Question

A container of hydrogen at 172 kpa was decreased to 85.0 kpa producing a new volume of 765 mL. What was the original volume

Answer 1

Answer:

$\boxed{\text{378 mL}}$

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rclrcl}p_{1}& =& \text{172 kPa}\qquad & V_{1} &= & \text{?} \\p_{2}& =& \text{85.0 kPa}\qquad & V_{2} &= & \text{765 mL}\\\end{array}$

Calculations:

$\begin{array}{rcl}172V_{1} & =& 85.0 \times 765\\172V_{1} & = & 65 025\\V_{1} & = &\textbf{378 mL}\\\end{array}\\\text{The original volume was } \boxed{\textbf{378 mL}}$