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

Answer:

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

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:  

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