Answer:
[tex]\large \boxed{\text{0.88 L}}[/tex]
Explanation:
The temperature and amount of gas are constant, so we can use Boyle’s Law.
[tex]p_{1}V_{1} = p_{2}V_{2}[/tex]
Data:
[tex]\begin{array}{rcrrcl}p_{1}& =& \text{1.34 atm}\qquad & V_{1} &= & \text{ 0.56 L} \\p_{2}& =& \text{0.85 atm}\qquad & V_{2} &= & ?\\\end{array}[/tex]
Calculations:
[tex]\begin{array}{rcl}\text{1.34 atm} \times \text{0.56 L} & =& \text{0.85 atm} \times V_{2}\\\text{0.750 L} & = & 0.85V_{2}\\V_{2} & = &\dfrac{0.750}{0.85}\\\\& = &\textbf{0.88 L}\\\end{array}\\\text{The balloon's new volume is $ \large \boxed{\textbf{0.88 L}}$}[/tex]
