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

`\boxed{\text{378 mL}}`

Step-by-step 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 * 765\\172V_(1) & = & 65 025\\V_(1) & = &\textbf{378 mL}\\\end{array}\\\text{The original volume was } \boxed{\textbf{378 mL}}`