Calculating Pressure of Air After a Bicycle Pump Stroke

[jack1234]

A bicycle pump contains air at STP. As the tire is pumped up, the volume of air decreases by 30% with each stroke. What is the new pressure of air (in atm) in the chamber after the first stroke, assuming no temperature change? Select the correct answer.

a. 1.4
b. 0.7
c. 2.6
d. 0.9
e. 2

What I have done is
T1=T2
P1*V1/(R*n1)=P2*V2/(R*n2)
P1*V1/(n1)=P2*V2/(n2)

But the question doesn't tell us number of mole of the air, how do we solve this question?

 

[mgb_phys]

Remember n doesn't change ( no air is lost or added) so n1=n2

 

[jack1234]

but the air has been pump out the the bicycle, some of the air molecules will be leaving the bicycle pump, so what is the reason that n1=n2?

 

[mgb_phys]

In that case it still doesn't matter.
Assume a valve between the tyre and pump opened only at this pressure.
Fractionally before the valve opened the pressure is is just 30% more, as soon as the valve opens the pressure in the pump and tyre are equal.

 

[jack1234]

Thanks for the reply, but I still not very clear...
if we pump once, the valve opened, some air molecules will go to the tyre, and then, the valve closed. Say after that pump, we stop. So air molecules in the bicycle pump will forever be lost. Hence by this reasoning n2<n1.
Anything wrong?

 

[mgb_phys]

In that case you don't have enough information. You would need to know the pressure difference with the tyre and the total volume of the tyre.
I suspect you are reading too much into the question - what level is this at?

 

[jack1234]

Hi mgb_phys, the question is at the entry level of physics.
If I assume nR is constant as you said, I get the correct answer, although it does not seems very logical to me, thanks though :)