# Find the pressure put into a box by a hand-squeezed pump

## Homework Statement

A spherical hand-held pump with a radius r is used to pump air into a cube with a length L. The pump and cube are connected together by rubber tubing. Express the pressure stored in the box as a function of the number of times the pump is fully squeezed.
Imagine the situation as that of a sphygomanometer without the cuff and the gauge replaced by a cube box. ## Homework Equations

I used Pcube*Vcube=Ppump*Vpump
Therefore: Pcube = (Ppump*Vpump) / Vcube

## The Attempt at a Solution

I am assuming the cube and pump are at the same temperature and that the pump holds 1 atm pressure when it is not compressed. My expression is the following:
P(x) = ((4/3)(pi)(r^3) * (1 atm) * (x)) / (L^3)
where x is the number of times the pump is squeezed.
I am not confident in my assumption that the initial pump pressure is 1 atm.
Anyone know how to express this situation?
Thanks

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The number of moles injected in each pumping is (1 atm)V/RT, where V is the volume of the sphere and T is the air temperature. So, when x pumping take place, the total number of moles in the cube is $n=\frac{(1\ atm)V}{RT}x$. This number of moles, at temperature T, is present in the cube of volume $L^3$. So the pressure in the cube is $$P=\frac{nRT}{L^3}=(1\ atm)\frac{xV}{L^3}$$This agrees with the answer you obtained. But, for this relationship to be correct, the cube needs to be under vacuum (empty) to begin with. If the cube has air in it to begin with, then the result is different.