Answer:
[tex]\large \boxed{\text{D.) 91 kPa}}[/tex]
Explanation:
We can use the Ideal Gas Law — pV = nRT
Data:
V = 15.0 L
n = 0.55 mol
T = 300 K
Calculation:
[tex]\begin{array}{rcl}pV & =& nRT\\p \times \text{15.0 L} & = & \text{0.55 mol} \times \text{8.31 kPa$\cdot$ L$\cdot$K$^{-1}$mol$^{-1}\times$ 300 K}\\15.0p & = & \text{1370 kPa}\\p & = & \textbf{91 kPa}\end{array}\\\text{The pressure in the container is $\large \boxed{\textbf{91 kPa}}$}[/tex]