Science, asked by Anonymous, 9 months ago

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?​

Answers

Answered by Anonymous
24

\large\red{\underline{{\boxed{\textbf{Brainliest\:Answer}}}}}

\rule{200}{2}

\red{\bold{{\underline{  Solution  \:    with\:  Explanation  \:  :}}}}

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

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}

\rule{200}{2}

\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}}$}

\rule{200}{2}

\boxed{\bold{\red{Keep\: Asking\: - \: Be \: Brainly}}}\\

Similar questions