Filling the wheel with air (ideal gas)

[songoku]

Homework Statement

A wheel has volume 1.2 x 10^-3 m³ when being full. A pump has working volume of 9 x 10^-5 m³. How many times do you need to push the pump to fill the air to the wheel (initially there is no air in the wheel) until the pressure is 3.0 x 10⁵ Pa? The atmospheric pressure is 1.0 x 10⁵ Pa and there is no change in temperature

Homework Equations

PV/T = constant

The Attempt at a Solution

I think my attempt is really really weird, but I'll post it anyway...

P₁.V₁ = P₂.V₂
1.0 x 10⁵.V₁ = 3.0 x 10⁵ . 1.2 x 10^-3
V₁ = 3.6 x 10^-3

ΔV = V₁ - V₂ = 2.4 x 10^-3

Number of pumping = 2.4 x 10^-3 / 9 x 10^-5 = 80 / 3 ≈ 27 times

Actually, my idea is to find the initial volume of air in the wheel. The difference between the initial and final volume is the volume needed to be pumped to the wheel. But I think the final volume should be greater (my calculation shows the opposite). At the beginning, I take the pressure of the wheel the same as atmospheric pressure.

Another thing that bothers me is the sentence "initially there is no air in the wheel". It means the initial volume is zero ?

 

[LawrenceC]

Hint:

Determine how much air (density*volume) is in the tire at specified pressure.

 

[parazit]

Hi. I have the same idea with you.
The volume of the wheel is 1,2E-3
The volume of the pump is 9E-5
The pressure requested on the wheel is 3E5 and
The output pressure of the air is 1E5
since the internal pressure have to be equal to the external pressure,
P₁V₁ = P₂V₂ and than
3E5*1,2E-3=1E5*V ==> from there we got V=3,6E-3 this have to be the volume of the pump to fill the wheel with air at just one push. However, our pump's volume is 9E-5 so we need to divide it.
(3,6E-3)/(9E-5)=40 That means we have to push our pump for 40 times to obtain the result.
I think you do not need to make any substraction since it says the wheel is initially empty.

 

[songoku]

Hint:

Determine how much air (density*volume) is in the tire at specified pressure.

At P = 1.0 x 10⁵, the volume is 3.6 x 10^-3. Taking density of air = 1 kg m^-3, the mass is 3.6 x 10^-3 kg

AT P = 3.0 x 10⁵, the volume is 1.2 x 10^-3 so the mass is 1.2 x 10^-3 kg

Then what should I do?

Hi. I have the same idea with you.
The volume of the wheel is 1,2E-3
The volume of the pump is 9E-5
The pressure requested on the wheel is 3E5 and
The output pressure of the air is 1E5
since the internal pressure have to be equal to the external pressure,
P₁V₁ = P₂V₂ and than
3E5*1,2E-3=1E5*V ==> from there we got V=3,6E-3 this have to be the volume of the pump to fill the wheel with air at just one push. However, our pump's volume is 9E-5 so we need to divide it.
(3,6E-3)/(9E-5)=40 That means we have to push our pump for 40 times to obtain the result.
I think you do not need to make any substraction since it says the wheel is initially empty.

3.6 x 10^-3 is the volume at 1.0 x 10⁵ Pa? How can that be the volume needed to be pumped to the wheel while the question states to find the number of pumping when the pressure is 3.0 x 10⁵?

When the wheel is empty, is the pressure the same as atmospheric pressure?

Thanks

 

[parazit]

ohh..I think i misunderstand something. due to rush, it may be wrong. sorry for wasting your time.

 

[LawrenceC]

Another hint:

Write an expression for the mass of air in the inflated tire. It will be a function of temperature and pressure. You do not know the temperature.

Write another expression for the mass of air in the pump cylinder at atmospheric pressure. It will be a function of temperature and pressure. You do not know the temperature.

The problem states the entire process is isothermal.

 

[songoku]

Another hint:

Write an expression for the mass of air in the inflated tire. It will be a function of temperature and pressure. You do not know the temperature.

Write another expression for the mass of air in the pump cylinder at atmospheric pressure. It will be a function of temperature and pressure. You do not know the temperature.

The problem states the entire process is isothermal.

Not sure how to do it but here's what I've done:
*inflated tire
P₁V₁ = n₁RT₁ ; n = m/M where M is molecular mass
P₁V₁M = m₁RT₁
m = P₁V₁M / RT₁ --> here the mass is the function of pressure, volume and temperature

*pump cylinder
got the same as above; m₂ = P₂V₂M / RT₂

Am I correct this far?

 

[LawrenceC]

Not sure how to do it but here's what I've done:
*inflated tire
P₁V₁ = n₁RT₁ ; n = m/M where M is molecular mass
P₁V₁M = m₁RT₁
m = P₁V₁M / RT₁ --> here the mass is the function of pressure, volume and temperature

*pump cylinder
got the same as above; m₂ = P₂V₂M / RT₂

Am I correct this far?

Yes, so far so good. Now you have an expression for the mass in the tire when inflated. You also have an expression for the mass that is delivered by a single pump stroke. You know the process is isothermal...

 

[songoku]

Yes, so far so good. Now you have an expression for the mass in the tire when inflated. You also have an expression for the mass that is delivered by a single pump stroke. You know the process is isothermal...

m₁ = P₁V₁M / RT₁
1 / T₁ = m₁ R / P₁ V₁ M

m₂ = P₂V₂M / RT₂
1 / T₂ = m₂ R / P₂ V₂ M

Isothermal : T₁ = T₂
m₁ R / P₁ V₁ M = m₂ R / P₂ V₂ M

m₁ / P₁ V₁ = m₂ / P₂ V₂

Is P₁ = 3 x 10⁵ , V₁ = 1.2 x 10^-3 , P₂ = 1.0 x 10⁵ , V₂ = 3.6 x 10^-3 ?

If yes, the number of pumping = m₁ / m₂ and I got 40 as the answer

 

[LawrenceC]

Your answer is correct but you typed

"V2 = 3.6 x 10-3"

when it should have been

V2=9.0 x 10-5