A pump contains 1.5 L of air at 175 kPa. You draw back on the piston of the pump, expanding the volume until the pressure reads 45 kPa. What is the new volume of air in the pump?

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Answer:

[tex]\large \boxed{\text{86.8 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}{rclrcl}p_{1}& =& \text{0.579 atm}\qquad & V_{1} &= & \text{150 L} \\p_{2}& =& \text{1.00 atm}\qquad & V_{2} &= & ?\\\end{array}[/tex]

Calculations:  

[tex]\begin{array}{rcl}0.579 \times 150 & =& 1.00V_{2}\\86.85 & = & 1.00V_{2}\\V_{2} & = &\dfrac{86.85}{1.00}\\\\& = &\textbf{86.8}\\\end{array}\\\text{The new volume of the gas is } \large \boxed{\textbf{86.8 L}}[/tex]