# Pressure of wheel after being pumped

## Homework Statement

A wheel has volume 500 cm3 and the initial pressure inside of it is 0.5 atm. The wheel is pumped using a pump having volume 50 cm3. The atmospheric pressure is 76 cmHg and there is no change in temperature. What is the pressure of the wheel after being pumped 4 times?

PV = nRT

## The Attempt at a Solution

I don't know how to set up the equation.

After being pumped 4 times, the final volume of wheel will be: 500 + 4 x 50 = 700 cm3??

There will be change of n because more molecules of gas inside the wheel after pumping so the equation will be:
P1.V1 / n1 = P2.V2 / n2 ??

There is no information about the moles and I don't know how to use information about given from the question.

Thanks

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TSny
Homework Helper
Gold Member
After being pumped 4 times, the final volume of wheel will be: 500 + 4 x 50 = 700 cm3??
No, the air is compressed by the pump as it enters the wheel (tire). I think the volume of the wheel can be considered as staying essentially constant as the air is pumped into the wheel.
There will be change of n because more molecules of gas inside the wheel after pumping
Yes.
so the equation will be:
P1.V1 / n1 = P2.V2 / n2 ??
OK, what does this tell you if the volume of the wheel is taken to be constant?

There is no information about the moles and I don't know how to use information about given from the question.
Before the pump is pumped, it contains a certain number of moles of air. How does this number of moles of air depend on volume, pressure, and temperature?

• songoku
OK, what does this tell you if the volume of the wheel is taken to be constant?
P1/n1 = P2/n2

Before the pump is pumped, it contains a certain number of moles of air. How does this number of moles of air depend on volume, pressure, and temperature?
I am not sure I get your hint. Maybe you mean: n = PV / RT?

P1 = 0.5 atm

number of moles going into the wheel each pump: n = PV / RT = (1 x 50) / RT ?

TSny
Homework Helper
Gold Member
P1/n1 = P2/n2
Yes. It might be helpful to solve this for P2 in order to see what you will need to know in order to determine P2.
I am not sure I get your hint. Maybe you mean: n = PV / RT?
Yes

P1 = 0.5 atm
OK. But you will need to know if the pressure given for P1 is absolute pressure or gauge pressure.

number of moles going into the wheel each pump: n = PV / RT = (1 x 50) / RT ?
Yes. Note that you are using non-SI units for P and V. This might be OK, but it's good to keep this in mind.

Try to put these ideas together to find P2.

• songoku
Yes. It might be helpful to solve this for P2 in order to see what you will need to know in order to determine P2.
I need to know n1 and n2 or at least the relation between them. Let say n1 = n so n2 = n1 + 4 x 50 / RT

OK. But you will need to know if the pressure given for P1 is absolute pressure or gauge pressure.
How to know whether it is absolute or gauge pressure? The question only states "the initial pressure inside the wheel".

Let me try it:
P1/n1 = P2 / n2
0.5 / n = P2 / (n + 200 / RT)
n . P2 = 0.5 n + 100 / RT
P2 = 0.5 + 100 / nRT
P2 = 0.5 + 100 / (P1 . V1)
P2 = 0.5 + 100 / (0.5 x 500)
P2 = 0.9 atm

Am I correct?

TSny
Homework Helper
Gold Member
I need to know n1 and n2 or at least the relation between them. Let say n1 = n so n2 = n1 + 4 x 50 / RT
Yes. (But it makes a lot of people around here nervous to see someone leave out the units that go with the numerical values.)

How to know whether it is absolute or gauge pressure? The question only states "the initial pressure inside the wheel".
If you poke a hole in a bicycle tire, it will deflate until the inside pressure equals the outer pressure of 1 atm. So, a "completely flat tire" would have a gauge pressure of 0 but an absolute pressure of 1 atm. An absolute pressure of 0.5 atm for a tire is very unnatural. You would have to hook the tire to a vacuum pump to get the absolute pressure below 1.0 atm. So, my guess is that the pressure of 0.5 atm is a gauge pressure.

Let me try it:
P1/n1 = P2 / n2
0.5 / n = P2 / (n + 200 / RT)
n . P2 = 0.5 n + 100 / RT
P2 = 0.5 + 100 / nRT
P2 = 0.5 + 100 / (P1 . V1)
P2 = 0.5 + 100 / (0.5 x 500)
P2 = 0.9 atm

Am I correct?
Yes, that looks correct. You might see what you get if you interpret the 0.5 atm as a gauge pressure. What would the final gauge pressure be?

I hope you see why you can get away with using cm3 for volume and atmospheres for pressures.

• songoku
Yes. (But it makes a lot of people around here nervous to see someone leave out the units that go with the numerical values.)

I hope you see why you can get away with using cm3 for volume and atmospheres for pressures.
I know why. I am usually careful with units :)

Thanks a lot

Nidum
Gold Member
A real pump will only discharge air into a receiver when the pressure in the pump cylinder exceeds the pressure in the receiver .

The pump cylinder pressure will only exceed the receiver pressure when the pump piston has travelled some distance down the cylinder during each working stroke .

The amount of air actually discharged is always less than the total amount of air in the pump cylinder at start of stroke .

TSny
Homework Helper
Gold Member
A real pump will only discharge air into a receiver when the pressure in the pump cylinder exceeds the pressure in the receiver .

The pump cylinder pressure will only exceed the receiver pressure when the pump piston has travelled some distance down the cylinder during each working stroke .
Yes, that's a good point. So, when the air in the pump begins entering the tire, the volume of the air in the pump will be less than 500 cm3 and the pressue will be greater than 1 atm. But at this instant, the number of moles of air in the pump will still be the same as when the pump had 500 cm3 of air at 1 atm.

The amount of air actually discharged is always less than the total amount of air in the pump cylinder at start of stroke .
It seems to me that the number of moles of air discharged will be essentially the full number of moles that was in the pump initially. I'm thinking of a simple bicycle pump where I push the piston all the way down. I guess there is a small loss if you take into account the air in the rubber hose that connects the pump to the tire. I could be overlooking something here.