Question

A 2.00 L container of nitrogen had a pressure of 3.20 atm. What volume would be
necessary to decrease the pressure to 1.03 atm?

Answer 1

Answer:

$\large \boxed{\text{6.21 L}}$

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:

$\begin{array}{rcrrcl}p_{1}& =& \text{3.20 atm}\qquad & V_{1} &= & \text{2.00 L} \\p_{2}& =& \text{1.03 atm}\qquad & V_{2} &= & ?\\\end{array}$

Calculation:  

$\begin{array}{rcl}3.20 \times 2.00 & =& 1.03V_{2}\\6.40 & = & 1.03V_{2}\\V_{2} & = &\dfrac{6.40}{1.03}\\\\& = &\textbf{6.21 L}\\\end{array}\\\text{The new volume must be } \large \boxed{\textbf{6.21 L}}$