Boyle`s law question, Grade 9

[LadyDiana1011]

A bicycle pump contains 200 cm³ o fair and is connected to a bicycle tyre.
The volume of the tyre is 800 cm³. The pressure of the air in the tyre (it is `flat´) is 1.0 atmosphere, the same as the air initially?
A) What is the total volume of the air initially? (=volume1)
B) What is the volume of air after the pump is pushed in? (=volume2)
C) What will be the pressure in the tyre then?


Can you please give me the answers?
If possible explain it?
Would be nice of you.
Thankyou!

Diana

 

[lewando]

Diana,

Sorry, but PF rules discourage "giving answers". Plus, where is the fun in that? We do, however, answer questions related to your problem solving process. Go ahead and attempt a solution--if you are confused about something, say so.

 

[tomcue]

,

I would be happy to provide a response to your questions about Boyle's Law. Boyle's Law states that the volume of a gas is inversely proportional to its pressure, assuming the temperature and amount of gas remain constant. This means that as the volume of a gas decreases, its pressure increases and vice versa.

Now, let's apply this law to the situation described in the question. We have a bicycle pump containing 200 cm³ of air, which is connected to a bicycle tire with a volume of 800 cm³. The pressure of the air in the tire is currently 1.0 atmosphere, which is the same as the initial pressure in the pump.

A) To answer the first question, we need to find the total volume of air in the pump and tire combined. This can be done by simply adding the volumes of the air in the pump and tire together. So, the total volume of air initially is 200 cm³ + 800 cm³ = 1000 cm³.

B) When the pump is pushed in, the volume of air in the pump decreases while the volume of air in the tire increases. This is because the air from the pump is transferred into the tire. Let's call the new volume of air in the pump and tire combined as volume2. According to Boyle's Law, the product of the initial volume and pressure of a gas is equal to the product of the final volume and pressure. Mathematically, this can be represented as P1V1 = P2V2. Since the pressure remains constant at 1.0 atmosphere, we can rearrange this equation to find volume2: V2 = (P1V1)/P2. Plugging in the values, we get V2 = (1.0 x 1000 cm³)/1.0 = 1000 cm³. Therefore, the volume of air after the pump is pushed in is still 1000 cm³.

C) Finally, to find the pressure in the tire after the pump is pushed in, we can use the same equation as above, but this time solving for pressure. So, P2 = (P1V1)/V2. Plugging in the values, we get P2 = (1.0 x 1000 cm³)/1000 cm³ = 1.0 atmosphere. This means that the pressure in the tire remains the same at 1.0 atmosphere after