# Homework Help: IGCSE Physics Boyle's law question

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1. Apr 30, 2015

### yohanes775

1. The problem statement, all variables and given/known data
The helium in the cylinder has a volume of 6.0 × 10^–3 m3^-1 (0.0060 m^3) and is at a pressure of 2.75 × 10^6 Pa.

(i) The pressure of helium in each balloon is 1.1 × 10^5 Pa. The volume of helium in an inflated balloon is 3.0 × 10^–3 (0.0030 m3). The temperature of the helium does not change. Calculate the number of balloons that were inflated.

2. Relevant equations

3. The attempt at a solution
so basically using P1V1=P2V2 :
=> [1.1 x 10^5] [Vᵢ ] = [2.75 x 10^6] [6x10^(-3)]

=> Vᵢ = [2.75 x 10^6] [6x10^(-3)] / [1.1 x 10^5] m³
so Vᵢ = 15 x 10^(-2) m³ or 0.15 m³

# of balloons = 15 x 10^(-2) m³ / vol. of 1 balloon = 15 x 10^(-2) m³ / [3 x 10^(-3)] m³

which should be 50 balloons

HOWEVER the answer is 48 balloons as one has to assume the cylinder doesn't fully empty and still holds helium at [1.1 x 10^5] Pa hence [6x10^(-3)] volume equivalent to 2 balloons would still remain in the cylinder at the balloon pressure hence Ans: is 50 - 2 = 48 balloons
but how can you assume the cylinder is still at 1.1 x 10^5 pascals? Basically whats the logic behind the last step?

Thanks!

2. Apr 30, 2015

### T Damen

Take into account that normal atmospheric pressure at sea level is also about 1.1 x 10^5 Pa.

3. May 20, 2015

### Bimal R

Hi,

The gas in the cylinder cannot be emptied to create vacuum. So, once the pressure inside the cylinder equals atmospheric pressure, it stops pushing gas to the balloon. Hence the last balloon may not be inflated to full. Considering this, the available pressure in the cylinder initially has to be initial cylinder pressure - atmospheric pressure which is
2.75 × 10^6 - 1.01 × 10^5 = 2.64 x 10^6 Pa.

Bimal