Question

The gas in a 250. mL piston experiences a change in pressure from 1.00 atm to 2.55 atm. What is the new volume (in mL) assuming the moles of gas and temperature are held constant?

Answer 1

`\large \boxed{\text{0.980 L}}`

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}{rcrrcl}p_(1)& =& \text{1.00 atm}\qquad & V_(1) &= & \text{250. mL} \\p_(2)& =& \text{2.55 atm}\qquad & V_(2) &= & ?\\\end{array}`

Calculations:

`\begin{array}{rcl}\text{1.00 atm} * \text{250. mL} & =& \text{2.55 atm} * V_(2)\\\text{250. mL} & = & 2.55V_(2)\\V_(2) & = &\frac{\text{250. mL}}{2.55}\\\\& = &\textbf{98.0 mL}\\\end{array}\\\text{The balloon's new volume is $ \large \boxed{\textbf{0.980 L}}$}`