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: Identify given values and formula

Given \( n=3.60\ \text{mol} \), \( V=5.73\ \text{L} \), \( T=301\ \text{K} \), \( R=0.0821\ \frac{\text{L·atm}}{\text{mol·K}} \); use \( PV=nRT \) → \( P=\frac{nRT}{V} \)

Step2: Substitute values into formula

\( P=\frac{3.60 \times 0.0821 \times 301}{5.73} \)

Step3: Calculate numerator first

\( 3.60 \times 0.0821 = 0.29556 \); \( 0.29556 \times 301 ≈ 88.96356 \)

Step4: Divide by volume

\( P≈\frac{88.96356}{5.73}≈15.5 \)

Answer:

15.5 atm