# Help on surface tension

Having trouble to understand a classical example of surface tension:
Two balloons are connected to each other with a valve. If the surface tension of the two balloons is the same but one balloon is bigger than the other, when the valve is lifted open so the air in the two balloons is now connected, what will happen to the two balloons?
The correct answer is the small balloon will get smaller and the big balloon gets bigger. This answer follows the application of Laplace's law where it says (something like the following) T(surface tension) = P(pressure) x R (radius). So the bigger the radius, the smaller the pressure. Thus when the valve is open, air will flow from high pressure (small balloon) to low pressure (big balloon), as always.
The confusion I have is, I assume a bigger balloon will have MORE air inside. If two balloons are made of the same material (so as to have the same surface tension) and I pump air into the balloons, shouldn't the bigger balloon contain more air, which means higher the pressure. Then the logical conclusion will be opposite to what Laplace's law predict: if I open the valve, air will flow from big balloon (more air) to small balloon (less air). What's wrong with my assumption?

Simon Bridge