Question

At constant temperature a bicycle tire pump contains 252mL of air at 995kPa pressure. The plunger of the pump is pushed down until the volume is 95.0mL. What is the new pressure of the air inside the pump?

Answer 1

Answer:

$p_2=2639.4kPa$

Explanation:

Hello,

In this case, we notice a relationship between the volume and pressure of a gas, accounting for the Boyle's law which has the following form:

$p_1V_1=p_2V_2$

In this case, we asked to find the pressure after the pump is pushed down, it means $p_2$ as shown below:

$p_2=\frac{p_1V_1}{V_2} =\frac{252mL*995kPa}{95.0mL}\\ \\p_2=2639.4kPa$

Such pressure increase accounts for a compression process at which the volume is decreased at the same time.

Best regards.

Answer 2

Answer:

The new pressure of the pump is 26.05 atm or 2639.4 kPa

Explanation:

Step 1: Data given

Volume of the bicycle tire pump = 252 mL = 0.252 L

Pressure of air = 995 kPa = 9.81989 atm

The volume of the pump is reduced to 95.0 mL = 0.095 L

Step 2: Calculate the new pressure

V1*P1 = V2*P2

⇒with V1 = the initial volume of the bicycle tire pump = 0.252 L

⇒with P1 = the initial pressure of the pump = 9.81989 atm = 995 kPa

⇒with V2 = the reduced volume of the pump = 0.095 L

⇒with P2 = the new pressure = TO BE DETERMINED

0.252 L * 9.81989 atm = 0.095 L * P2

P2 = 26.05 atm

The new pressure is 26.05 atm

OR

0.252 L * 995 = 0.095 L * P2

P2 = 2639.4 kPa

The new pressure of the pump is 26.05 atm or 2639.4 kPa