Question

A bicycle pump contains 250 mL of air at a pressure of 102 kPa. If the pump is closed and the pressure increases to 210 kPa, the new volume becomes ________ mL.

Answer 1

Q155788

A bicycle pump contains 250 mL of air at a pressure of 102 kPa. If the pump is closed and the pressure increases to 210 kPa, the new volume becomes ________ mL.

Solution:

We are given the initial volume of the pressure of the air inside the bicycle pump.

P₁ = 102 kPa and V₁ = 250 mL.

After the pump is closed the pressure is 210 kPa and the volume of air inside the pump is unknown. Here we assume that no air is given out of the pump. The air is still inside the

pump but it occupies less volume compared to the volume when the pump was not closed

So P₂ = 210 kPa and V₂ = unknown.

We assume that no air is given out of the pump and the temperature of the air inside the pump remains the same.

So we can use the Boyles Law here.

Boyle's Law: The pressure of an ideal gas is inversely proportional to the volume at constant temperature and moles of gas.

two-point equation of Boyle's law is given as

$P_1V_1 = P_2V_2 ;$

Substitute the given information in this formula we have

 $102 kPa * 250 mL = 210 kPa * V_2 ;$

divide both the side by 210 kPa, we have

$\frac{102 kPa \space* \space 250 mL }{210 kPa} = \frac{210 kPa\space * \space V_2}{210 kPa}$

$\frac{25500mL}{210} = V_2 ;$

$V_2 = 121.43 mL ;$

In question, the quantity with the least number of the significant figure is 250mL and 210 kPa. Both are in 2 significant figures, so our final answer must also be in 2 significant figures.

Hence the volume of air inside the pump after it is closed is 120 mL.