 K, R=0.0821 L·atm/(mol·K)

Step2: Rearrange ideal gas law for P

$P = \frac{nRT}{V}$

Step3: Calculate numerator

$3.60 \times 0.0821 \times 301 ≈ 3.60×24.7121 ≈ 88.9636$

Step4: Divide by volume

$P ≈ \frac{88.9636}{5.73} ≈ 15.5$

Answer:

15.5 atm