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Maslowich
4 years ago
13

At constant temperature a bicycle tire pump contains 252mL of air at 995kPa pressure. The plunger of the pump is pushed down unt

il the volume is 95.0mL. What is the new pressure of the air inside the pump?
Chemistry
2 answers:
Trava [24]4 years ago
7 0

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.

Novay_Z [31]4 years ago
4 0

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

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